MCQ Questions on Coordinate Geometry for Competitive Exams

$$\eqalign{ & 3x + 4y - 24 = 0 \cr & \Rightarrow 3x + 4y = 24 \cr & \Rightarrow \frac{{3x}}{{24}} + \frac{{4y}}{{24}} = 1 \cr & \Rightarrow \frac{x}{8} + \frac{y}{6} = 1 \cr & {\text{Area of }}\Delta OAB = \frac{1}{2} \times 6 \times 8 = 24{\text{ sq}}{\text{. units}} \cr & {\text{And,}} \cr & x + y + 4 = 0 \cr & \Rightarrow x + y = - 4 \cr & \Rightarrow \frac{x}{{\left( { - 4} \right)}} + \frac{y}{{\left( { - 4} \right)}} = 1 \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta OCD = \frac{1}{2} \times 4 \times 4 = 8{\text{ sq}}{\text{. units}} \cr & \therefore {\text{Area of }}\Delta OCD = \frac{1}{3}{\text{Area of }}\Delta OAB \cr} $$

Distance formula between two points
Distance = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$
⇒ 13 = $$\sqrt {{{\left( {k - 4} \right)}^2} + {{\left( { - 4 - 8} \right)}^2}} $$
⇒ 169 = k² + 16 - 8k + 144
⇒ k² - 8k - 9 = 0
⇒ k² - 9k + k - 9 = 0
⇒ k(k - 9) + 1(k - 9) = 0
⇒ (k + 1)(k - 9) = 0
⇒ k = -1 and 9
∴ k = -1 (According to options)

$$\eqalign{ & {\text{Slope}} \Rightarrow m = - 1 \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & - 1 = \frac{{5 - \left( { - 1} \right)}}{{x - 2}} \cr & - 1\left( {x - 2} \right) = 6 \cr & - x + 2 = 6 \cr & x = 2 - 6 \cr & x = - 4 \cr} $$

$$\eqalign{ & 17x + my + 102 = 0 \cr & 23x + 299y + 138 = 0 \cr & {\text{Infinite solution}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \cr & \frac{{17}}{{23}} = \frac{m}{{299}} \cr & m = \frac{{17 \times 299}}{{23}} = 221 \cr} $$

$$\eqalign{ & {\text{For coincident lines}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} \cr & \therefore \frac{k}{3} = \frac{2}{1} = \frac{2}{1} \cr & {\text{Hence, }}k = 3 \times 2 \cr & k = 6 \cr} $$

$$\eqalign{ & {\text{At }}x{\text{ - axis}},\,y = 0 \cr & 8x + 3y = 24 \cr & x = 3 \cr & 2x + 8 = y \cr & x = - 4 \cr & \,\,\,8x + 3y = 24 \to \left( {\text{i}} \right) \cr & \,\,\,8x - 4y = 32 \to \left( {{\text{ii}}} \right) \cr & \underline {\, - \,\,\,\,\, + \,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,} \cr & 7y = 56 \cr & y = 8 \cr & {\text{Area}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 7 \times 8 \cr & = 28 \cr} $$

Slope (m₁) of line which passes through two points (-3, 4) and (0, 3)
$$\eqalign{ & {m_1} = \left( {\frac{{3 - 4}}{{0 + 3}}} \right)\,\,\,\,\,\left[ {\because m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right] \cr & \Rightarrow {m_1} = \frac{{ - 1}}{3} \cr} $$
Similarly, slope (m₂) of line which passes through the two points (5, 7) and (4, x)
$${m_2} = \frac{{x - 7}}{{4 - 5}} = - \left( {x - 7} \right)$$
∵ These lines perpendicular to each other,
$$\eqalign{ & \therefore {m_1} \times {m_2} = - 1 \cr & \frac{{ - 1}}{3} \times \left[ { - \left( {x - 7} \right)} \right] = - 1 \cr & x - 7 = - 3 \cr & x = 4 \cr} $$

Given that (2, 0) is solution of equation 2x + 3y = k
∴ On putting value of x & y in the above equation,
2 × 2 + 3 × 0 = k
k = 4

$$\eqalign{ & 59x + 14y - 112 = 0 \cr & \Rightarrow 59x + 14y = 112 \cr & \Rightarrow \frac{{59x}}{{112}} + \frac{{14y}}{{112}} = 1 \cr & \Rightarrow \frac{x}{{\frac{{112}}{{59}}}} + \frac{y}{8} = 1 \cr & \therefore y{\text{ - intercept of the line is }}8 \cr} $$

$$\eqalign{ & 2x + 3y = 11{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & x - 2y = - 12{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\left( {{\text{ii}}} \right)_{ \times 2}} \cr & \,2x - 4y = - 24 \cr & \underline {\,2x + 3y = 11\,} \cr & \,\,\,\,\,\,\,\,\,\,\,\,7y = 35 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 5 \cr & {\text{From equation}}\left( {\text{i}} \right) \cr & 2x + 15 = 11 \cr & x = - 2 \cr & \left( {{x_1},\,{y_1}} \right) = \left( { - 2,\,5} \right) \cr & {\text{At }}x{\text{ - axis}},\,y = 0 \cr & x - 2 \times 0 = - 12 \cr & x = - 12 \cr & \left( {{x_2},\,{y_2}} \right) = \left( { - 12,\,0} \right) \cr & {x_1} - {x_2} + {y_1} + {y_2} = - 2 + 12 + 5 + 0 = 15 \cr} $$