Master Coordinate Geometry with these 10 fully solved examples covering Distance Formula, Section Formula, and Area of Triangles. Afterward, test your skills with 50 practice questions provided with an answer key.
Part 1: 10 Fully Solved Important Questions
Q1 - Distance Formula
Find the distance between the points $A(2, 3)$ and $B(4, 1)$.
Find the coordinates of the midpoint of the line segment joining $P(-5, 7)$ and $Q(-1, 3)$.
Solution:
Using the midpoint formula $M(x, y) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$:
$$x = \frac{-5 + (-1)}{2} = \frac{-6}{2} = -3$$
$$y = \frac{7 + 3}{2} = \frac{10}{2} = 5$$
The midpoint is $(-3, 5)$.
Q3 - Section Formula
Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2:3$.
Solution:
Let the points be $A(-1, 7)$ and $B(4, -3)$. Ratio $m_1:m_2 = 2:3$.
Using Section Formula:
$$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1+m_2}$$
$$x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$$
$$y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$
The required point is $(1, 3)$.
Q4 - Value of k
Find the value of $k$ if the points $A(2, 3)$, $B(4, k)$, and $C(6, -3)$ are collinear.
Solution:
For collinear points, the area of the triangle formed by them is 0.
$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$$
$$2(k - (-3)) + 4(-3 - 3) + 6(3 - k) = 0$$
$$2(k + 3) + 4(-6) + 18 - 6k = 0$$
$$2k + 6 - 24 + 18 - 6k = 0$$
$$-4k = 0 \Rightarrow k = 0$$
Q5 - Equidistant Points
Find a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.
Solution:
Let the point on y-axis be $P(0, y)$.
Given $PA = PB$, so $PA^2 = PB^2$.
$$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$$
$$36 + 25 + y^2 - 10y = 16 + 9 + y^2 - 6y$$
$$61 - 10y = 25 - 6y$$
$$36 = 4y \Rightarrow y = 9$$
The point is $(0, 9)$.
Q6 - Ratio Finding
In what ratio does the point $(-4, 6)$ divide the line segment joining the points $A(-6, 10)$ and $B(3, -8)$?
Solution:
Let the ratio be $k:1$. Using the Section Formula for the x-coordinate:
$$-4 = \frac{k(3) + 1(-6)}{k+1}$$
$$-4(k+1) = 3k - 6$$
$$-4k - 4 = 3k - 6$$
$$2 = 7k \Rightarrow k = \frac{2}{7}$$
Therefore, the ratio is $2:7$.
Q7 - Parallelogram Vertex
If $(1, 2)$, $(4, y)$, $(x, 6)$, and $(3, 5)$ are vertices of a parallelogram taken in order, find $x$ and $y$.
Solution:
Diagonals of a parallelogram bisect each other.
Midpoint of AC = Midpoint of BD.
$$\left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{4+3}{2}, \frac{y+5}{2}\right)$$
$$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$$
$$\frac{8}{2} = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$$
So, $x=6, y=3$.
Q8 - Centroid of Triangle
Find the centroid of the triangle formed by the vertices $(3, -5)$, $(-7, 4)$, and $(10, -2)$.
Find the coordinates of a point A, where AB is the diameter of a circle whose center is $(2, -3)$ and B is $(1, 4)$.
Solution:
Let $A = (x, y)$. The center $C(2, -3)$ is the midpoint of $AB$.
$$\frac{x+1}{2} = 2 \Rightarrow x+1=4 \Rightarrow x=3$$
$$\frac{y+4}{2} = -3 \Rightarrow y+4=-6 \Rightarrow y=-10$$
Point A is $(3, -10)$.
Q10 - Area of Triangle
Find the area of a triangle whose vertices are $(1, -1)$, $(-4, 6)$, and $(-3, -5)$.
