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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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  

\displaystyle \\

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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