MCQ Questions on Coordinate Geometry for Competitive Exams

Point (a, b) = (3, -5)
Slope = 2
The general equation of a straight line
(y - b) = m(x - a)
(y + 5) = 2(x - 3)
y + 5 = 2x - 6
2x - y - 11 = 0

25x + 75y = 225 . . . . . . (i)
x = 9 . . . . . . (ii)
For meeting point, solving the equation (i) & (ii)
⇒ 25 × 9 + 75y = 225
⇒ 75y = 225 - 225
⇒ y = 0
∴ Meeting point = (9, 0)

In given rectangle
Let $$\square $$ ABCD
Angle of rectangle = 90°
AB ⊥ DA
∴ Slope $$\left[ {{{\text{m}}_1} = \frac{{ - 1}}{{{{\text{m}}_2}}}} \right]$$   (when lines are perpendicular to each other)
Slope of side DA (m₁) = $$\frac{5}{3}$$
Slope of side AB (m₂) = $$\frac{{ - 3}}{5}$$

Co-ordinates of mid point (D) which lies on the line BC = $$\left( {\frac{{3 + 5}}{2},\,\frac{{0 - 2}}{2}} \right) = \left( {4,\, - 1} \right)$$
Now, Co-ordinates of line AD = A(2, -4), D(4, -1)
∴ Equation of the line which passes through the two point (x₁, y₁), & (x₂, y₂)
⇒ y - y₁ = m(x - x₁)
∴ required equation of the line
= y + 4 = m(x - 2)
$$\eqalign{ & \left[ {\because m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right] \cr & \therefore m = \frac{{ - 1 + 4}}{{4 - 2}} = \frac{3}{2} \cr & \Rightarrow y + 4 = \frac{3}{2}\left[ {x - 2} \right] \cr & \Rightarrow 2y + 8 = 3x - 6 \cr & \Rightarrow 3x - 2y = 14 \cr} $$

$$\eqalign{ & 9x - 12y = 108 \cr & \Rightarrow \frac{{9x}}{{108}} - \frac{{12y}}{{108}} = 1 \cr & \Rightarrow \frac{x}{{12}} - \frac{y}{9} = 1 \cr} $$

$$\eqalign{ & {\text{Length of Intercept}} = \sqrt {{{12}^2} + {9^2}} \cr & = \sqrt {144 + 81} \cr & = \sqrt {225} \cr & = 15{\text{ units}} \cr} $$

Standard equation of the line y = mx + c
∴ m = tanθ = tan45° (θ = 45°)
Given
m = 1
∴ c = $$ - \frac{3}{4}$$
New required equation of the line ⇒ y = mx + c
⇒ y = 1 × x $$ - \frac{3}{4}$$
⇒ 4y = 4x - 3
⇒ 4x - 4y = 3

$$\eqalign{ & {\text{Given,}} \cr & 4x - y = 1 \cr & \therefore y = 4x - 1 \cr & 5x - ky = 2 \cr & \therefore y = \frac{{5x}}{k} - \frac{2}{k} \cr & y = {m_1}x + C \cr & {m_1} = \frac{5}{k} \cr} $$
If the lines are ⊥ then the product of their slope is -1
\[\begin{array}{l} {m_1} \times {m_2} = - 1\\ 4 \times \frac{5}{k} = - 1\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} {m_1} = 4\\ {m_2} = \frac{5}{k} \end{array} \right.\\ \therefore k = - 20 \end{array}\]

2x - 3y = 6
Cut Y-axis at x = 0
∴ 0 - 3y = 6
$$\boxed{y = - 2}$$
Hence, point will be (0, -2)

Equation of given line
$$\eqalign{ & \Rightarrow 20x + 5y = 3 \cr & \Rightarrow 5y = 3 - 20x \cr & \Rightarrow y = - 4x + \frac{3}{5} \cr} $$
Slope of line, m₁ = -4
If two are lines are ⊥ then the product of their slope = -1
$$\eqalign{ & {m_1} \times {m_2} = - 1 \cr & - 4 \times {m_2} = - 1 \cr & {m_2} = \frac{1}{4} \cr} $$
Lines passing through the points (-2, 5) and (6, b)
Therefore,
$$\eqalign{ & {\text{Slope}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{b - 5}}{{6 + 2}} = \frac{{b - 5}}{8} \cr & {\text{According to the question,}} \cr & \frac{{b - 5}}{8} = \frac{1}{4} \cr & b - 5 = 2 \cr & \therefore b = 7 \cr} $$

$$\eqalign{ & 2x - 3y = 11 \cr & 3y = 2x - 11 \cr & y = \frac{2}{3}x - \frac{{11}}{3} \cr & \left( {y = mx + c,{\text{ where }}m{\text{ is slope}}} \right) \cr & {\text{Slope}} = {m_1} = \frac{2}{3} \cr & 3x + ky = - 4 \cr & ky = - 3x - 4 \cr & y = - \frac{3}{k}x - \frac{4}{k} \cr & {m_2} = - \frac{3}{k} \cr} $$
Relation between slope of perpendicular lines
$$\eqalign{ & {m_1}{m_2} = - 1 \cr & \Rightarrow \left( {\frac{2}{3}} \right) \times \left( { - \frac{3}{k}} \right) = 1 \cr & \Rightarrow k = 2 \cr} $$