Practice Questions
Test your analytical skills with these coordinate geometry questions.
Find the distance between points P(2, -3) and Q(10, 3).
$d = \sqrt{(10-2)^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = 10$.
What is the midpoint of the line segment joining (4, 5) and (6, 9)?
$x = (4+6)/2 = 5$ and $y = (5+9)/2 = 7$.
Find the slope of a line that passes through the origin and (4, 8).
$m = (8-0) / (4-0) = 8/4 = 2$.
What is the area of a triangle whose vertices are (0,0), (4,0), and (0,3)?
Area = $0.5 \times |0(0-3) + 4(3-0) + 0(0-0)| = 0.5 \times |12| = 6$.
Find the coordinates of the point which divides the line joining (2,4) and (8,10) internally in the ratio 1:2.
$x = (1 \cdot 8 + 2 \cdot 2)/(1+2) = 12/3 = 4$, $y = (1 \cdot 10 + 2 \cdot 4)/(1+2) = 18/3 = 6$.
Determine whether the points (1,2), (3,6), and (5,10) are collinear.
Slope = $(6-2)/(3-1) = 2$ and $(10-6)/(5-3) = 2$. Equal slopes $\Rightarrow$ collinear.
Find the equation of the line passing through (2,3) and perpendicular to $3x + 4y = 7$.
Slope of given = $-3/4$. Perpendicular slope = $4/3$. Equation: $y-3 = 4/3(x-2) \Rightarrow 4x-3y+1=0$.
Find the area of the triangle formed by (0,0), (4,0), and (4,5).
Area = $0.5 \times |0(0-5) + 4(5-0) + 4(0-0)| = 0.5 \times 20 = 10$ sq units.
If the distance between (x,2) and (3,6) is 5 units, find x.
$\sqrt{(x-3)^2 + (2-6)^2} = 5 \Rightarrow (x-3)^2 + 16 = 25 \Rightarrow (x-3)^2 = 9 \Rightarrow x = 0$ or $6$.
Find the centroid of triangle with vertices (1,2), (3,4), and (5,0).
Centroid = $((1+3+5)/3, (2+4+0)/3) = (9/3, 6/3) = (3,2)$.
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