MCQ Questions on Coordinate Geometry for Competitive Exams

At y-axis, x = 0
0 - 7y = -42
y = 6
(α, β) = (0, 6)
At x-axis, y = 0
6x + 0 = 15
x = $$\frac{5}{2}$$
(γ, δ) = $$\left( {\frac{5}{2},\,0} \right)$$
α + β + γ + δ = 0 + 6 + $$\frac{5}{2}$$ + 0 = $$\frac{{17}}{2}$$

$$\eqalign{ & {m_1}\left( {{\text{slope}}} \right) = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & \Rightarrow \frac{{ - 4 - \left( { - 1} \right)}}{{4 - 5}} \cr & \Rightarrow \frac{{ - 3}}{{ - 1}} = \left( 3 \right) \cr & \therefore {m_1} = {m_2}\left( {{\text{In parallel condition}}} \right) \cr & \therefore {m_2} = 3 \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta ABC \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {2 - 14} \right) + 8\left( {14 - 2} \right) + 8\left( {2 - 2} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times \left( { - 12} \right) + 8\left( {12} \right) + 8\left( 0 \right)} \right] \cr & = \frac{1}{2}\left[ { - 48 + 96} \right] \cr & = \frac{1}{2} \times 48 \cr & = 24{\text{ sq}}{\text{. units}} \cr & {\text{Area of }}\Delta ACD \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {14 - 10} \right) + 8\left( {10 - 2} \right) + 4\left( {2 - 14} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times 4 + 8 \times 8 + 4 \times \left( { - 12} \right)} \right] \cr & = \frac{1}{2}\left[ {16 + 64 - 48} \right] \cr & = \frac{1}{2} \times 32 \cr & = 16{\text{ sq}}{\text{. units}} \cr & {\text{Hence Area of Quadrilateral }}ABCD \cr & = {\text{Area of }}\left( {\Delta ABC + \Delta ACD} \right) \cr & = 24 + 16 \cr & = 40{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & 3x - 6y = 4 \cr & \Rightarrow 6y = 3x - 4 \cr & \Rightarrow y = \frac{{3x}}{6} - \frac{4}{6} \cr & \Rightarrow y = \frac{1}{2}x - \frac{2}{3} \cr & y = {m_1}x + c \cr & {\text{If lines are parallel then their slopes are equal}} \cr & \therefore {m_1} = {m_2} = \frac{1}{2} \cr} $$

$$\eqalign{ & 2x + 5y = 12 \cr & x{\text{ - axis}}:y = 0\,;\,x = 6 \cr & x + y = 3 \cr & x{\text{ - axis}}:y = 0\,;\,x = 3 \cr & 2x + 5y = 12\,\,\,\,\,*1 \cr & \underline {\,x + y = 3\,\,\,\,\,\,\,\,\,\,\,\,*2\,} \cr & \,\,\,\,\,2x + 5y = 12 \cr & \,\,\,\,\,2x + 2y = 6 \cr & \underline {\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, - \,\,\,\,} \cr & \,\,\,\,\,3y = 6 \cr & \,\,\,\,\,\,y = 2 \cr & \,\,\,\,\,\,x = 3 - 2 = 1 \cr} $$

$$\eqalign{ & {\text{Area of traingle}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 3 \times 2 \cr & = 3 \cr} $$

Since ordinate = abscissa
∴ k = 2k - 1
k = 1

Reflection of point (4, -6) in origin is (-4, 6)

Go through option,
Points (4, 3) and (-2, 1) will satisfy the equation from option B, x² + y² - 2x - 4y = 5

$$\eqalign{ & 5x - 2y = - 1\,.\,.\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \times 2 \cr & 4y - 3x = - 5\,.\,.\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & 10x - 4y = - 2 \cr & \underline {\, - 3x + 4y = - 5\,} \cr & 7x = - 7 \cr & x = - 1,\,\,y = - 2 \cr & 2\alpha - 3\beta \cr & = 2\left( { - 1} \right) - 3\left( { - 2} \right) \cr & = - 2 + 6 \cr & = 4 \cr} $$

Area of Δ formed by the graph x - y = 0, x + y = 2 and x-axis

Intersection point of lines x - y = 0 and x + y = 2 will be (1, 1)
∴ Area bounded by lines x - y = 0, x - y = 2 and x-axis
= $$\frac{1}{2}$$ × 2 × 1
= 1 sq unit