Unit - IV Coordinate Geometry
Exercise 1
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°.
$d) √3x -y +2= 0 and x -√3y +3 = 0$.
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°.
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.
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.
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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Author: D. R. Simkhada
Editor: I. R. Simkhada
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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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