Class 10: Equation of a Line – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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]