MCQ Questions on Coordinate Geometry for Competitive Exams

A(3, -2), B(1, 4), C(-2, x)
Condition for collinearity of three points
0 = x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)
0 = 3(4 - x) + 1(x + 2) + (-2)(-2 - 4)
0 = 12 - 3x + x + 2 + 12
2x = 26
x = 13

Alternate Solution:
For collinear condition slope will be same
$$\eqalign{ & \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} \cr & \frac{{4 + 2}}{{1 - 3}} = \frac{{x - 4}}{{ - 2 - 1}} \cr & \frac{6}{{ - 2}} = \frac{{x - 4}}{{ - 3}} \cr & 9 = x - 4 \cr & x = 13 \cr} $$

2x - 3y + 6 = 0
y = 0
⇒ 2x - 3 × 0 = -6
2x = -6
x = -3
y = 0
y = 0 ⇒ 4x + 0 = 16
x = 4 ; y = 0
$$\eqalign{ & 2x - 3y = - 6\,\,\, * 2 \cr & \underline {4x + y = 16\,\,} \,\,\,\,\, * 1 \cr} $$
4x - 6y = -12 . . . . . . (i)
4x + y = 16 . . . . . . (ii)
Solve equation (i) and (ii)
y = 4 ; x = 3

$$\Delta {\text{ABC}} = \frac{1}{2} \times \left( {4 + 3} \right) \times 4 = 14$$

$$\eqalign{ & \left( {x,\,y} \right) = \left( {\frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\,\frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right) \cr & 5 = \frac{{2 \times 0 + 5 \times x}}{{2 + 5}} \cr & 35 = 5x \cr & x = 7 \cr & - 2 = \frac{{2 \times y + 5 \times 0}}{{2 + 5}} \cr & - 14 = 2y \cr & y = - 7 \cr & x = 7,\,y = - 7 \cr} $$

Given,
Equation of the lines:
x - y = -1 . . . . . . (i)
3x - 2y = 0 . . . . . . (ii)
From equation (i) & (ii)
Intersecting co-ordinate (x₁, y₁) = (2, 3)
m = $$\frac{{ - 1}}{2}$$ given
Now,
Required equation of the line
= y - y₁ = m(x - x₁)
= y - 3 = $$\frac{{ - 1}}{2}$$ (x - 2)
x + 2y = 8

$$\eqalign{ & {\text{O is mid point of AB}} \cr & x = \frac{{5 + 0}}{2} = \frac{5}{2} \cr & y = \frac{{ - 3 + 2}}{2} = \frac{{ - 1}}{2} \cr & {\text{Slope of AB }}\left( {{m_1}} \right) \cr & = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & = \frac{{2 - \left( { - 3} \right)}}{{0 - 5}} \cr & = \frac{5}{{ - 5}} \cr & = - 1 \cr} $$
Slope of (m₂) CD = 1 [lines are perpendicular to each other]
m₁ × m₂ = -1
Equation of line
$$\eqalign{ & \left( {y + \frac{1}{2}} \right) = 1\left( {x - \frac{5}{2}} \right) \cr & \Rightarrow 2\left( {x - y} \right) = 6 \cr & \Rightarrow x - y = 3 \cr} $$

(x - a)² + (y - b)² = r²
Origin point = (0, 0)
(x - 0)² + (y - 0)² = 6²
x² + y² = 36
x² + y² - 36 = 0

$$\eqalign{ & x = \frac{{4 \times \frac{7}{2} + 1 \times 6}}{{4 + 1}} \Rightarrow 4 \cr & y = \frac{{4 \times 6 + 1 \times 1}}{{4 + 1}} \Rightarrow 5 \cr & {\text{A}}\left( {4,\,5} \right) \cr} $$

$$\eqalign{ & {\text{By distance formula, }} \cr & {\text{PQ}} = \sqrt {{{\left( {0 - 2} \right)}^2} + {{\left( {\frac{3}{2} - 0} \right)}^2}} \cr & = \sqrt {4 + \frac{9}{4}} \cr & = \frac{{\sqrt {25} }}{2} \cr & = \frac{5}{2} \cr & = 2.5{\text{ cm}} \cr} $$

$$\eqalign{ & \left( {y - {y_1}} \right) = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right) \cr & \Rightarrow \left( {y - 6} \right) = - \frac{1}{3} \times \left( x \right) \cr & \Rightarrow 3y - 18 = - x \cr & \Rightarrow x + 3y = 18 \cr} $$

$$\eqalign{ & \,\,\,\,4x - 2y = 10 \cr & \,\,\,\,4x + ky = 2 \cr & \underline {\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \,\,\,\,} \cr & \left( { - k - 2} \right)y = 8 \cr & {\text{The quardinate of }}y{\text{ is}} = 4 \cr & \left( { - k - 2} \right) \times 4 = 8 \cr & - k - 2 = 2 \cr & k = - 4 \cr} $$