MCQ Questions on Coordinate Geometry for Competitive Exams

$$\eqalign{ & \Rightarrow {\text{mid point co - ordinate}}\left( {\frac{{ - 5 + x}}{2},\,\frac{{y + 3}}{2}} \right) \cr & \Rightarrow \frac{{ - 5 + x}}{2} = - 2 \cr & \Rightarrow - 5 + x = - 4 \cr & \Rightarrow x = 1 \cr} $$

$$\eqalign{ & \frac{{k:1}}{{{\text{A}}\left( { - 2,\,1} \right)\,\,\,\,\,\,\,\,\,\,{\text{C}}\left( {x,\,y} \right)\,\,\,\,\,\,\,\,\,\,{\text{B}}\left( {2,\,3} \right)}} \cr & \Rightarrow x = \frac{{ - 2k + 2}}{{k + 1}} \cr & {\text{At }}y{\text{ - axis}}\,\boxed{x = 0} \cr & \therefore 0 = \frac{{ - 2k + 2}}{{k + 1}} \cr & \Rightarrow \boxed{k = 1} \cr & {\text{Ratio}} = 1:1 \cr} $$

3x + 4y = 10 . . . . . . . (i)
-x + 2y = 0 . . . . . . . (ii)
On solving both the equation
3(2y) + 4y = 10
10y = 10
y = 1
∴ x = 2 × 1 = 2
∴ Solution (a, b) = (2, 1)
∴ a + b = 2 + 1 = 3

Slope (m₁) for the line PQ $$ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{0 - 3}}{{4 + 1}} = \frac{{ - 3}}{5}$$
Now, required equation of the line passing through (-1, 3) and having x-intercept of 4
$$\eqalign{ & y = m\left( {x - a} \right) \cr & \Rightarrow y = - \frac{3}{5}\left( {x - 4} \right) \cr & \Rightarrow 5y = - 3x + 12 \cr & \Rightarrow 3x + 5y = 12 \cr} $$

Parallelogram ABCD
AD || BC
Slope of line AD (m) = Slope of line BC
Slope of line BC $$ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 3}}{{ - 1 + 2}} = 1$$
Slope of AD = 1
Equation of line AD ⇒
y - y₁ = m(x - x₁)
x₁ = 5, y₁ = 0
y - 0 = 1(x - 5)
y = x - 5

$$\eqalign{ & ax - 4y = - 6 \cr & \Rightarrow 4y = ax + 6 \cr} $$

$$\eqalign{ & y = \frac{a}{4}x + \frac{6}{4} \cr & \frac{a}{4} = \frac{{ - 3}}{2} \cr & \boxed{a = - 6} \cr} $$

2x + 5y - 12 = 0
x + y = 3
at y = 0
(3, 0), (6, 0)
at x = 0
(0, 2.4)(0, 3)
Area of Δ = $$\frac{1}{2}$$ × 3 × 2 = 3

Slope of the line passing through the point (-5, 1) & (-2, 0)
$${m_1} = \frac{{0 - 1}}{{ - 2 - \left( { - 5} \right)}} = \frac{{ - 1}}{3}\,\,\,\,\,\left[ {m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right]$$
Slope of the 1^st line $$ = \frac{{ - 1}}{{{m_1}}} = \frac{{ - 1}}{{\left( {\frac{{ - 1}}{3}} \right)}} = 3$$
(∴ If 2 lines are ⊥ then their slope m₁ × m₂ = -1)

18x - 72y + 13 = 0
7x - my - 17 = 0
There is no solution, means they are parallel
$$\eqalign{ & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \cr & \Rightarrow \frac{{18}}{7} = \frac{{72}}{m} \cr & \Rightarrow m = 7 \times 4 \cr & \Rightarrow m = 28 \cr} $$