Class 10 - Angle Between Two Lines Solved Exercises | Readmore Optional Mathematics

Unit - IV Coordinate Geometry

Exercise 1

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1. Find the angle between the following lines:

$b) 2x -3y = 8$ and $3x +2y= -5$

Solution:

Given,

Equation of line 1 = $2x -3y = 8$

Slope of line 1 $(m_1)$ = $- \dfrac{coefficient of x}{coefficient of y}$

= $- \dfrac{2}{-3}$

= $\dfrac{2}{3}$

Equation of line 2 = $3x+2y= -5$

Slope of line 2 $(m_2)$ = $- \dfrac{coefficient of x}{coefficient of y}$

= $- \dfrac{3}{2}$

Using formula;

Angle between two lines A and B is;

$tan \theta = \left ( \pm \dfrac{m_1 - m_2}{1 + m_1.m_2} \right )$

$or, \theta = tan^{-1} \left ( \pm \dfrac{\dfrac{2}{3} - \left ( - \dfrac{3}{2} \right ) }{1 + \dfrac{2}{3}. \left ( - \dfrac{3}{2} \right )} \right )$

$or, \theta = tan^{-1} \left ( \pm \dfrac{\dfrac{2}{3} - \left ( - \dfrac{3}{2} \right ) }{0} \right )$

[ Anything divided by 0 is undefined ]

$or, \theta = tan^{-1} \left ( \pm undefined \right )$

Taking positive sign,

$or, \theta = tan^{-1} undefined$

$\therefore, \theta = 90°$

Taking negative sign,

$or, \theta = tan^{-1} (-undefined)$

$\therefore, \theta = 90°$

Therefore, the required angle between the given two lines is either 90° or 90°.

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$d) √3x -y +2= 0 and x -√3y +3 = 0$.

d) 

Solution:

Given,

Equation of line (i) = $√3x -y +2=0$

Slope of line (i)  $(m_1) = - \dfrac{ coefficient \; of \;x}{coefficient \; of \;y}$

$= - \dfrac{√3}{-1}$

$= √3$

$= tan 60°$

Equation of line (ii) = $x - √3y +3 = 0$

Slope of line (ii) $(m_2) = -\dfrac{coefficient \; of \;x}{coefficient \; of \;y}$

$= -\dfrac{1}{-√3}$

$= -\dfrac{1}{√3}$

$= tan 30°$

Now,

Angle between two lines is given by:

$tan \theta = \left ( \pm \dfrac{m_1 -m_2}{1 +m_1×m_2} \right )$

$or, tan \theta = \left ( \pm \dfrac{tan60° -tan30°}{1+ tan60°×tan30°} \right )$

$or, tan \theta = \pm tan (60°-30°)$

$or, tan \theta = \pm tan 30°$

Taking positive sign,

$or, tan \theta = tan30°$

$\therefore, \theta = 30°$

Taking negative sign,

$or, tan \theta = - tan 30°$

$or, tan \theta = tan (180° -30°)$

$\therefore, \theta = 150°

Therefore, the angle between the given pair of lines is either 30° or 150°.

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3 b) Lines ax-3y=5 and 4x -2y = 3 are parallel to each other. Find the value of a.

Solution:

Given,

Equation of line (i) = $ax -3y =5$

Slope of line (i), $(m_1) = - \dfrac{coefficient\; of \;x}{coefficient\;of\;y}$

$= - \dfrac{a}{-3}$

$= \dfrac{a}{3}$

Equation of line (ii) = $4x -2y = 3$

Slope of line (ii), $(m_2) = -\dfrac{coefficient\; of \;x}{coefficient\;of\;y}$

$= - \dfrac{4}{-2}$

$= 2$

We know,

When two lines are parallel, their slopes are equal

i.e. $m_1 = m_2$

$or, \dfrac{a}{3} = 2$

$or, a = 2×3$

$\therefore a = 6$

Hence, the value of a is 6 in the above equation of line.

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4 b) Lines (a-1)x+y = 5 and (a+1)x -3y = 15 are perpendicular to each other. Find the value of a. 

5 b) Find the value of k so that 5x + ky = 20 makes an angle of 60° with x-axis.

5 d) Find the value of k so that kx + 5y = 10 makes an angle of 30° with Y-axis.

6 a) The angle between two lines is 45°. If the slope of one of them is 1/2, find the slope of the other. 

6 c) The angle between two lines is 30°. If the slope of one of them is 2, find the slope of the other.

7 a) If the line passing through (3,-4) and (-2,a) is parallel to the line given by the equation y+2x+3=0, find the value of a.

8 a) For what value of k, the line kx-3y+6=0 is perpendicular to the line joining (4,3) and (5,-3)?

9 a) Find the equation of a straight line parallel to the line y=8x+3 and passing through the point (2,-8).

9 b) Find the equation of a straight line passing through (2,-3) and parallel to the line 14x +13y +9 = 0.

10 a) Find the equation of a straight line perpendicular to the line 5x+6y+4=0 and passing through (-9,0).

10 c) Find the equation of a straight line perpendicular to the straight line 6x-4y+14=0 and passing through (-3,3).

11 a) Find the equation of the perpendicular bisector of the straight line segment joining the points (3,2) and (7,6).

12 a) A(0,6) and B(10,0) are two points and O is the origin. Find the equation of the median OM and altitude OD of triangle AOB.

13 a) In the figure, point A divides the line segment CD in the ratio of 3:2 and BA _|_ CD. Find the equation of the line AB.

14 a) The equation of a diagonal of a square is 3x -4y +5 = 0 and one of its vertices is (1,2). Find the equation of the side of the square that meet at this point.

15 a) In the given figure, ABCD is a rhombus. Find the equation of diagonals AC and BD if diagonal AC passes through the point (8,-4).

16 a) Find the equation of the straight lines which pass through the point (1,3) and make an angle of 45° with the line x - 3y +4 = 0.

16 c) In the adjoining figure, ABCD is a square. If the equation of its diagonal BD is x -3y = 2, find the equations of AB and BC.

17 a) Find the equation of the straight lines which pass through the point (0,0) and making an angle of 60° with the line x+y+3=0. 

18 a) Find the equation of the line passing through the point of intersection of the lines 3x +y = 7 and 3y = 4x -5 and parallel to the line 2x -y = 3.

19 a) Find the equation of straight line which passes through the point of intersection of the lines 2x -3y = 1 and x +2y -4 = 0 and  perpendicular to the line 3x -4y +5 = 0.

20 a) If the line x/a + y/b = 1 passes through the point of intersection of the lines x+y = 3 and 2x -3y =1 and parallel to the line y = x -6, then find the values of a and b.

21 a) Find the equation of straight line y = mx +c which is perpendicular to the line 3x -4y +5 = 0 and passes through the point of intersection of the lines 2x +y = 5 and x - y = 1.

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Class 10 - Angle Between Two Lines Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to Angle between two lines of Coordinate Geometry for Nepal's Secondary Education Examination (SEE) appearing students.

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