MCQ Questions on Coordinate Geometry for Competitive Exams

Equation of line perpendicular to line 3x - 2y = 6 is 2x + 3y + p = 0

As, the line passes through origin (0, 0)
∴ 2 × 0 + 3 × 0 + p = 0
∴ p = 0
Now, the equation is 2x + 3y = 0
∴ 2x = -3y
x = $$\frac{{ - 3}}{2}$$ y
By putting this value in equation 3x - 2y = 6
$$\eqalign{ & \Rightarrow 3\left( {\frac{{ - 3}}{2}} \right)y - 2y = 6 \cr & \Rightarrow \frac{{ - 9}}{2}y - 2y = 6 \cr & \Rightarrow \frac{{ - 13y}}{2} = 6 \cr & \Rightarrow y = - \frac{{12}}{{13}} \cr & \therefore x = \frac{{ - 3}}{2} \times \left( {\frac{{ - 12}}{{13}}} \right) \cr & x = \frac{{18}}{{13}} \cr & \therefore {\text{Co - ordinate of point}} \cr & {\text{M}} = \left[ {\frac{{18}}{{13}},\, - \frac{{12}}{{13}}} \right] \cr} $$

Slope of line y = mx + c is m
y = 2x + c, slope (m₁) = 2
Lines are perpendicular to each other $${m_1} = \frac{{ - 1}}{{{m_2}}}$$
Slope of perpendicular line $$\left( {{m_2}} \right) = \frac{{ - 1}}{2}$$
on x axis y = 0
Equation of line which passes through (-5, 0)
(y - y₁) = slope (m₂)(x - x₁)
y - 0 = $$\frac{{ - 1}}{2}$$ (x + 5)
2y = -x - 5
x + 2y = -5

$$\eqalign{ & 3x + 4y = 12 \cr & \Rightarrow \frac{{3x}}{{12}} + \frac{{4y}}{{12}} = 1 \cr & \Rightarrow \frac{x}{4} + \frac{y}{3} = 1 \cr & \therefore {\text{Length of intercept AB}} = \sqrt {{4^2} + {3^3}} \cr & = \sqrt {25} \cr & = 5{\text{ units}} \cr} $$

Slope of line PR
$$\eqalign{ & \Rightarrow {m_1} = \frac{{0 - 2}}{{ - 2 - 4}} \cr & \Rightarrow {m_1} = \frac{{ - 2}}{{ - 6}} \cr & \Rightarrow {m_1} = \frac{1}{3} \cr} $$
∴ Slope of line QS = -3 = m₂ {As it is perpendicular to PR}
Coordinates of point O
$$ \Rightarrow \left[ {\frac{{ - 2 + 4}}{2},\,\frac{{2 + 0}}{2}} \right] \Rightarrow \left( {1,\,1} \right)$$
∴ Equation of line QS which passes through point O(1, 1)
⇒ y - 1 = m₂(x - 1)
⇒ y - 1 = -3(x - 1)
⇒ y - 1 = -3x + 3
⇒ 3x + y = 4

$$\eqalign{ & 2x + 3y = - 6 \cr & y = - \frac{2}{3}x - \frac{6}{3} \cr & \therefore {\text{Slope}} = - \frac{2}{3} \cr} $$
⇒ Two lines are perpendicular if m₁ × m₂ = -1
Y intercept = 3
⇒ y = mx + c
⇒ m₂ = $$\frac{3}{2}$$
The equation of line with slope $$\frac{3}{2}$$ is:-
y - y₁ = m₂(x - x₁)
where, (x₁ y₁) = (0, 3)
y - 3 = $$\frac{3}{2}$$(x - 0)
2y - 6 = 3x
3x - 2y = -6

Given,
Equation of line x - 6y = 8
we write it as
$$y = \frac{x}{6} - \frac{8}{6}$$
∴ y = mx (where m is a slope)
∴ m₁ = $$\frac{1}{6}$$
If lines are perpendicular then product of their slopes is equal to -1
m₁ × m₂ = -1
∴ $$\frac{1}{6}$$ × m₂ = -1
m₂ = - 6
∴ Slope of perpendicular line = - 6
Perpendicular line which passes through the points (5, a) and (4, 3)
$$\eqalign{ & \therefore \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = {m_2} \cr & \Rightarrow \frac{{3 - a}}{{4 - 5}} = - 6 \cr & \Rightarrow 3 - a = + 6 \times + 1 \cr & \therefore a = - 3 \cr} $$

Equation of line which intercepts x-axis and y-axis are given below:-
$$\eqalign{ & \Rightarrow \frac{x}{a} + \frac{y}{b} = 1 \cr & {\text{where, }}a = \frac{3}{4},\,b = \frac{{ - 2}}{3}\,\,\left[ {{\text{given}}} \right] \cr & \Rightarrow \frac{x}{{\frac{3}{4}}} + \frac{y}{{\frac{{ - 2}}{3}}} = 1 \cr & \Rightarrow \frac{{4x}}{3} - \frac{{3y}}{2} = 1 \cr & \Rightarrow 8x - 9y = 6 \cr} $$

$$\eqalign{ & 3x + y = 5 \cr & \underline {2x - y = 5} \cr & 5x = 10 \cr & x = 2,\,y = - 1 \cr & \left( {\alpha ,\,\beta } \right) = \left( {2,\, - 1} \right) \cr & 3\alpha + \beta = 3 \times 2 - 1 = 5 \cr} $$

Slope (m₁) for the line which passes through the points (-3, 1) and (x, 5) $$ = \frac{{5 - 1}}{{x + 3}} = \frac{4}{{x + 3}}$$
Similarly, slope (m₂) for the line which passes through the points (-2, -1) and (6, 3) $$ = \frac{{3 + 1}}{{6 + 2}} = \frac{4}{8} = \frac{1}{2}$$
If two lines are parallel to each other.
$$\eqalign{ & {\text{Then, }}{m_1} = {m_2} \cr & \Rightarrow \frac{4}{{x + 3}} = \frac{1}{2} \cr & \Rightarrow x + 3 = 8 \cr & \Rightarrow x = 5 \cr} $$

Given,

$$\eqalign{ & m = \frac{{ - 2}}{3} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & \Rightarrow \frac{{ - 2}}{3} = \frac{{2 + 3}}{{5 - x}} \cr & \Rightarrow \frac{{ - 2}}{3} = \frac{5}{{5 - x}} \cr & \Rightarrow - 10 + 2x = 15 \cr & \Rightarrow 2x = 25 \cr & \Rightarrow x = 12.5 \cr} $$