Question 1: A line intersects \displaystyle x-axis at point \displaystyle (-2, 0) and cuts off an intercept of \displaystyle 3 units from the positive side of \displaystyle y-axis . Find the equation of the line. [1992]

\displaystyle \text{Answer:}

\displaystyle \text{ x-intercept }  = (-2, 0)
\displaystyle \text{ y-intercept }  = (0, 3)
Equation of line
\displaystyle y-3 = \frac{3-0}{0-(-2)} (x-0)
\displaystyle y-3 = \frac{3}{2} x
\displaystyle 2y -6 = 3x
\displaystyle 2y = 3x+6

Question 2: Find the equation of a line passing through the point \displaystyle (2, 3) and having the \displaystyle \text{ x-intercept }  of \displaystyle 4 units. [2002]

\displaystyle \text{Answer:}

\displaystyle \text{ x-intercept }  = (4, 0)
Equation of line passing through \displaystyle (2,3) \text{ and }  (4, 0)
\displaystyle y - 0 = \frac{0-3}{4-2} (x-2)
\displaystyle y = - \frac{3}{2} (x-4)
\displaystyle 2y = -3x+12
\displaystyle 2y + 3x = 12

Question 3: The given figure (not drawn to scale) shows two straight lines \displaystyle AB\text{ and } CD . If equation of the line \displaystyle AB is \displaystyle y =x+1 and equation of \displaystyle CD is \displaystyle y = \sqrt{3}x-1 . Write down the inclination of lines \displaystyle AB\text{ and } CD ; also, find the angle between \displaystyle AB\text{ and } CD [1989]

\displaystyle \text{Answer:}

\displaystyle AB: y = x+1
\displaystyle CD: y = \sqrt{3}x - 1
\displaystyle \text{Slope of } AB = 1
\displaystyle \tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^o
\displaystyle \text{Slope of } CD = \sqrt{3}
\displaystyle \tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^o
\displaystyle \text{Therefore } 45^o+(180^o-60^o) + \theta = 180^o
\displaystyle \Rightarrow \theta = 15^o

Question 4: Write down the equation of the line whose gradian is \displaystyle \frac{2}{3} and which passes through \displaystyle P , where \displaystyle P divides the line segment joining \displaystyle A (-2, 6)\text{ and } B (3, -4) in the ratio \displaystyle 2 : 3 [2001]

\displaystyle \text{Answer:}

Given \displaystyle P divides the line segment joining \displaystyle A (-2, 6)\text{ and } B (3, -4) in the ratio \displaystyle 2 : 3
\displaystyle \text{Let the coordinates of } P = (x, y)
Therefore
\displaystyle x = \frac{2 \times 3 + 3 \times (-2)}{2+3} = \frac{6-6}{5} = 0
\displaystyle y = \frac{2 \times (-4) + 3 \times 6}{2+3} = \frac{-8+18}{5} = 2
\displaystyle \therefore P(0,2)
\displaystyle \text{Equation of a line passing through } P(0,2) \text{ with slope } \frac{3}{2}
\displaystyle y-2 = \frac{3}{2} (x-0)
\displaystyle 2y - 4 = 3x \Rightarrow 2y = 3x+4

Question 5: Point \displaystyle A\text{ and } B have coordinates \displaystyle (7, 3)\text{ and } (1, 9) respectively. Find:
\displaystyle \text{(i)  The Slope of } AB  
(ii) The equation of perpendicular bisector of the line segment \displaystyle AB  
(iii) The value of \displaystyle p \text{ of }  (-2, p) lies on it [2008]
Answer:
\displaystyle \text{(i)  Slope of } AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2
\displaystyle \text{(ii)  Midpoint of } AB = ( \frac{7+1}{2} , \frac{-3+9}{2} ) = (4,3)
\displaystyle \text{Therefore equation of line passing through } (4,3) \text{ and slope } \frac{1}{2} \text{ is }
\displaystyle y - 3 = \frac{1}{2} (x-4)
\displaystyle 2y -6 = x-4
\displaystyle 2y= x+2
\displaystyle \text{(iii) } p \text{ of }  (-2, p) \text{ lies on it }
\displaystyle \text{Therefore } 2(p) = (-2)+2 \Rightarrow p = 0

Question 6: \displaystyle A\text{ and } B are two points on the \displaystyle x-axis\text{ and } y-axis respectively. \displaystyle P (2, -3) is the mid-point of \displaystyle AB . Find the
(i) coordinates of \displaystyle A\text{ and } B  
(ii) Slope of line \displaystyle AB  
(iii) Equation of line \displaystyle AB [2010]

\displaystyle \text{Answer:}

\displaystyle \text{(i) Let } A(x,0) \text{ and }  B(0, y)
\displaystyle P(2, -3) is the mid point
\displaystyle \text{(i) Therefore } 2 = \frac{0+x}{2} \Rightarrow x = 4
\displaystyle -3 = \frac{y+0}{2} \Rightarrow y = -6
Hence \displaystyle A(4,0) \text{ and }  B(0, -6)
\displaystyle \text{(ii)  Slope of } AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}
(iii) Equation of \displaystyle AB
\displaystyle y = 0 = \frac{3}{2} (x-4)
\displaystyle 2y = 3x-12

Question 7: The equation of a line is \displaystyle 3x + 4y - 7 = 0 . Find:
(i) Slope of the line.
(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines \displaystyle x -y + 2 = 0 \text{ and }  3x + y- 10 = 0 [2010]

\displaystyle \text{Answer:}

\displaystyle \text{(i) Slope } = - \frac{3}{4}
\displaystyle \text{(ii) Slope of perpendicular } = \frac{4}{3}
For point of intersection solve \displaystyle x -y + 2 = 0 \text{ and }  3x + y- 10 = 0
\displaystyle y = 4 \text{ and }  x = 2
Therefore intersection \displaystyle = (2, 4)
Therefore equation of line
\displaystyle y-4 = \frac{4}{3} (x-2)
\displaystyle 3y-12 = 4x-8
\displaystyle 3y = 4x+4

Question 8: \displaystyle ABCD is a parallelogram where \displaystyle A (x, y), B (5, 8), C (4, 7)\text{ and } D (2, -4) . Find:
(i) coordinates of \displaystyle A  
(ii) Equation of diagonal \displaystyle BD [2011]
\displaystyle \text{Answer:}
\displaystyle \text{(i)  Midpoint of } BD = ( \frac{5+2}{2} , \frac{-4+8}{2} )= ( \frac{7}{2} , 2)
\displaystyle \text{Therefore we have } A(x, 0), O( \frac{7}{2} , 2) \text{ and }  C(4, 7)
\displaystyle O \text{ is the  Midpoint of } AC \text{ as well (diagonals of a parallelogram bisect each other) }
\displaystyle \text{Hence } \frac{x+4}{2} = \frac{7}{2} \Rightarrow x = 3
\displaystyle \text{and } \frac{y+7}{2} = 2 \Rightarrow y = -3
Hence \displaystyle A ( 3, -3)
Equation of \displaystyle \text{(ii) } BD
\displaystyle y - 8 = \frac{-4-8}{2-5} (x-5)
\displaystyle y-8 = 4(x-5)
\displaystyle y - 8 = 4x - 20
\displaystyle y + 12 = 4x

Question 9: From the given figure, find:

(i) The coordinates of \displaystyle A, B,\text{ and } C .

(ii) The equation of the line through \displaystyle A and parallel to \displaystyle BC . [2004]

\displaystyle \text{Answer:}

\displaystyle \text{Slope of } BC = \frac{0-2}{3-(-1)} = \frac{-2}{4} = - \frac{1}{2}
The equation of line parallel to \displaystyle BC and passing through \displaystyle A(2,3)
\displaystyle y-3 = - \frac{1}{2} (x-2)
\displaystyle 2y-6 = -x+2
\displaystyle 2y+x=8

Question 10: \displaystyle P (3, 4), Q (7, -2)\text{ and } R (-2, -1) are the vertices of triangle \displaystyle PQR . Write down the equation of the median of the triangle through \displaystyle R . [2005]
\displaystyle \text{Answer:}
\displaystyle \text{Midpoint of } PQ = ( \frac{3+7}{2} , \frac{4-2}{2} ) = (5,1)
Therefore equation passing through \displaystyle (5,1) \text{ and }  P(-2,-1) is
\displaystyle y-1 = \frac{-1-1}{-2-5} (x-5)
\displaystyle y-1 = \frac{2}{7} (x-5)
\displaystyle 7y-7 = 2x-10
\displaystyle 7y-2x+3=0
\displaystyle \\
Question 11: Find the value of \displaystyle k for which the lines \displaystyle kx - 5y + 4 = 0 \text{ and }  5x - 2y + 5 = 0 are perpendicular to each other. [2003]
\displaystyle \text{Answer:}
\displaystyle \text{Slope of } kx - 5y + 4 = 0 \Rightarrow m_1 = \frac{k}{5}
\displaystyle \text{Slope of } 5x - 2y + 5 = 0 \Rightarrow m_2 = \frac{5}{2}
Since the two lines are perpendicular, \displaystyle m_1. m_2 = -1
\displaystyle \Rightarrow \frac{k}{5} . \frac{5}{2} = -1
\displaystyle \Rightarrow k = -2
\displaystyle \\
Question 12: A straight line passes through the points \displaystyle P (-1, 4)\text{ and } Q (5, -2) . It intersects the co-ordinate axes at points \displaystyle A\text{ and } B . \displaystyle M is the mid-point of the line segment \displaystyle AB . Find:
The equation of the line
The coordinates of \displaystyle A\text{ and } B  
The coordinates of \displaystyle M [2003]
\displaystyle \text{Answer:}
\displaystyle P (-1, 4)\text{ and } Q (5, -2)
The equation of the line:
\displaystyle y - 4 = \frac{-2-4}{5-(-1)} (x+1)
\displaystyle \Rightarrow y-4 = -1 (x+1)
\displaystyle \Rightarrow y+x=3
\displaystyle \text{ The x-intercept }  A = (3,0) \text{ and the } \text{ y-intercept }  B = (0,3)
\displaystyle \text{The coordinate of } M = ( \frac{3+0}{2} , \frac{0+3}{2} ) = ( \frac{3}{2} , \frac{3}{2} )
\displaystyle \\
Question 13: If the lines \displaystyle y = 3x + 7\text{ and } 2y + px = 3 are perpendicular to each other, find the value of \displaystyle p . [2006]
\displaystyle \text{Answer:}
Given equation is \displaystyle y = 3x + 7
\displaystyle \Rightarrow \text{ Slope } (m_1) = 3
Given equation is \displaystyle 2y + px = 3
\displaystyle \Rightarrow 2y=-px+3
\displaystyle \Rightarrow y = \frac{-p}{2} x+ \frac{3}{2}
\displaystyle \Rightarrow \text{ Slope } (m_2) = \frac{-p}{2}
Since they are perpendicular, \displaystyle m_1 . m_2 =-1
\displaystyle \Rightarrow 3. \frac{-p}{2} =-1
\displaystyle \Rightarrow p= \frac{2}{3}
\displaystyle \\
Question 14: The line through \displaystyle A (-2, 3)\text{ and } B (4, b) is perpendicular to the line \displaystyle 2x - 4y = 5 . Find the value of \displaystyle b [2012]
\displaystyle \text{Answer:}
\displaystyle \text{Slope of } AB = \frac{b-3}{4-(-2)} = \frac{b-3}{6}
Given equation is \displaystyle 2x - 4y = 5
\displaystyle \Rightarrow 4y=2x-5
\displaystyle \Rightarrow y = \frac{1}{2} x- \frac{5}{4}
\displaystyle \Rightarrow \text{ Slope } (m_2) = \frac{1}{2}
Since they are perpendicular, \displaystyle m_1 . m_2 =-1
\displaystyle \Rightarrow \frac{b-3}{6} . \frac{1}{2} =-1
\displaystyle \Rightarrow p=-9
\displaystyle \\
Question 15: i) Find the equation of the line passing through \displaystyle (5, -3) and parallel to \displaystyle x - 3y = 4 .
ii) Find the equation of the line parallel to the line \displaystyle 3x + 2y = 8 and passing through the point \displaystyle (0, 1) [2007]
\displaystyle \text{Answer:}
i) Given Point \displaystyle (x_1, y_1)=(5,-3)
Given equation is \displaystyle x - 3y = 4
\displaystyle \Rightarrow 3y=x-4
\displaystyle \Rightarrow y = \frac{1}{3} x-4
\displaystyle \Rightarrow \text{ Slope } (m) = \frac{1}{3}
Equation of a line with slope \displaystyle m and passing through \displaystyle (x_1, y_1) is
\displaystyle y-y_1=m(x-x_1)
\displaystyle \Rightarrow y-(-3)= \frac{1}{3} (x-5)
\displaystyle \Rightarrow 3y+9=x-5
\displaystyle \Rightarrow x-3y-14=0
ii) Given Point \displaystyle (x_1, y_1)=(0,1)
Given equation is \displaystyle 3x+2y=8
\displaystyle \Rightarrow 2y=-3x+8
\displaystyle \Rightarrow y = \frac{-3}{2} x+4
\displaystyle \Rightarrow \text{ Slope } (m) = \frac{-3}{2}
Equation of a line with slope \displaystyle m and passing through \displaystyle (x_1, y_1) is
\displaystyle y-y_1=m(x-x_1)
\displaystyle \Rightarrow y-1= \frac{-3}{2} (x-0)
\displaystyle \Rightarrow 2y-2=-3x
\displaystyle \Rightarrow 2y+3x=2
\displaystyle \\
Question 16: i) Write down the equation of the line \displaystyle AB , through \displaystyle (3, 2) and perpendicular to the line \displaystyle 2y = 3x + 5 .
ii) \displaystyle AB meets the \displaystyle x-axis at A and the \displaystyle y-axis at \displaystyle B . write down the coordinates of \displaystyle A\text{ and } B . Calculate the area of triangle \displaystyle OAB , where \displaystyle O is origin. [1995]
\displaystyle \text{Answer:}
i) Given Point \displaystyle (x_1, y_1)=(3,2)
Given equation is \displaystyle 2y=3x+5
\displaystyle \Rightarrow y = \frac{3}{2} x+5
\displaystyle \Rightarrow \text{ Slope } (m) = \frac{3}{2}
\displaystyle \text{Therefore slope of the new line } = m = \frac{-1}{\frac{3}{2}} = -\frac{2}{3}
Equation of a line with slope \displaystyle m and passing through \displaystyle (x_1, y_1) is \displaystyle y-y_1=m(x-x_1)
\displaystyle \Rightarrow y-2= - \frac{2}{3} (x-(-2))
\displaystyle \Rightarrow 3y-6=-2x+6
\displaystyle \Rightarrow 3y+2x=12
ii) Equation of \displaystyle AB is \displaystyle \Rightarrow 3y+2x=12
When \displaystyle y = 0, x = 6 . \displaystyle \text{Therefore } A (6,0)
When \displaystyle c = 0, y =4 . \displaystyle \text{Therefore } B(0,4)
\displaystyle \text{Area of the triangle } = \frac{1}{2} \times 6 \times 4 = 12 \text{sq. units.}
\displaystyle \\
Question 17: Find the value of a for the points \displaystyle A (a, 3), B (2, 1)\text{ and } C (5, a) are collinear. Hence, find the equation of the line. [2014]
\displaystyle \text{Answer:}
Given points \displaystyle A (a, 3), B (2, 1)\text{ and } C (5, a)
\displaystyle \text{Slope of } AB = \frac{1-3}{2-a} = \frac{-2}{2-a}
\displaystyle \text{Slope of } BC = \frac{a-1}{5-2} = \frac{a-1}{3}
Because \displaystyle A, B,\text{ and } C are collinear:
\displaystyle \frac{-2}{2-a} = \frac{a-1}{3}
\displaystyle -6=(2-a)(a-1)
\displaystyle -6=2a-2-a^2+a
\displaystyle -6=3a-a^2-2
\displaystyle a^2-3a-4=0
\displaystyle (a-4)(a+1)=0 \Rightarrow a=4, or -1
\displaystyle \\
Question 18: In, \displaystyle A = (3, 5), B = (7, 8)\text{ and } C = (1, -10) . Find the equation of the median through \displaystyle A . [2013]
\displaystyle \text{Answer:}
\displaystyle A = (3, 5), B = (7, 8)\text{ and } C = (1, -10)
\displaystyle \text{Let } D be the \displaystyle \text{Midpoint of } BC . Therefore the coordinates of \displaystyle D are
\displaystyle D =( \frac{1+7}{2} , \frac{-10+8}{2} )=(4, -1)
\displaystyle \text{Slope of } AD = m = \frac{-1-5}{4-3} = \frac{-6}{1} =-6
Equation of \displaystyle AD:
\displaystyle y-(-1)=-6(x-4)
\displaystyle \Rightarrow y+1=-6x+24
\displaystyle \Rightarrow y+6x=23
\displaystyle \\
Question 19: The line through \displaystyle P (5, 3) intersects \displaystyle y-axis at Q .
i) Write the slope of the line.
ii) Write the equation of the line.
iii) Find the coordinates of \displaystyle Q [2012]
\displaystyle \text{Answer:}
Given points \displaystyle P(-2, 0)\text{ and } Q(0, y)
Slope \displaystyle = m = \tan 45^{\circ} = 1
Equation of line:
\displaystyle y-3=1(x-5)
\displaystyle \Rightarrow y = x-2
When \displaystyle x=0, y = -2
Hence the coordinates of \displaystyle Q = (0, -2)
\displaystyle \\

Question 20: \displaystyle A (1, 4), B (3, 2)\text{ and } C (7, 5) are vertices of a triangle \displaystyle ABC . Find:

i) The coordinates of the centroid of a triangle \displaystyle ABC .

ii) The equation of a line through the centroid and parallel to \displaystyle AB . [2002]

\displaystyle \text{Answer:}

\displaystyle \text{Let } O be the centroid. Therefore the coordinates of \displaystyle O are:
\displaystyle O=( \frac{1+3+7}{3} , \frac{4+2+5}{3} )=( \frac{11}{3} , \frac{11}{3} )
\displaystyle \text{Slope } m= \frac{2-4}{3-1} = \frac{-2}{2} = -1
\displaystyle \text{Therefore the equation of a line parallel to } AB \text{will pass through } ( \frac{11}{3} , \frac{11}{3} )
Equation of the line:
\displaystyle y- \frac{11}{3} =-1(x- \frac{11}{3} )
\displaystyle \Rightarrow 3y-11=-(3x-11)
\displaystyle \Rightarrow 3y+3x=22

\displaystyle \textbf{Question 21:}~ABC\text{ is a triangle whose vertices are }A(1,-1),~B(0,4)\text{ and } \\ C(-6,4).~D\text{ is the midpoint of }BC.~\text{Find: (i) coordinates of }D.~\text{(ii) equation of} \\ \text{the median }AD.~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, vertices are }A(1,-1),~B(0,4)\text{ and }C(-6,4)
\displaystyle \text{(i) Mid-point of }BC=\left(\frac{0-6}{2},\frac{4+4}{2}\right)=(-3,4)

\displaystyle \Rightarrow \text{Coordinates of }D\text{ are }(-3,4)
\displaystyle \text{(ii) Equation of the median }AD
\displaystyle \text{Equation of the line joining two points is }
\displaystyle \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}
\displaystyle A(1,-1)\text{ and }D(-3,4)
\displaystyle \Rightarrow \frac{y-(-1)}{4-(-1)}=\frac{x-1}{-3-1}
\displaystyle \Rightarrow \frac{y+1}{5}=\frac{x-1}{-4}
\displaystyle \Rightarrow -4(y+1)=5(x-1)
\displaystyle \Rightarrow -4y-4=5x-5
\displaystyle \Rightarrow 5x+4y=1
\displaystyle \text{Hence, equation of median }AD\text{ is }5x+4y=1

\displaystyle \textbf{Question 22:}~A(4,2),~B(6,8)\text{ and }C(8,4)\text{ are the vertices of a triangle } \\ ABC.~\text{Write down the equation of the median of the triangle through }A.~\text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the mid-point of }BC
\displaystyle D=\left(\frac{6+8}{2},\frac{8+4}{2}\right)=(7,6)
\displaystyle \text{Equation of median of the triangle through }A=\text{Equation of }AD
\displaystyle \Rightarrow \frac{6-2}{7-4}(x-4)=(y-2)
\displaystyle \Rightarrow \frac{4}{3}(x-4)=y-2
\displaystyle \Rightarrow 4(x-4)=3(y-2)
\displaystyle \Rightarrow 4x-16=3y-6
\displaystyle \Rightarrow 4x-3y=10

\displaystyle \textbf{Question 23:}~\text{In the diagram given below, equation of }AB\text{ is } x-\sqrt{3}y+1=0 \\ \text{ and }\text{equation of }AC\text{ is }x-y-2=0.~\text{(i) Write down the angles that the lines} \\ AC\text{ and }AB\text{ make} \text{with the positive direction of} \text{X-axis. (ii) Find }\angle BAC.~\text{[ICSE 2017]}

\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\theta_1\text{ be the angle which }AB\text{ makes with positive X-axis}
\displaystyle \text{Let slope of }AB=m_1=\tan\theta_1
\displaystyle \text{Given equation of }AB:x-\sqrt{3}y+1=0
\displaystyle \Rightarrow \sqrt{3}y=x+1
\displaystyle \Rightarrow y=\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\quad\text{(i)}
\displaystyle \text{On comparing Eq. (i) with equation of line }y=mx+c

\displaystyle m_1=\frac{1}{\sqrt{3}}=\tan\theta_1
\displaystyle \Rightarrow \theta_1=30^\circ
\displaystyle \text{Similarly, slope of line }AC:x-y-2=0
\displaystyle \Rightarrow y=x-2
\displaystyle m_2=1=\tan\theta_2
\displaystyle \Rightarrow \theta_2=45^\circ
\displaystyle \text{(ii) We have, }\angle ABC=30^\circ
\displaystyle \angle ACB=180^\circ-45^\circ=135^\circ
\displaystyle \therefore \angle BAC=180^\circ-30^\circ-135^\circ=15^\circ
\displaystyle \text{Hence, }\angle BAC=15^\circ

\displaystyle \textbf{Question 24:}~\text{The slope of a line joining }P(6,k)\text{ and }Q(1-3k,3)\text{ is }\frac{1}{2}.~\text{Find (i) the} \\ \text{value of }k.~\text{(ii) mid-point of }PQ,~\text{using the value of }k\text{ found in (i). [ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, points are }P(6,k)\text{ and }Q(1-3k,3)
\displaystyle \text{(i) Since, slope of line joining these two points is }\frac{1}{2}
\displaystyle \therefore \frac{3-k}{(1-3k)-6}=\frac{1}{2}\qquad\left[\text{slope}=\frac{y_2-y_1}{x_2-x_1}\right]
\displaystyle \Rightarrow \frac{3-k}{-3k-5}=\frac{1}{2}
\displaystyle \Rightarrow 2(3-k)=-3k-5
\displaystyle \Rightarrow 6-2k=-3k-5
\displaystyle \Rightarrow 6+k=-5
\displaystyle \Rightarrow k=-11
\displaystyle \text{(ii) When }k=-11,\text{ the coordinates of }P\text{ and }Q\text{ are }(6,-11)\text{ and }(34,3)
\displaystyle \text{Now, mid-point of }PQ=\left(\frac{6+34}{2},\frac{-11+3}{2}\right)
\displaystyle =\left(\frac{40}{2},\frac{-8}{2}\right)=(20,-4)
\displaystyle \text{Hence, the mid-point of }PQ\text{ is }(20,-4)

\displaystyle \textbf{Question 25:}~\text{Find the equation of a line with x-intercept }=5\text{ and passing} \\ \text{through the point }(4,-7).~\text{[ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Since, x-intercept }=5,\text{ therefore the line will intersect X-axis at }(5,0)
\displaystyle \text{Thus, the line will pass through }(5,0)\text{ and }(4,-7)
\displaystyle \text{Hence, the equation of line is}
\displaystyle y-0=\frac{-7-0}{4-5}(x-5)\qquad\left[\text{equation of a line passing through two points}\right]
\displaystyle \Rightarrow y=\frac{-7}{-1}(x-5)
\displaystyle \Rightarrow y=7(x-5)
\displaystyle \Rightarrow y=7x-35
\displaystyle \Rightarrow 7x-y-35=0
\displaystyle \text{which is the required equation of a line}

\displaystyle \textbf{Question 26:}~\text{Write down the equation of the line, whose gradient is }\frac{3}{2}\text{ and which} \\ \text{passes through }P,~\text{where }P\text{ divides the line segment joining }A(-2,6)\text{ and }B(3,-4) \\ \text{in the ratio of }2:3.~\text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of }P\text{ be }(x,y),\text{ then the point }P(x,y)
\displaystyle \text{divides the line joining }A(-2,6)\text{ and }B(3,-4)\text{ in the ratio }2:3
\displaystyle \therefore P(x,y)=\left(\frac{2\times3+3\times(-2)}{2+3},\frac{2\times(-4)+3\times6}{2+3}\right)
\displaystyle =\left(\frac{6-6}{5},\frac{-8+18}{5}\right)=\left(0,\frac{10}{5}\right)=(0,2)
\displaystyle \text{Now, the equation of a line, whose slope }m=\frac{3}{2}\text{ and passing through }P(0,2)\text{ is}
\displaystyle y-2=\frac{3}{2}(x-0)
\displaystyle \Rightarrow 2(y-2)=3x
\displaystyle \Rightarrow 2y-4=3x
\displaystyle \Rightarrow 2y=3x+4
\displaystyle \Rightarrow y=\frac{3}{2}x+2

\displaystyle \textbf{Question 27:}~A(2,5),~B(-1,2)\text{ and }C(5,8)\text{ are the vertices of a }\triangle ABC,~M\text{ is a point on } \\ AB\text{ such that }AM:MB=1:2.~\text{Find the coordinates of }M.~\text{Hence, find the equation} \\ \text{of the line passing through the points }C\text{ and }M.~\text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A(2,5),~B(-1,2)\text{ and }C(5,8)
\displaystyle \text{Since, }AM:MB=1:2
\displaystyle \therefore \text{Coordinates of }M=\left(\frac{1\times(-1)+2\times2}{1+2},\frac{1\times2+2\times5}{1+2}\right)
\displaystyle =\left(\frac{-1+4}{3},\frac{2+10}{3}\right)=\left(1,4\right)
\displaystyle \text{Now, we have }M(1,4)\text{ and }C(5,8)
\displaystyle \therefore \text{Equation of line joining }C\text{ and }M\text{ is given by}
\displaystyle y-4=\frac{8-4}{5-1}(x-1)
\displaystyle \Rightarrow y-4=\frac{4}{4}(x-1)
\displaystyle \Rightarrow y-4=x-1
\displaystyle \Rightarrow y=x+3

\displaystyle \textbf{Question 28:}~\text{Prove that }A(2,1),~B(0,3)\text{ and }C(-2,1)\text{ are the three vertices of} \\ \text{an isosceles right angled triangle.}~\text{Hence find the coordinates of a point }D,~\text{if } \\ ABCD\text{ is a square.}~\text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, coordinates are }A(2,1),~B(0,3),~C(-2,1)
\displaystyle \text{We have by distance formula}
\displaystyle AB=\sqrt{(0-2)^2+(3-1)^2}=\sqrt{(-2)^2+2^2}=\sqrt{4+4}=2\sqrt{2}
\displaystyle BC=\sqrt{(-2-0)^2+(1-3)^2}=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=2\sqrt{2}
\displaystyle CA=\sqrt{(2+2)^2+(1-1)^2}=\sqrt{4^2+0^2}=4
\displaystyle \text{Here, }AB=BC\text{ and }AB^2+BC^2=CA^2

\displaystyle \therefore A(2,1),~B(0,3),~C(-2,1)\text{ are the three vertices of an isosceles right angled triangle}
\displaystyle \text{For coordinates of point }D\text{: Let the coordinates of }D\text{ be }(x,y)
\displaystyle \because ABCD\text{ is a square}
\displaystyle \text{The coordinates of }O\text{ by mid-point formula}
\displaystyle O=\left(\frac{-2+2}{2},\frac{1+1}{2}\right)=(0,1)
\displaystyle \text{Mid-point of }AC=\text{Mid-point of }BD

\displaystyle (0,1)=\left(\frac{0+x}{2},\frac{3+y}{2}\right)
\displaystyle \Rightarrow \frac{x}{2}=0\Rightarrow x=0
\displaystyle \Rightarrow \frac{3+y}{2}=1\Rightarrow 3+y=2\Rightarrow y=-1
\displaystyle \therefore \text{The coordinates of point }D\text{ are }(0,-1)

\displaystyle \textbf{Question 29:}~\text{In the given figure, }ABC\text{ is a triangle and }BC\text{ is parallel} \\ \text{to the Y-axis.}~AB\text{ and }AC\text{ intersect the Y-axis at }P\text{ and }Q\text{ respectively.}~\text{[ICSE 2015]}

\displaystyle \text{(i) Write the coordinates of }A.
\displaystyle \text{(ii) Find the length of }AB\text{ and }AC.
\displaystyle \text{(iii) Find the ratio in which }Q\text{ divides }AC.
\displaystyle \text{(iv) Find the equation of the line }AC.
\displaystyle \text{Answer:}
\displaystyle \text{(i) From the given graph, the coordinates of }A\text{ are }(4,0)
\displaystyle \text{(ii) Clearly, the length of }AB=\sqrt{(4+2)^2+(0-3)^2}=\sqrt{6^2+(-3)^2}
\displaystyle =\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\text{ units}
\displaystyle \text{and the length of }AC=\sqrt{(4+2)^2+(0+4)^2}=\sqrt{6^2+4^2}
\displaystyle =\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\text{ units}
\displaystyle \text{(iii) Let }Q\text{ divides }AC\text{ in the ratio }K:1
\displaystyle \text{Then, coordinates of }Q=\left(\frac{-2K+4}{K+1},\frac{-4K+0}{K+1}\right)
\displaystyle \text{But }Q\text{ lies on Y-axis, therefore x-coordinate of }Q\text{ will be }0
\displaystyle \Rightarrow \frac{-2K+4}{K+1}=0
\displaystyle \Rightarrow -2K+4=0\Rightarrow 2K=4\Rightarrow K=2
\displaystyle \therefore \text{Required ratio is }2:1
\displaystyle \text{(iv) Equation of line }AC\text{ is given by}
\displaystyle y-0=\frac{-4-0}{-2-4}(x-4)
\displaystyle \Rightarrow y=\frac{-4}{-6}(x-4)=\frac{2}{3}(x-4)
\displaystyle \Rightarrow 3y=2x-8
\displaystyle \Rightarrow 2x-3y-8=0

\displaystyle \textbf{Question 30:}~\text{The line through }P(5,3)\text{ intersects Y-axis at }Q.

\displaystyle \text{(i) Write the slope of the line.}
\displaystyle \text{(ii) Write the equation of the line.}
\displaystyle \text{(iii) Find the coordinates of }Q.~\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) From the given figure, }\theta=45^\circ
\displaystyle \text{Slope of the line }PQ=m=\tan45^\circ=1
\displaystyle \text{(ii) The equation of line passing through }P(5,3)\text{ having slope }1
\displaystyle y-3=1(x-5)
\displaystyle \Rightarrow y-3=x-5
\displaystyle \Rightarrow y=x-2
\displaystyle \text{(iii) Since, }Q\text{ lies on Y-axis, therefore }x=0
\displaystyle \Rightarrow y=0-2=-2
\displaystyle \therefore \text{Coordinates of }Q\text{ are }(0,-2)

\displaystyle \textbf{Question 31:}~\text{Given equation of line }L_1\text{ is }y=4.

\displaystyle \text{(i) Write the slope of line }L_2,~\text{if }L_2\text{ is the bisector of }\angle O.
\displaystyle \text{(ii) Write the coordinates of point }P.
\displaystyle \text{(iii) Find the equation of }L_2.~\text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }L_2\text{ is a bisector of }\angle O
\displaystyle \text{Since, }\angle XOY=90^\circ
\displaystyle \therefore \angle POX=\angle POY=\frac{90^\circ}{2}=45^\circ=\theta\text{ (say)}
\displaystyle \text{Then, slope of }L_2=\tan45^\circ=1
\displaystyle \text{(ii) Let line }L_1\text{ intersects Y-axis at }M.\text{ Draw a line from }P,
\displaystyle \text{parallel to X-axis. Then, we get}

\displaystyle NP=OM=4\qquad\text{(i)}
\displaystyle \text{Also, in }\triangle ONP,
\displaystyle \angle PON+\angle ONP+\angle NPO=180^\circ
\displaystyle \Rightarrow 45^\circ+90^\circ+\angle NPO=180^\circ
\displaystyle \Rightarrow \angle NPO=180^\circ-90^\circ-45^\circ=45^\circ
\displaystyle \Rightarrow ON=NP\qquad[\text{sides opposite to equal angles are equal}]
\displaystyle \Rightarrow ON=4\qquad[\text{using Eq. (i)}]
\displaystyle \text{So, the coordinates of }N\text{ are }(4,0)\text{ and coordinates of }M\text{ are }(0,4)
\displaystyle \text{Hence, the coordinates of }P\text{ are }(4,4)
\displaystyle \text{(iii) Equation of line }L_2,\text{ which is passing through origin }(0,0)
\displaystyle \text{and having slope }1\text{ is}
\displaystyle y-0=1(x-0)
\displaystyle \Rightarrow y=x
\displaystyle \text{Hence, the equation of }L_2\text{ is }y=x

\displaystyle \textbf{Question 32:}~A\text{ and }B\text{ are two points on the X-axis and Y-axis respectively.} \\ \text{If }P(2,-3)\text{ is the mid-point of }AB,\text{ then find }

\displaystyle \text{(i) coordinates of }A\text{ and }B.
\displaystyle \text{(ii) slope of line }AB.
\displaystyle \text{(iii) equation of line }AB.~\text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Here, the line is intersecting the X-axis at }A
\displaystyle \text{and Y-axis at }B,\text{ so, let the coordinates of }A\text{ and }B\text{ be }(x_1,0)
\displaystyle \text{and }(0,y_1),\text{ respectively}
\displaystyle \text{So, the coordinates of mid-point of }A\text{ and }B\text{ are }\left(\frac{x_1+0}{2},\frac{0+y_1}{2}\right)
\displaystyle \left[\text{mid-point formula }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]
\displaystyle \text{But given coordinates of mid-point are }(2,-3)
\displaystyle \therefore (2,-3)=\left(\frac{x_1}{2},\frac{y_1}{2}\right)
\displaystyle \text{On equating the x and y-coordinates of both sides, we get}
\displaystyle 2=\frac{x_1}{2}\Rightarrow x_1=4\qquad\text{and}\qquad -3=\frac{y_1}{2}\Rightarrow y_1=-6
\displaystyle \text{Hence, the coordinates of }A\text{ are }(4,0)\text{ and }B\text{ are }(0,-6)
\displaystyle \text{(ii) Now, slope of line }AB,~m=\frac{y_2-y_1}{x_2-x_1}=\frac{-6-0}{0-4}=\frac{-6}{-4}=\frac{3}{2}
\displaystyle \text{(iii) Equation of line passing through the points }A(4,0)\text{ and }B(0,-6)\text{ is given by}
\displaystyle y-0=\frac{-6-0}{0-4}(x-4)
\displaystyle \Rightarrow y=\frac{-6}{-4}(x-4)
\displaystyle \Rightarrow y=\frac{3}{2}(x-4)
\displaystyle \Rightarrow 2y=3(x-4)\Rightarrow 2y=3x-12
\displaystyle \Rightarrow 3x-2y=12
\displaystyle \text{Hence, the equation of line }AB\text{ is }3x-2y=12

\displaystyle \textbf{Question 33:}~\text{Find the equation of a line passing through the point }(-2,3)\text{ and having the} \\ \text{x-intercept }4\text{ units.}~\text{[ICSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{Since x-intercept }=4,\text{ therefore the line meets the X-axis at }(4,0).
\displaystyle \text{Thus, the line passes through the points }(4,0)\text{ and }(-2,3).
\displaystyle \text{Using equation of a line passing through two points,}
\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)
\displaystyle \Rightarrow y-0=\frac{3-0}{-2-4}(x-4)
\displaystyle \Rightarrow y=\frac{3}{-6}(x-4)
\displaystyle \Rightarrow y=-\frac{1}{2}(x-4)
\displaystyle \Rightarrow y=-\frac{1}{2}x+2
\displaystyle \Rightarrow x+2y-4=0
\displaystyle \text{Hence, the required equation of the line is }x+2y-4=0.

\displaystyle \textbf{Question 34:}~\text{A straight line passes through the points }P(-1,4)\text{ and }Q(5,-2).\text{ It} \\ \text{intersects the coordinate axes at points }A\text{ and }B.\text{ If }M\text{ is the mid-point of the} \\ \text{segment }AB,\text{ then find }


\displaystyle \text{(i) the equation of the line.}
\displaystyle \text{(ii) the coordinates of }A\text{ and }B.
\displaystyle \text{(iii) the coordinates of }M.~\text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) The equation of line passing through }P(-1,4)\text{ and }Q(5,-2)\text{ is}
\displaystyle y-4=\frac{-2-4}{5-(-1)}(x+1)
\displaystyle \Rightarrow y-4=\frac{-6}{6}(x+1)=\frac{-6}{6}(x+1)
\displaystyle \Rightarrow y-4=-1(x+1)
\displaystyle \Rightarrow y-4=-x-1
\displaystyle \Rightarrow y+x-3=0
\displaystyle \text{which is the required equation of line}
\displaystyle \text{(ii) Here, the line }x+y-3=0\text{ meets the X and Y-axes}
\displaystyle \text{Putting }y=0\text{ in Eq. (i), we get }x+0-3=0\Rightarrow x=3
\displaystyle \text{Thus, the line meets the X-axis at }A(3,0)
\displaystyle \text{Now, putting }x=0\text{ in Eq. (i), we get }0+y-3=0\Rightarrow y=3
\displaystyle \text{Thus, the line meets the Y-axis at }B(0,3)
\displaystyle \text{Hence, the coordinates of }A\text{ are }(3,0)\text{ and }B\text{ are }(0,3)
\displaystyle \text{(iii) The coordinates of mid point of }A(3,0)\text{ and }B(0,3)
\displaystyle =\left(\frac{3+0}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right)
\displaystyle \text{Hence, the coordinates of }M\text{ are }\left(\frac{3}{2},\frac{3}{2}\right)

\displaystyle \textbf{Question 35:}~\text{If the lines }7y=ax+4\text{ and }2y=3-x\text{ are parallel to each other,} \\ \text{then the value of }a\text{ is}~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, lines are }7y=ax+4\text{ and }2y=3-x
\displaystyle \Rightarrow y=\frac{a}{7}x+\frac{4}{7}\text{ and }y=\frac{3}{2}-\frac{x}{2}
\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2\text{ respectively, we get}
\displaystyle m_1=\frac{a}{7}\text{ and }m_2=-\frac{1}{2}
\displaystyle \text{Since, the lines are parallel, }m_1=m_2
\displaystyle \Rightarrow \frac{a}{7}=-\frac{1}{2}
\displaystyle \Rightarrow a=-\frac{7}{2}

\displaystyle \textbf{Question 36:}~\text{If two lines are perpendicular to one another, then the relation between} \\ \text{their slopes }m_1\text{ and }m_2\text{ is}~\text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(d) If two lines having slopes }m_1\text{ and }m_2\text{ are perpendicular,}
\displaystyle \text{then the angle between them is }90^\circ
\displaystyle \text{Also, }m_1\times m_2=-1

\displaystyle \textbf{Question 37:}~\text{Line }AB\text{ is perpendicular to }CD.\text{ Coordinates of }B,~C\text{ and }D\text{ are respectively } \\ (4,0),~(0,-1)\text{ and }(4,3).\text{ Find:}

\displaystyle \text{(i) slope of }CD.
\displaystyle \text{(ii) Equation of }AB.~\text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Slope of }CD,~m=\frac{y_2-y_1}{x_2-x_1}
\displaystyle \Rightarrow m=\frac{3-(-1)}{4-0}=\frac{4}{4}=1
\displaystyle \text{(ii) Equation of }AB
\displaystyle y-y_1=-\frac{1}{m}(x-x_1)\qquad[\because AB\perp CD]
\displaystyle y-0=-\frac{1}{1}(x-4)
\displaystyle \Rightarrow y=-x+4
\displaystyle \Rightarrow x+y-4=0

\displaystyle \textbf{Question 38:}~\text{Find }a,\text{ if }A(2a+2,3),~B(7,4)\text{ and }C(2a+5,2)\text{ are collinear.} \\ \text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, three points }A(2a+2,3),~B(7,4)\text{ and }C(2a+5,2)\text{ are collinear}
\displaystyle \therefore \text{Slope of }AB=\text{Slope of }BC
\displaystyle \Rightarrow \frac{4-3}{7-(2a+2)}=\frac{2-4}{(2a+5)-7}
\displaystyle \Rightarrow \frac{1}{5-2a}=\frac{-2}{2a-2}
\displaystyle \Rightarrow 2a-2=-2(5-2a)
\displaystyle \Rightarrow 2a-2=-10+4a
\displaystyle \Rightarrow 2a=8
\displaystyle \Rightarrow a=4

\displaystyle \textbf{Question 39:}~\text{(i) If the lines }kx-y+4=0\text{ and }2y=6x+7\text{ are perpendicular} \\ \text{to each other, then find the value of }k.
\displaystyle \text{(ii) Find the equation of a line parallel to }2y=6x+7\text{ and passing through }(-1,1). \\ \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }kx-y+4=0\qquad\text{and}\qquad 2y=6x+7
\displaystyle \text{From Eq. (i), }y=kx+4
\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m_1=k
\displaystyle \text{From Eq. (ii), }y=3x+\frac{7}{2}
\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m_2=3
\displaystyle \text{Since the lines are perpendicular, }m_1\times m_2=-1
\displaystyle \Rightarrow k\times 3=-1
\displaystyle \Rightarrow k=-\frac{1}{3}
\displaystyle \text{(ii) Let }6x-2y+\lambda=0\text{ be the equation of line parallel to }2y=6x+7
\displaystyle \text{and passing through }(-1,1)
\displaystyle \Rightarrow 6(-1)-2(1)+\lambda=0
\displaystyle \Rightarrow -6-2+\lambda=0
\displaystyle \Rightarrow \lambda=8
\displaystyle \text{Hence, the equation of line is }6x-2y+8=0
\displaystyle \text{or }2y=6x+8

\displaystyle \textbf{Question 40:}~\text{Find the equation of a line parallel to the line }2x+y-7=0\text{ and passing through the intersection of the lines }x+y-4=0\text{ and }2x-y=8.~\text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Given lines are }2x+y-7=0\qquad\text{(i)}
\displaystyle x+y-4=0\qquad\text{(ii)}
\displaystyle 2x-y-8=0\qquad\text{(iii)}
\displaystyle \text{From Eq. (i), }y=-2x+7
\displaystyle \Rightarrow \text{Slope, }m=-2
\displaystyle \text{Now, solving Eqs. (ii) and (iii), we get}
\displaystyle x+y-4=0\Rightarrow y=4-x
\displaystyle \text{Substituting in Eq. (iii): }2x-(4-x)-8=0
\displaystyle \Rightarrow 2x-4+x-8=0
\displaystyle \Rightarrow 3x-12=0\Rightarrow x=4
\displaystyle \Rightarrow y=4-4=0
\displaystyle \text{Thus, }x_1=4\text{ and }y_1=0
\displaystyle \text{Equation of required line }y-y_1=m(x-x_1)
\displaystyle y-0=-2(x-4)
\displaystyle \Rightarrow y=-2x+8
\displaystyle \Rightarrow 2x+y=8

\displaystyle \textbf{Question 41:}~\text{Find the value of }p\text{ if the lines }5x-3y+2=0\text{ and} \\ 6x-py+7=0 \text{ are perpendicular to each other. Hence, find the equation of a line} \\ \text{passing through }(-2,-1)\text{ and parallel to }6x-py+7=0.~\text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of lines are }5x-3y+2=0\text{ and }6x-py+7=0
\displaystyle \Rightarrow -3y=-5x-2\Rightarrow y=\frac{5}{3}x+\frac{2}{3}
\displaystyle \Rightarrow -py=-6x-7\Rightarrow y=\frac{6}{p}x+\frac{7}{p}
\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2\text{ respectively}
\displaystyle m_1=\frac{5}{3}\text{ and }m_2=\frac{6}{p}
\displaystyle \text{Since, given lines are perpendicular}
\displaystyle m_1\times m_2=-1
\displaystyle \Rightarrow \frac{5}{3}\times\frac{6}{p}=-1
\displaystyle \Rightarrow \frac{10}{p}=-1
\displaystyle \Rightarrow p=-10
\displaystyle \text{When }p=-10,\text{ slope of line }6x-py+7=0
\displaystyle m=\frac{6}{-10}=-\frac{3}{5}
\displaystyle \text{Equation of a line passing through }(-2,-1)\text{ and parallel to }6x-py+7=0
\displaystyle y+1=-\frac{3}{5}(x+2)
\displaystyle \Rightarrow 5y+5=-3x-6
\displaystyle \Rightarrow 3x+5y+11=0

\displaystyle \textbf{Question 42:}~\text{From the above figure, find:}

\displaystyle \text{(i) coordinates of points }P,~Q,~R.
\displaystyle \text{(ii) equation of the line through }P\text{ and parallel to }QR.~\text{[ICSE 2020]}
\displaystyle \text{Answer:}

\displaystyle \text{(i) From the graph, }P(4,-3),~Q(-2,-2)\text{ and }R(2,0)
\displaystyle \text{(ii) Slope of line joining }QR=\frac{0-(-2)}{2-(-2)}=\frac{2}{4}=\frac{1}{2}
\displaystyle \text{Since, required line is parallel to }QR,\text{ slope }=\frac{1}{2}
\displaystyle \text{Equation of line passing through }P(4,-3)\text{ and having slope }\frac{1}{2}
\displaystyle y-(-3)=\frac{1}{2}(x-4)
\displaystyle \Rightarrow y+3=\frac{1}{2}(x-4)
\displaystyle \Rightarrow 2y+6=x-4
\displaystyle \Rightarrow x-2y-10=0

\displaystyle \textbf{Question 43:}~\text{Find the equation of the perpendicular bisector of line segment} \\ \text{joining }A(4,2)\text{ and }B(-3,-5).~\text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A(4,2)\text{ and }B(-3,-5)
\displaystyle \text{Mid-point of line segment joining }A\text{ and }B
\displaystyle =\left(\frac{4+(-3)}{2},\frac{2+(-5)}{2}\right)=\left(\frac{1}{2},-\frac{3}{2}\right)
\displaystyle \left[\text{mid-point formula }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]
\displaystyle \text{Slope of line joining }A(4,2)\text{ and }B(-3,-5)
\displaystyle =\frac{-5-2}{-3-4}=\frac{-7}{-7}=1
\displaystyle \left[\text{slope }=\frac{y_2-y_1}{x_2-x_1}\right]
\displaystyle \text{Since, required line is perpendicular to }AB
\displaystyle \text{Slope of required line }=-\frac{1}{1}=-1
\displaystyle \text{Equation of line passing through }(x_0,y_0)\text{ and slope }m\text{ is}
\displaystyle y-y_0=m(x-x_0)
\displaystyle \text{Required line passes through }\left(\frac{1}{2},-\frac{3}{2}\right)\text{ and slope }-1
\displaystyle \Rightarrow y+\frac{3}{2}=-1\left(x-\frac{1}{2}\right)
\displaystyle \Rightarrow 2y+3=-2x+1
\displaystyle \Rightarrow 2x+2y+2=0\Rightarrow x+y+1=0

\displaystyle \textbf{Question 44:}~\text{The vertices of }\triangle ABC\text{ are }A(3,8),~B(-1,2)\text{ and }C(6,-6).
\displaystyle \text{(i) slope of }BC.
\displaystyle \text{(ii) equation of a line perpendicular to }BC\text{ and passing through }A.~\text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given vertices of }\triangle ABC\text{ are }A(3,8),~B(-1,2)\text{ and }C(6,-6)
\displaystyle \text{(i) Now, slope of }BC,~m=\frac{-6-2}{6-(-1)}=\frac{-8}{7}
\displaystyle \text{(ii) Equation of line passing through }A(3,8)\text{ and perpendicular to }BC
\displaystyle \text{Slope of perpendicular line }=\frac{7}{8}
\displaystyle y-8=\frac{7}{8}(x-3)
\displaystyle \Rightarrow 8y-64=7x-21
\displaystyle \Rightarrow 7x-8y+43=0

\displaystyle \textbf{Question 45:}~\text{If the straight lines }3x-5y=7\text{ and }4x+ay+9=0\text{ are perpendicular} \\ \text{to one another, then find the value of }a.~\text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of straight lines are }3x-5y=7\text{ and }4x+ay+9=0
\displaystyle \Rightarrow 5y=3x-7\Rightarrow y=\frac{3}{5}x-\frac{7}{5}\qquad\text{(i)}
\displaystyle \Rightarrow ay=-4x-9\Rightarrow y=-\frac{4}{a}x-\frac{9}{a}\qquad\text{(ii)}
\displaystyle \text{On comparing Eq. (i) by }y=m_1x+c_1\text{ and Eq. (ii) by }y=m_2x+c_2
\displaystyle m_1=\frac{3}{5}\text{ and }m_2=-\frac{4}{a}
\displaystyle \text{Since, given straight lines are perpendicular}
\displaystyle m_1\times m_2=-1
\displaystyle \Rightarrow \frac{3}{5}\times\left(-\frac{4}{a}\right)=-1
\displaystyle \Rightarrow -\frac{12}{5a}=-1
\displaystyle \Rightarrow 5a=12\Rightarrow a=\frac{12}{5}

\displaystyle \textbf{Question 46:}~\text{A line }AB\text{ meets X-axis at }A\text{ and Y-axis at }B\text{ and } \\ P(4,-1)\text{ divides }AB\text{ in the ratio }1:2.\text{ Find:}


\displaystyle \text{(i) the coordinates of }A\text{ and }B.
\displaystyle \text{(ii) the equation of a line through }P\text{ and perpendicular to }AB.~\text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let coordinates of }A\text{ on X-axis be }(x,0)\text{ and }B\text{ on Y-axis be }(0,y)

\displaystyle \text{Given }P(4,-1)\text{ divides }AB\text{ in ratio }1:2
\displaystyle \text{By section formula, }x_P=\frac{1\times0+2\times x}{1+2}
\displaystyle 4=\frac{2x}{3}\Rightarrow x=6
\displaystyle \text{Similarly, }y_P=\frac{1\times y+2\times0}{1+2}
\displaystyle -1=\frac{y}{3}\Rightarrow y=-3
\displaystyle \text{Hence, coordinates of }A\text{ and }B\text{ are }(6,0)\text{ and }(0,-3)
\displaystyle \text{(ii) Clearly, slope of line }AB=\frac{-3-0}{0-6}=\frac{-3}{-6}=\frac{1}{2}
\displaystyle \text{Let }m\text{ be slope of desired line perpendicular to }AB
\displaystyle \frac{1}{2}\times m=-1\Rightarrow m=-2
\displaystyle \text{Equation of line passing through }(4,-1)\text{ and slope }-2
\displaystyle y+1=-2(x-4)
\displaystyle \Rightarrow y+1=-2x+8
\displaystyle \Rightarrow 2x+y-7=0

\displaystyle \textbf{Question 47:}~\text{Find the value of }a\text{ for which the following points }A(a,3),~B(2,1)\text{ and } \\ C(5,a)\text{ are collinear. Hence, find the equation of the line.}~\text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{If three points }A,~B\text{ and }C\text{ are collinear, then use the relation}
\displaystyle \text{slope of }AB=\text{slope of }BC
\displaystyle \text{where slope of a line is given by }\frac{y_2-y_1}{x_2-x_1}
\displaystyle \text{Further, use equation }(y-y_1)=m(x-x_1)\text{ to get the required}
\displaystyle \text{equation of line}
\displaystyle \text{Given, points }A(a,3),~B(2,1)\text{ and }C(5,a)\text{ are collinear}
\displaystyle \text{So, slope of }AB=\text{slope of }BC
\displaystyle \therefore \frac{1-3}{2-a}=\frac{a-1}{5-2}
\displaystyle \Rightarrow \frac{-2}{2-a}=\frac{a-1}{3}
\displaystyle \Rightarrow -2\times3=(a-1)(2-a)
\displaystyle \Rightarrow -6=2a-a^2-2+a
\displaystyle \Rightarrow -6=-a^2+3a-2
\displaystyle \Rightarrow a^2-3a-4=0
\displaystyle \Rightarrow a^2-4a+a-4=0\qquad[\text{splitting the middle term}]
\displaystyle \Rightarrow a(a-4)+1(a-4)=0
\displaystyle \Rightarrow (a+1)(a-4)=0
\displaystyle \Rightarrow a=-1,~4
\displaystyle \text{So, the given points become either }A(-1,3),~B(2,1),~C(5,-1)
\displaystyle \text{or }A(4,3),~B(2,1),~C(5,4)
\displaystyle \text{Now, equations of line is:}
\displaystyle \text{Case I: When collinear points are }A(-1,3),~B(2,1)\text{ and }C(5,-1)
\displaystyle \text{Then, equation of line passing through }A(-1,3)\text{ and }B(2,1)\text{ is}
\displaystyle y-3=\frac{1-3}{2-(-1)}(x+1)
\displaystyle \Rightarrow y-3=\frac{-2}{3}(x+1)
\displaystyle \Rightarrow 3(y-3)=-2(x+1)
\displaystyle \Rightarrow 3y-9=-2x-2
\displaystyle \Rightarrow 3y+2x=7
\displaystyle \text{Case II: When collinear points are }A(4,3),~B(2,1)\text{ and }C(5,4)
\displaystyle \text{Then, equation of line passing through }A(4,3)\text{ and }B(2,1)\text{ is}
\displaystyle y-3=\frac{1-3}{2-4}(x-4)
\displaystyle \Rightarrow y-3=\frac{-2}{-2}(x-4)
\displaystyle \Rightarrow y-3=x-4
\displaystyle \Rightarrow y=x-1
\displaystyle \text{Hence, equations of the line is }3y+2x=7\text{ or }y=x-1

\displaystyle \textbf{Question 48:}~\text{The line through }A(-2,3)\text{ and }B(4,b)\text{ is perpendicular to the line } \\ 2x-4y=5.\text{ Find the value of }b.~\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, points are }A(-2,3)\text{ and }B(4,b)
\displaystyle \text{Now, slope of line joining the points }A\text{ and }B\text{ is}
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{b-3}{4-(-2)}=\frac{b-3}{6}\qquad\text{(i)}
\displaystyle \text{Also, given equation of line is }2x-4y=5
\displaystyle \Rightarrow 4y=2x-5
\displaystyle \Rightarrow y=\frac{2x-5}{4}\qquad[\text{dividing both sides by }4]
\displaystyle \Rightarrow y=\frac{1}{2}x-\frac{5}{4}
\displaystyle \text{On comparing with }y=m_2x+c\text{ we get }m_2=\frac{1}{2}\qquad\text{(ii)}
\displaystyle \text{According to the question, the lines are perpendicular, therefore}
\displaystyle m_1\times m_2=-1
\displaystyle \Rightarrow \frac{b-3}{6}\times\frac{1}{2}=-1
\displaystyle \Rightarrow b-3=-12
\displaystyle \Rightarrow b=-9
\displaystyle \text{Hence, the value of }b\text{ is }-9

\displaystyle \textbf{Question 49:}~\text{The equation of a line is }3x+4y-7=0.\text{ Find:}
\displaystyle \text{(i) the slope of the line.}
\displaystyle \text{(ii) the equation of a line perpendicular to the given line and passing through the} \\ \text{intersection of the lines }x-y+2=0\text{ and }3x+y-10=0.~\text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given equation of line is }3x+4y-7=0
\displaystyle \Rightarrow 4y=-3x+7
\displaystyle \Rightarrow y=-\frac{3}{4}x+\frac{7}{4}\qquad[\text{dividing both sides by }4]
\displaystyle \text{On comparing with }y=mx+c,\text{ slope }(m)=-\frac{3}{4}
\displaystyle \text{Hence, the slope of line is }-\frac{3}{4}
\displaystyle \text{(ii) Now, slope of required line perpendicular to given line is}
\displaystyle m_1=-\frac{1}{m}=-\frac{1}{-\frac{3}{4}}=\frac{4}{3}\qquad[\because m_1m_2=-1]
\displaystyle \text{Also, given lines are }x-y+2=0\qquad\text{(i)}
\displaystyle \text{and }3x+y-10=0\qquad\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 4x-8=0\Rightarrow x=2
\displaystyle \text{On putting }x=2\text{ in Eq. (i), we get }2-y+2=0
\displaystyle \Rightarrow y=4
\displaystyle \text{So, point of intersection of lines (i) and (ii) is }(2,4)
\displaystyle \text{Equation of line passing through }(2,4)\text{ and slope }\frac{4}{3}\text{ is}
\displaystyle y-4=\frac{4}{3}(x-2)
\displaystyle \Rightarrow 3(y-4)=4x-8
\displaystyle \Rightarrow 3y-12=4x-8
\displaystyle \Rightarrow 3y=4x+4
\displaystyle \Rightarrow 4x-3y+4=0
\displaystyle \text{Hence, equation of required line is }3y=4x+4\text{ or }4x-3y+4=0

\displaystyle \textbf{Question 50:}~\text{Find the value of }p\text{ for which the lines }2x+3y-7=0\text{ and } \\ 4y-px-12=0\text{ are perpendicular to each other.}~\text{[ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{If lines having slopes }m_1\text{ and }m_2\text{ are perpendicular, then }m_1m_2=-1
\displaystyle \text{Given lines are }2x+3y-7=0\text{ and }4y-px-12=0
\displaystyle \Rightarrow 3y=-2x+7\Rightarrow y=-\frac{2}{3}x+\frac{7}{3}
\displaystyle \Rightarrow 4y=px+12\Rightarrow y=\frac{p}{4}x+3
\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2
\displaystyle m_1=-\frac{2}{3}\text{ and }m_2=\frac{p}{4}
\displaystyle \text{Since, lines are perpendicular}
\displaystyle m_1m_2=-1
\displaystyle \Rightarrow -\frac{2}{3}\times\frac{p}{4}=-1
\displaystyle \Rightarrow -\frac{2p}{12}=-1
\displaystyle \Rightarrow -\frac{p}{6}=-1
\displaystyle \Rightarrow p=6
\displaystyle \text{Hence, value of }p\text{ is }6

\displaystyle \textbf{Question 51:}~\text{Find the equation of the line parallel to }3x+2y=8\text{ and passing} \\ \text{through the point }(0,1).~\text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Given equation of line is }3x+2y=8
\displaystyle \Rightarrow 2y=-3x+8\Rightarrow y=-\frac{3}{2}x+4
\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m=-\frac{3}{2}
\displaystyle \text{Since, required line is parallel, slope remains }-\frac{3}{2}
\displaystyle \text{Equation passing through }(0,1)\text{ is}
\displaystyle y-1=-\frac{3}{2}(x-0)
\displaystyle \Rightarrow 2(y-1)=-3x
\displaystyle \Rightarrow 2y-2=-3x
\displaystyle \Rightarrow 3x+2y-2=0

\displaystyle \textbf{Question 52:}~\text{If the lines }y=3x+7\text{ and }2y+px=3\text{ are perpendicular to each} \\ \text{other, then find the value of }p.~\text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Given lines are }y=3x+7\text{ and }2y+px=3
\displaystyle \text{First line is already in the form }y=mx+c
\displaystyle \therefore m_1=3
\displaystyle \text{From second equation, }2y= -px+3
\displaystyle \Rightarrow y=-\frac{p}{2}x+\frac{3}{2}
\displaystyle \therefore m_2=-\frac{p}{2}
\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1
\displaystyle \Rightarrow 3\left(-\frac{p}{2}\right)=-1
\displaystyle \Rightarrow -\frac{3p}{2}=-1
\displaystyle \Rightarrow -3p=-2
\displaystyle \Rightarrow p=\frac{2}{3}
\displaystyle \text{Hence, the value of }p\text{ is }\frac{2}{3}.

\displaystyle \textbf{Question 53:}~\text{In the given figure, }ABC\text{ is a triangle, where }B(4,-4)\text{ and }C(-4,-2). \\ D\text{ is a point on }AC.

\displaystyle \text{(i) Write down the coordinates of }A\text{ and }D.
\displaystyle \text{(ii) Find the coordinates of the centroid of }\triangle ABC.
\displaystyle \text{(iii) If }D\text{ divides }AC\text{ in the ratio }k:1,\text{ then find the value of }k.
\displaystyle \text{(iv) Find the equation of the line }BD.~\text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Coordinates of }A\text{ are }(0,6)\text{ and }D\text{ are }(-3,0)
\displaystyle \text{(ii) Coordinates of vertices of }\triangle ABC\text{ are }A(0,6),~B(4,-4),~C(-4,-2)
\displaystyle \text{Centroid of }\triangle ABC=\left(\frac{0+4-4}{3},\frac{6-4-2}{3}\right)
\displaystyle =\left(\frac{0}{3},\frac{0}{3}\right)=(0,0)
\displaystyle \text{Thus, origin is the centroid}
\displaystyle \text{(iii) Given }D\text{ divides }AC\text{ in ratio }k:1
\displaystyle (-3,0)=\left(\frac{k(-4)+1(0)}{k+1},\frac{k(-2)+1(6)}{k+1}\right)
\displaystyle \text{On comparing y-coordinates, }0=\frac{-2k+6}{k+1}
\displaystyle \Rightarrow -2k+6=0
\displaystyle \Rightarrow 2k=6
\displaystyle \Rightarrow k=3
\displaystyle \text{(iv) The line }BD\text{ passes through the points }(4,-4)\text{ and }(-3,0)
\displaystyle \text{Slope of }BD=\frac{0-(-4)}{-3-4}=\frac{4}{-7}=-\frac{4}{7}
\displaystyle \text{Equation of line }BD\text{ is}
\displaystyle y+4=-\frac{4}{7}(x-4)
\displaystyle \Rightarrow 7(y+4)=-4(x-4)
\displaystyle \Rightarrow 7y+28=-4x+16
\displaystyle \Rightarrow 4x+7y+12=0
\displaystyle \text{which is the required equation of line }BD

\displaystyle \textbf{Question 54:}~A\text{ and }B\text{ are two points on the X-axis and Y-axis, respectively.}

\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.
\displaystyle \text{(ii) }P\text{ is a point on }AB\text{ such that }AP:PB=3:1.\text{ Using section formula, find} \\ \text{the coordinates of point }P.
\displaystyle \text{(iii) Find the equation of a line passing through }P\text{ and perpendicular to }AB.~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Coordinates of }A=(4,0)\text{ and coordinates of }B=(0,4)
\displaystyle \text{(ii) Given that, }P\text{ divides }AB\text{ internally in ratio }3:1
\displaystyle \text{By section formula, }P=\left(\frac{3\times0+1\times4}{3+1},\frac{3\times4+1\times0}{3+1}\right)
\displaystyle =\left(\frac{4}{4},\frac{12}{4}\right)=(1,3)
\displaystyle \text{(iii) Equation of line }AB
\displaystyle y-0=\frac{4-0}{0-4}(x-4)
\displaystyle \Rightarrow y=\frac{4}{-4}(x-4)
\displaystyle \Rightarrow y=-(x-4)
\displaystyle \Rightarrow y=4-x
\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m=-1
\displaystyle \text{Slope of line perpendicular to }AB\text{ is }1
\displaystyle \text{Equation passing through }P(1,3)
\displaystyle y-3=1(x-1)
\displaystyle \Rightarrow y=x+2

\displaystyle \textbf{Question 55:}~ABCD\text{ is a square, where }B(1,3)\text{ and }D(3,2)\text{ are the end points} \\ \text{of the diagonal }BD.\text{ Find:}
\displaystyle \text{(i) the coordinates of point of intersection of the diagonals }AC\text{ and }BD.
\displaystyle \text{(ii) the equation of the diagonal }AC.~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Diagonal }AC\text{ is perpendicular bisector of }BD

\displaystyle \text{Coordinates of }O=\left(\frac{1+3}{2},\frac{3+2}{2}\right)=\left(2,\frac{5}{2}\right)
\displaystyle \text{Hence, coordinates of intersection of diagonals }AC\text{ and }BD\text{ are }\left(2,\frac{5}{2}\right)
\displaystyle \text{(ii) Slope of }BD=\frac{2-3}{3-1}=-\frac{1}{2}
\displaystyle \text{Since, }AC\perp BD,\text{ slope of }AC=2
\displaystyle y-\frac{5}{2}=2(x-2)
\displaystyle \Rightarrow y-\frac{5}{2}=2x-4
\displaystyle \Rightarrow 2y-5=4x-8
\displaystyle \Rightarrow 4x-2y-3=0

\displaystyle \textbf{Question 56:}~\text{From the given figure:}

\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.
\displaystyle \text{(ii) If }P\text{ divides }AB\text{ in the ratio }2:3,\text{ then find the coordinates of point }P.
\displaystyle \text{(iii) Find the equation of a line parallel to line }AB\text{ and passing through origin.}~\text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) From figure, line }AB\text{ intersects X-axis at }(5,0)\text{ and Y-axis at }(0,3)
\displaystyle \text{Hence, coordinates of }A(5,0)\text{ and }B(0,3)
\displaystyle \textbf{(ii) }P\text{ divides }AB\text{ in the ratio }2:3
\displaystyle \text{Let }m_1=2\text{ and }m_2=3
\displaystyle \text{Coordinates of }P=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)
\displaystyle \text{Here, }(x_1,y_1)=(5,0)\text{ and }(x_2,y_2)=(0,3)
\displaystyle \Rightarrow P=\left(\frac{2\times0+3\times5}{2+3},\frac{2\times3+3\times0}{2+3}\right)
\displaystyle =\left(\frac{15}{5},\frac{6}{5}\right)=\left(3,\frac{6}{5}\right)
\displaystyle \text{Hence, coordinates of }P\text{ are }\left(3,\frac{6}{5}\right)
\displaystyle \textbf{(iii) Now, slope of line }AB=\frac{3-0}{0-5}=-\frac{3}{5}
\displaystyle \text{Let equation of line parallel to }AB\text{ be }y=mx+c
\displaystyle \text{Since line passes through origin, }0=m\cdot0+c
\displaystyle \Rightarrow c=0\Rightarrow y=mx
\displaystyle \text{Slope }m=-\frac{3}{5}
\displaystyle \therefore \text{Equation of line is }y=-\frac{3}{5}x

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.