MCQ Questions on Coordinate Geometry for Competitive Exams

The point in 4^th quadrant that is 6 unit away from x-axis and 7 unit away from y-axis is (7, -6).

By the midpoint formula, $$\left( {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right)$$
Midpoint of line AC $$ = \left( {\frac{{7 + 1}}{2},\,\frac{{ - 8 + 4}}{2}} \right) = \left( {4,\, - 2} \right)$$
O(x₃, y₃) = (4, -2)
(M₁) Slope of line AC,
$${M_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - \left( { - 8} \right)}}{{1 - 7}} = - 2$$
If the lines are ⊥, then M₁ × M₂ = -1
- 2 × M₂ = -1
M₂ = $$\frac{1}{2}$$
∴ Equation of line BD
y - y₃ = M₂(x - x₃)
y - (-2) = $$\frac{1}{2}$$(x - 4)
y + 2 = $$\frac{1}{2}$$(x - 4)
x - 2y = 8

The equation whose graph passes through origin must satisfy point (0, 0) in the equation means if we put value of x & y equal to zero, the equation must be equal to zero.
∴ 2x + 3y = 0

If P₂(4, -3) is reflected through y-axis then if becomes P₁(-4, -3) and if it is reflected through origin then it becomes P(4, 3).

$$\eqalign{ & {\text{Coordinates of centroid of a }}\Delta \cr & \Rightarrow \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right) \cr & \Rightarrow \left( {\frac{{2 + \left( { - 4} \right) + 5}}{3},\,\frac{{5 + 0 + 4}}{3}} \right) \cr & \Rightarrow \left( {1,\,3} \right) \cr} $$

$$\eqalign{ & 4x + \frac{1}{3}y = \frac{8}{3} \cr & 12x + y = 8{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0 \cr & 2x + 3y = - 10{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Solve equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 12x + y = 8 \cr & \underline {2x + 3y = - 10} \,\,\,\,\, * 6 \cr & 12x + y = 8 \cr & 12x + 18y = - 60 \cr & \underline { - \,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,} \cr & - 17y = 68 \cr & y = - 4 \cr & x = 1 \cr & P = \left( {1,\, - 4} \right) \cr & {\text{Only option }}\left( {\text{D}} \right){\text{satisfy in this point }}P \cr & 3x - y - 7 = 0 \cr & 3 \times 1 - \left( { - 4} \right) - 7 = 0 \cr & 0 = 0\,\,\,\left[ {{\text{satisfy}}} \right] \cr} $$

Given,
P(1, 2), Q(2, -1) and R(0, 5)
Slope of line QR (m₁) $$ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{5 - \left( { - 1} \right)}}{{0 - 2}} = - 3$$
If the lines are perpendicular then product of slopes is equal to -1
m₁ × m₂ = -1
-3 × m₂ = -1
m₂ = $$\frac{1}{3}$$
∴ Equation of line passes through the point (1, 2) whose slope (m₂) = $$\frac{1}{3}$$
y - y₁ = m₂(x - x₁)
y - 2 = $$\frac{1}{3}$$(x - 1)
y - $$\frac{1}{3}$$x = 2 - $$\frac{1}{3}$$
3y - x = $$\frac{5}{3}$$ × 3
x - 3y = -5

If Q₂ (-5, 3) is first reflected in x-axis then it goes in III^rd quadrant Q₁(-5, -3)
If Q₁ is reflected in y-axis to Q(5, -3)

y = 3x must pass through the point (0, 0) because only this point satisfies the equation.