Class 10: Equation of a Line – Exercise 14(e) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Point $\displaystyle P$ divides the line segment joining the points $\displaystyle A (8, 0) \text{ and } B (16, -8)$ in the $\displaystyle\text{Ratio: } 3 : 5 .$ Find the co-ordinates of point $\displaystyle P .$ Also, find the equation of the line through $\displaystyle P$ and parallel to $\displaystyle 3x + 5y = 7 .$

Answer:

$\displaystyle\text{Given } P$ divides $\displaystyle A (8, 0) \text{ and } B (16, -8)$ in the $\displaystyle\text{Ratio: } 3 : 5$

$\displaystyle\text{Ratio: } m_1:m_2 = 3 : 5$

$\displaystyle\text{Let the coordinates of the point } P \text{ be } (x, y) .$ Therefore

$\displaystyle x = \frac{3 \times 16+5 \times 8}{3+5} = \frac{88}{8} = 11$

$\displaystyle y = \frac{3 \times (-8)+5 \times 0}{3+5} = \frac{-24}{8} = -3$

$\displaystyle\text{Therefore } P = (11, -3)$

$\displaystyle\text{Equation of the line given : } 3x+5y=7 \Rightarrow \text{ Slope } = m = \frac{-3}{5}$

Therefore the required equation is

$\displaystyle y - y_1 = m(x-x_1)$

$\displaystyle y -(-3) = \frac{-3}{5} (x-11) \text{ or } 5y +3x=18$

$\displaystyle \\$

Question 2: The line segment joining the points $\displaystyle A (3, -4) \text{ and } B (-2, 1)$ is divided in the $\displaystyle\text{Ratio: } 1 : 3$ at point $\displaystyle P$ in it. Find the co-ordinates of $\displaystyle P .$ Also, find the equation of the line through $\displaystyle P$ and perpendicular to the line $\displaystyle 5x - 3y = 4 .$

Answer:

$\displaystyle\text{Given } P \text{ divides } A (3,-4) \text{ and } B (-2,1)$ in the $\displaystyle\text{Ratio: } 1:3$

$\displaystyle\text{Ratio: } m_1:m_2 = 1:3$

$\displaystyle\text{Let the coordinates of the point } P \text{ be } (x, y) . \text{Therefore}$

$\displaystyle x = \frac{1 \times (-2)+3 \times 3}{1+3} = \frac{7}{4}$

$\displaystyle y = \frac{1 \times 1+3 \times (-4)}{1+3} = \frac{-11}{4}$

$\displaystyle\text{Therefore } P = ( \frac{7}{4} , \frac{-11}{4} )$

$\displaystyle\text{Equation of the line given : } 5x - 3y = 4 \Rightarrow \text{ Slope } = m_1 = \frac{5}{3}$

$\displaystyle\text{Therefore the slope of the line perpendicular to the above line } = \frac{-3}{5}$

Therefore the required equation is

$\displaystyle y - y_1 = m(x-x_1)$

$\displaystyle y - \frac{-11}{4} = \frac{-3}{5} (x- \frac{7}{4} ) \text{ or } 10y +6x+17 = 0$

$\displaystyle \\$

Question 3: A line $\displaystyle 5x + 3y + 15 = 0$ meets $\displaystyle y-axis$ at point $\displaystyle P .$ Find the co-ordinates of point $\displaystyle P .$ Find the equation of a line through $\displaystyle P$ and perpendicular to $\displaystyle x - 3y + 4 = 0 .$

Answer:

At $\displaystyle y-axis, x = 0 \Rightarrow y = -5$

Therefore the coordinate of $\displaystyle P = (0, -5)$

$\displaystyle x - 3y + 4 = 0 \Rightarrow \text{ Slope } = \frac{1}{3}$

Therefore the slope of a like perpendicular to this line $\displaystyle = -3$

Hence the line passing through $\displaystyle (0, -5) \text{ with a Slope of } -3 \text{ is}$

$\displaystyle y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0$

$\displaystyle \\$

Question 4: 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:

$\displaystyle\text{Slope of } kx - 5y + 4 = 0 \Rightarrow m_1 = \frac{k}{5}$

$\displaystyle\text{Slope of } 5x - 2y + 5 = 0 \Rightarrow m_2 = \frac{5}{2}$

$\displaystyle \text{Since the two lines are perpendicular, } m_1. m_2 = -1$

$\displaystyle \Rightarrow \frac{k}{5} . \frac{5}{2} = -1$

$\displaystyle \Rightarrow k = -2$

$\displaystyle \\$

Question 5: 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 co-ordinates of $\displaystyle A \text{ and } B$

The co-ordinates of $\displaystyle M$ [2003]

Answer:

$\displaystyle P (-1, 4) \text{ and } Q (5, -2)$

The equation of the line:

$\displaystyle y - 4 = \frac{-2-4}{5-(-1)} (x+1)$

$\displaystyle \Rightarrow y-4 = -1 (x+1)$

$\displaystyle \Rightarrow y+x=3$

$\displaystyle \text{The x-intercept} A = (3,0) \text{ and the y-intercept} B = (0,3)$

$\displaystyle \text{The coordinate of } M = ( \frac{3+0}{2} , \frac{0+3}{2} ) = ( \frac{3}{2} , \frac{3}{2} )$

$\displaystyle \\$

Question 6: $\displaystyle (1, 5) \text{ and } (-3, -1)$ are the co-ordinates of vertices $\displaystyle A \text{ and } C$ respectively of rhombus $\displaystyle ABCD .$ Find the equations of the diagonals $\displaystyle AC \text{ and } BD .$

Answer:

$\displaystyle\text{Midpoint } M = ( \frac{-3+1}{2} , \frac{-1+5}{2} ) = (-1, 2)$

$\displaystyle\text{Slope of } AC = \frac{-1-5}{-3-1} = \frac{3}{2}$

Equation of $\displaystyle AC$ :

$\displaystyle y - 5 = \frac{3}{2} (x-1) \Rightarrow 3x-2y+7=0$

$\displaystyle\text{Slope of } BD = - \frac{2}{3}$

$\displaystyle \text{Therefore equation of } BD$ :

$\displaystyle y - 2 = - \frac{2}{3} (x+1) \Rightarrow 3y+2x=4$

$\displaystyle \\$

Question 7: Show that $\displaystyle A (3, 2), B (6, -2) \text{ and } C (2, -5)$ can be vertices of a square. Find the coordinates of its fourth vertex $\displaystyle D$ , if $\displaystyle ABCD$ is a square. Without using the coordinates of vertex $\displaystyle D$ , find the equation of side $\displaystyle AD$ of the square and the equation of diagonal $\displaystyle BD .$

Answer:

$\displaystyle\text{Midpoint of } AC = ( \frac{3+2}{2} , \frac{-5+2}{2} ) = ( \frac{5}{2} , \frac{-3}{2} )$

$\displaystyle \text{Let the coordinate of } D \text{ be } (x, y)$

In a square, the diagonals bisect each other. Therefore

$\displaystyle \frac{x+6}{2} = \frac{5}{2} \Rightarrow x = -1$

$\displaystyle \frac{y-2}{2} = \frac{-3}{2} \Rightarrow y = -1$

$\displaystyle \text{Hence } D \text{ is } (-1, -1)$

$\displaystyle \text{ Slope of } AB = \frac{-2-2}{6-3} = \frac{-4}{3}$

$\displaystyle \text{Since } AB \perp AD \text{, Slope of } AD = \frac{3}{4}$

$\displaystyle \text{Hence the equation of } AD$ :

$\displaystyle y - 2 = \frac{3}{4} (x-3) \Rightarrow 4y-3x+1 = 0$

$\displaystyle \text{Slope of } BD = \frac{-2+1}{6+1} = - \frac{1}{7}$

$\displaystyle \text{Hence the equation of } BD$ :

$\displaystyle y +2 = - \frac{1}{7} (x-6) \Rightarrow 7y+x+8=0$

$\displaystyle \\$

Question 8: A line through origin meets the line $\displaystyle x = 3y + 2$ at right angles at point $\displaystyle X .$ find the coordinates of point $\displaystyle X .$

Answer:

$\displaystyle\text{Given } x = 3y + 2 .$ .. … … … (i)

$\displaystyle \text{Slope of line } x = 3y + 2 \text{ is } \frac{1}{3}$

$\displaystyle \text{Slope of perpendicular } = -3$

The equation of a line passing through $\displaystyle (0,0)$ and having slope $\displaystyle 3$ is

$\displaystyle y - 0 = -3 (x-0) \Rightarrow y = -3x .$ .. … … … (i)

Solving equations (i) and (ii)

$\displaystyle x= 3 (-3x) + 2 \Rightarrow x = \frac{1}{5}$

$\displaystyle \text{Hence } y = - \frac{3}{5}$

$\displaystyle \\$

Question 9: A straight line passes through the point $\displaystyle (3, 2)$ and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer:

Let y-intercept be $\displaystyle (0, y)$ and x-intercept be $\displaystyle (x, 0)$

$\displaystyle\text{Given } (3,2)$ is the $\displaystyle\text{Midpoint of } (0,y) \text{ and } (x,0) .$ Therefore:

$\displaystyle \text{y-intercept} = (0, 4)$

$\displaystyle \text{x-intercept} = (6, 0)$

$\displaystyle \text{Slope of line } = \frac{0-4}{6-0} = - \frac{2}{3}$

Equation of line:

$\displaystyle y-0 = - \frac{2}{3} (x-6) \Rightarrow 3y+2x=12$

$\displaystyle \\$

Question 10: Find the equation of the line passing through the point of intersection of $\displaystyle 7x + 6y = 71 \text{ and } 5x - 8y = -23$ ; and perpendicular to the line $\displaystyle 4x -2y = 1 .$

Answer:

Solve equations

$\displaystyle 7x + 6y = 71 .$ .. … … … (i)

$\displaystyle 5x - 8y = -23 .$ .. … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

$\displaystyle 43x = 215 \Rightarrow x = 5$

$\displaystyle \text{Substituting } x = 5$ in (i) we get

$\displaystyle 7(5)+6y=71 \Rightarrow y = 6$

Therefore the intercept is $\displaystyle (5, 6)$

$\displaystyle \text{Slope of line } 4x -2y = 1 \text{ is } 2$

$\displaystyle \text{Therefore the slope of perpendicular } = - \frac{1}{2}$

Hence the equation of the perpendicular:

$\displaystyle y - 6 = - \frac{1}{2} (x-5)$

$\displaystyle 2y -12 = -x+5$

$\displaystyle 2y+x=17$

$\displaystyle \\$

Question 11: Find the equation of the line which is perpendicular to the line $\displaystyle \frac{x}{a} - \frac{y}{b} =1$ at the point where this line meets $\displaystyle y-axis .$

Answer:

$\displaystyle \text{Slope of line } \frac{x}{a} - \frac{y}{b} =1 \text{ is } \frac{b}{a}$

$\displaystyle \text{Therefore slope of line perpendicular to given line } = - \frac{a}{b}$

$\displaystyle \text{y-intercept} = b$

$\displaystyle\text{Slope of } - \frac{a}{b} \text{ is }$

$\displaystyle \text{Therefore the equation of line passing through (0,b) and having } \\ \\ y - b = - \frac{a}{b} (x-0)$

$\displaystyle \Rightarrow by-b^2 = -ax$

$\displaystyle \Rightarrow ax+by=b^2$

$\displaystyle \\$

Question 12: $\displaystyle O (0, 0), A (3, 5) \text{ and } B (-5, -3)$ are the vertices of a triangle $\displaystyle OAB .$ Find:

(i) The equation of the median of triangle $\displaystyle OAB$ through vertex $\displaystyle O$

(ii) The equation of altitude of triangle $\displaystyle OAB$ through vertex $\displaystyle B$

Answer:

$\displaystyle\text{Midpoint of } AB = ( \frac{-5+3}{2} , \frac{-3+5}{2} ) = (-1, 1)$

Therefore the equation of median of $\displaystyle \triangle OAB$ through $\displaystyle O$ is

$\displaystyle \text{(i) } y - 0 = \frac{1-0}{-1-0} (x-0) \Rightarrow y = -x \text{ or } y+x = 0$

$\displaystyle \text{(ii) } \text{Slope of } OA = \frac{5-0}{3-0} = \frac{5}{3}$

$\displaystyle \text{Slope of line perpendicular to } OA = - \frac{3}{5}$

Therefore the equation of altitude of $\displaystyle \triangle OAB$ through $\displaystyle B$

$\displaystyle y - (-3) = - \frac{3}{5} (x-(-5))$

$\displaystyle y+3 = - \frac{3}{5} (x+5)$

$\displaystyle 5y+15 = -3x-15$

$\displaystyle 5y+3x+30 = 0$

$\displaystyle \\$

Question 13: Determine whether the line through points $\displaystyle (-2, 3) \text{ and } (4, 1)$ is perpendicular to the line $\displaystyle 3x = y + 1 .$ Does line $\displaystyle 3x = y +1$ bisect the line segment joining the two given points?

Answer:

$\displaystyle \text{Slope of line passing through } (-2,3) \text{ and } (4,1) = \frac{1-3}{4-(-2)} = - \frac{1}{3}$

$\displaystyle\text{Slope of } 3x = y + 1 \text{ is } 3$

$\displaystyle \text{Slope of perpendicular } = - \frac{1}{3}$

Therefore line passing through $\displaystyle (-2, 3) \text{ and } (4, 1)$ is perpendicular to $\displaystyle 3x = y + 1$

$\displaystyle\text{Midpoint of } (-2, 3) \text{ and } (4, 1) = ( \frac{4-2}{2} , \frac{1+3}{2} ) = (1,2)$

$\displaystyle \text{Substituting } (1, 2)$ in $\displaystyle 3x=y+1$ we get that it satisfies the equation. $\displaystyle\text{Therefore } 3x=y+1$ bisects the line joining $\displaystyle (-2, 3) \text{ and } (4, 1)$

$\displaystyle \\$

Question 14: Given a straight line $\displaystyle x \cos 30^{\circ} +y \sin 30^{\circ}=2 .$ Determine the equation of the other line which is parallel to its and passes through $\displaystyle (4, 3) .$

Answer:

$\displaystyle\text{Given } x \cos 30^{\circ} + y \sin 30^{\circ} = 2$

$\displaystyle \text{Slope of this line } = - \frac{ \cos 30^{\circ}}{ \sin 30^{\circ}} = -\sqrt{3}$

Equation of line with slope $\displaystyle -\sqrt{3}$ and passing through $\displaystyle (4, 3)$ is

$\displaystyle y-3 = -\sqrt{3} (x-4)$

$\displaystyle y+\sqrt{3} x= 4 \sqrt{3}+3$

$\displaystyle \\$

Question 15: Find the value of $\displaystyle k$ such that the line $\displaystyle (k-2)x+(k+3)y-5=0$ is:

(i) Perpendicular to the line $\displaystyle 2x - y + 7 = 0$ (ii) Parallel to it.

Answer:

$\displaystyle\text{Given } (k-2)x+(k+3)y-5=0$

$\displaystyle \text{Slope of this line } = - \frac{k-2}{k+3}$

$\displaystyle \text{Slope of line } 2x-y+7 = 0 \text{ is } 2$

$\displaystyle \text{Slope of line perpendicular to this line } = - \frac{1}{2}$

(i) If perpendicular

$\displaystyle - \frac{1}{2} = - \frac{k-2}{k+3}$

$\displaystyle k+3 = 2k-4$

$\displaystyle k= 7$

(ii) If parallel

$\displaystyle 2 = - \frac{k-2}{k+3}$

$\displaystyle 2k+6 = -k +2$

$\displaystyle 3k = -4 \Rightarrow k = - \frac{4}{3}$

$\displaystyle \\$

Question 16: The vertices of a triangle $\displaystyle ABC$ are $\displaystyle A (0, 5), B (-1, -2) \text{ and } C (11, 7) .$ Write down the equation of $\displaystyle BC .$ Find:

(i) The equation of the line through $\displaystyle A$ and perpendicular to $\displaystyle BC .$

(ii) The coordinates of the point $\displaystyle P$ , where the perpendicular through $\displaystyle A$ , as obtained in (i.), meets $\displaystyle BC .$

Answer:

$\displaystyle \text{(i) } \text{Slope of } BC = \frac{7-(-2)}{11-(-1)} = \frac{9}{4} = \frac{3}{4}$

$\displaystyle \text{Slope of line perpendicular to } BC = - \frac{4}{3}$

Therefore equation of line passing through $\displaystyle A (0,5)$ with slope $\displaystyle - \frac{4}{3}$ is:

$\displaystyle y-5 = - \frac{4}{3} (x-0)$

$\displaystyle 3y-15 = -4x$

$\displaystyle 3y+4x=15 .$ .. … … … (i)

(ii) Equation of $\displaystyle BC$

$\displaystyle y-7 = \frac{3}{4} (x-11)$

$\displaystyle 4y - 28 = 3x - 33$

$\displaystyle 4y-3x= -5 .$ .. … … … (ii)

Solving (i) and (ii) we get $\displaystyle x = 3 \text{ and } y = 1 .$

$\displaystyle\text{Therefore } P \text{ is } (3, 1)$

$\displaystyle \\$

Question 17: From the given figure, find:

(i) The co-ordinates 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 18: $\displaystyle P (3, 4), Q (7, -2) \text{ and } R (-2, -1)$ 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 19: $\displaystyle A ((8, -6), B (-4, 2) \text{ and } C (0, -10)$ are vertices of a triangle $\displaystyle ABC .$ If $\displaystyle P$ is the mid-point of $\displaystyle AC$ , use co-ordinate geometry to show that $\displaystyle PQ$ is parallel to $\displaystyle BC .$ Give a special name to quadrilateral $\displaystyle PBCQ .$

Answer:

$\displaystyle \text{Coordinates of } P = ( \frac{-4+8}{2} , \frac{2-6}{2} ) = (2, -2)$

$\displaystyle \text{Coordinates of } Q = ( \frac{8+0}{2} , \frac{-6-10}{2} ) = (4,-8)$

$\displaystyle\text{Slope of } PQ = \frac{-8-(-2)}{4-2} = \frac{-6}{2} = -3$

$\displaystyle\text{Slope of } BC = \frac{-10-2}{0-(-4)} = \frac{-12}{4} = -3$

$\displaystyle\text{Therefore } PB \parallel BC .$

$\displaystyle PQBC$ is a trapezoid.

$\displaystyle \\$

Question 20: A line $\displaystyle AB$ meets the $\displaystyle x-axis$ at point $\displaystyle A \text{ and } y-axis$ at point $\displaystyle B .$ The point $\displaystyle P (-4, -2)$ divides the line segment $\displaystyle AB$ internally such that $\displaystyle AP : PB = 1 : 2 .$ Find:

(i) The co-ordinates of $\displaystyle A \text{ and } B .$

(ii) Equation of the line through $\displaystyle P$ and perpendicular to $\displaystyle AB .$

Answer:

$\displaystyle AP:PB = 1:2$

$\displaystyle \text{(i) Let } A (0,y) \text{ and } B(x,0)$

$\displaystyle\text{Therefore } -4 = \frac{1 \times x+2 \times 0}{3} \Rightarrow x = -6$

Similarly, $\displaystyle -2 = \frac{1 \times 0+2 \times y}{3} \Rightarrow y= -3$

$\displaystyle\text{Therefore } B(-6, 0) \text{ and } A (0, -3)$

$\displaystyle \text{(ii) } \text{Slope of } AB = \frac{-3-0}{0-(-6)} = \frac{-3}{6} = \frac{-1}{2}$

Slope of line perpendicular to $\displaystyle AB = 2$

Therefore the equation of line passing through $\displaystyle P(-4, -2)$ with slope $\displaystyle 2$ :

$\displaystyle y - (-2) = 2(x-(-4))$

$\displaystyle y+2 = 2(x+4)$

$\displaystyle y = 2x+6$

$\displaystyle \\$

Question 21: 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 22: 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 23: The given figure (not drawn to scale) shows two straight lines $\displaystyle AB \text{ and } CD .$ If equation of the line $\displaystyle AB is y =x+1$ and equation of $\displaystyle CD \text{ is } y = \sqrt{3}x-1 .$ Write down the inclination of lines $\displaystyle AB \text{ and } CD$ ; also, find the angle between $\displaystyle AB \text{ and } CD$ [1989]

Answer:

$\displaystyle AB: y = x+1$

$\displaystyle CD: y = \sqrt{3}x - 1$

$\displaystyle\text{Slope of } AB = 1$

$\displaystyle \tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^{\circ}$

$\displaystyle\text{Slope of } CD = \sqrt{3}$

$\displaystyle \tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^{\circ}$

$\displaystyle\text{Therefore } 45^{\circ}+(180^{\circ}-60^{\circ}) + \theta = 180^{\circ}$

$\displaystyle \Rightarrow \theta = 15^{\circ}$

$\displaystyle \\$

Question 24: 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 $\displaystyle\text{Ratio: } 2 : 3$ [2001]

Answer:

$\displaystyle\text{Given } P$ divides the line segment joining $\displaystyle A (-2, 6) \text{ and } B (3, -4)$ in the $\displaystyle\text{Ratio: } 2 : 3$

Let the coordinates of $\displaystyle P = (x, y)$

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 25: The ordinate of a point lying on the line joining points $\displaystyle (6, 4) \text{ and } (7, -5) is -23 .$ Find the co-ordinates of that point.

Answer:

$\displaystyle \text{Let the ordinate of a point lying on the line joining points } (6, 4) \text{ and } \\ \\ (7, -5) \text{ is } -23 \text{ be } (x, -23)$

Equation of line passing through $\displaystyle (6,4) \text{ and } (7, -5)$

$\displaystyle y - 4 = \frac{-5-4}{7-6} (x-6)$

$\displaystyle y-4 = -9(x-6)$

$\displaystyle y+9x=58$

Therefore if $\displaystyle y = -23$ , then

$\displaystyle -23+9x=58 \Rightarrow x = 9$

Therefore the point is $\displaystyle (9, -23)$

$\displaystyle \\$

Question 26: Point $\displaystyle A \text{ and } B$ have co-ordinates $\displaystyle (7, 3) \text{ and } (1, 9)$ respectively. Find:

(i) The $\displaystyle\text{Slope of } AB$

(ii) The equation of perpendicular bisector of the line segment $\displaystyle AB$

(iii) The value of $\displaystyle p of (-2, p)$ lies on it [2008]

Answer:

$\displaystyle \text{(i) } \text{Slope of } AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2$

$\displaystyle \text{(ii) } \text{Midpoint of } AB = ( \frac{7+1}{2} , \frac{-3+9}{2} ) = (4,3)$

Therefore equation of line passing through $\displaystyle (4,3)$ and slope $\displaystyle \frac{1}{2}$ 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 27: $\displaystyle A \text{ and } B$ 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) Co-ordinates of $\displaystyle A \text{ and } B$

$\displaystyle \text{(ii) } \text{Slope of line } AB$

(iii) Equation of line $\displaystyle AB$ [2010]

Answer:

$\displaystyle \text{Let } A(x,0) \text{ and } B(0, y)$

$\displaystyle P(2, -3)$ is the mid point

$\displaystyle \text{(i) } \text{Therefore } 2 = \frac{0+x}{2} \Rightarrow x = 4$

$\displaystyle -3 = \frac{y+0}{2} \Rightarrow y = -6$

$\displaystyle \text{Hence } A(4,0) \text{ and } B(0, -6)$

$\displaystyle \text{(ii) } \text{Slope of } AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}$

(iii) Equation of $\displaystyle AB$

$\displaystyle y = 0 = \frac{3}{2} (x-4)$

$\displaystyle 2y = 3x-12$

$\displaystyle \\$

Question 28: 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 $\displaystyle x -y + 2 = 0 \text{ and } 3x + y- 10 = 0$

$\displaystyle y = 4 \text{ and } x = 2$

$\displaystyle \text{Therefore intersection } = (2, 4)$

Therefore equation of line

$\displaystyle y-4 = \frac{4}{3} (x-2)$

$\displaystyle 3y-12 = 4x-8$

$\displaystyle 3y = 4x+4$

$\displaystyle \\$

Question 29: $\displaystyle ABCD$ is a parallelogram where $\displaystyle A (x, y), B (5, 8), C (4, 7) \text{ and } D (2, -4) .$ Find: (i) Co-ordinates of $\displaystyle A$ (ii) Equation of diagonal $\displaystyle BD$ [2011]