Coordinate Geometry: Complete Guide for Class 10
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 $B(4, 1)$.
Solution:
Let $A(x_1, y_1) = (2, 3)$ and $B(x_2, y_2) = (4, 1)$.
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(4 - 2)^2 + (1 - 3)^2}$$ $$d = \sqrt{(2)^2 + (-2)^2}$$ $$d = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units.}$$
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(4 - 2)^2 + (1 - 3)^2}$$ $$d = \sqrt{(2)^2 + (-2)^2}$$ $$d = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units.}$$
Q2 - Midpoint Formula
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 $(1, 3)$.
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 $(0, 9)$.
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 $(10, -2)$.
Solution:
Centroid $G(x,y) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$$x = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2$$
$$y = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1$$
Centroid is $(2, -1)$.
Q9 - Circle Diameter
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 $(-3, -5)$.
Solution:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$$= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$$
$$= \frac{1}{2} |1(11) - 4(-4) - 3(-7)|$$
$$= \frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24 \text{ sq units.}$$
Part 2: 50 Practice Questions
- Find the distance between points (0, 0) and (36, 15).
- Find the distance between (a, b) and (-a, -b).
- Calculate the distance of point P(6, -6) from the origin.
- Find x if the distance between (x, 7) and (1, 15) is 10 units.
- Find the midpoint of the line segment joining (3, 4) and (5, 2).
- The midpoint of line segment joining (2a, 4) and (-2, 2b) is (1, 2a+1). Find a and b.
- Find the centroid of the triangle with vertices (1, 4), (-1, -1), and (3, -2).
- Find the coordinates of the point dividing the line joining (-1, 3) and (4, -7) in ratio 3:4.
- Determine the ratio in which the line segment joining (1, -5) and (-4, 5) is divided by the x-axis.
- Find the coordinates of the point of trisection of the line segment joining (2, -2) and (-7, 4).
- Find the area of the triangle formed by vertices (2, 3), (-1, 0), (2, -4).
- Find the value of k if points (7, -2), (5, 1), and (3, k) are collinear.
- If the distance between (4, p) and (1, 0) is 5, find p.
- Check if the points (5, -2), (6, 4), and (7, -2) form an isosceles triangle.
- Find a point on the x-axis which is equidistant from (2, -5) and (-2, 9).
- Find the perimeter of the triangle with vertices (0, 0), (3, 0), and (0, 4).
- Find the fourth vertex of the rectangle with three vertices (0,0), (2,0), and (0,3).
- The coordinates of one end of a diameter of a circle are (2, 3) and the center is (-2, 5). Find the other end.
- Find the value of y for which the distance between P(2, -3) and Q(10, y) is 10 units.
- Show that points (1, 7), (4, 2), (-1, -1) are vertices of a square (Distance check).
- Find the ratio in which the y-axis divides the line segment joining (5, -6) and (-1, -4).
- If (1, 2), (4, y), (x, 6), and (3, 5) are vertices of a parallelogram, find x+y.
- Find the distance of point (2, 3) from the x-axis.
- Find the distance of point (-5, 4) from the y-axis.
- What is the distance between the points ($a \cos \theta, 0$) and ($0, a \sin \theta$)?
- Find the centroid of a triangle with vertices (3, -7), (-8, 6), and (5, 10).
- If the origin is the centroid of the triangle with vertices (x, 1), (y, -2), (2, 3), find x and y.
- The line segment joining (2, -3) and (5, 6) is divided by the x-axis in what ratio?
- Find the coordinates of a point on the x-axis which is equidistant from (5, 4) and (-2, 3).
- The area of a triangle with vertices (a, 0), (0, b), and (1, 1) is collinear. Find the relation between a and b.
- Name the type of triangle formed by (3, 2), (-2, -3), (2, 3).
- If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1, find a and b.
- Find the distance between A(2a, 6a) and B(2a + \sqrt{3}a, 5a).
- If A(-2, 1), B(a, 0), C(4, b), and D(1, 2) are vertices of a parallelogram, find a and b.
- Find the coordinates of the point which is equidistant from the three vertices of $\Delta$ AOB where A=(0,2y), O=(0,0), B=(2x,0).
- Find the area of the triangle formed by (0, 0), (4, 0), and (0, 3).
- Points A(4, 3), B(6, 4), C(5, -6) and D(-3, 5) are vertices of a parallelogram? (True/False).
- Find the perpendicular distance of A(5, 12) from the origin.
- Find the value of k for which A(-5, 1), B(1, k), and C(4, -2) are collinear.
- The midpoint of (3p, 4) and (-2, 2q) is (2, 6). Find p + q.
- Find the coordinates of the circumcenter of the triangle formed by (0, 0), (4, 0), and (0, 4).
- In what ratio does the point P(2, -5) divide the line joining A(-3, 5) and B(4, -9)?
- The distance between points (5, 3) and (x, -1) is 5. Find x.
- Find the area of the quadrilateral ABCD with vertices A(-5, 7), B(-4, -5), C(-1, -6), and D(4, 5).
- Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.
- Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = 2/5 AB.
- Find the relation between x and y such that the point (x, y) is equidistant from (7, 1) and (3, 5).
- Find the coordinates of the point of intersection of the medians of a triangle with vertices (-1, 0), (5, -2), and (8, 2).
- If the area of a triangle formed by (x, 2x), (-2, 6), and (3, 1) is 5 sq units, find x.
- Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Answer Key
| Q.No | Answer | Q.No | Answer | Q.No | Answer | Q.No | Answer | Q.No | Answer |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 39 | 11 | 10.5 sq units | 21 | 5:1 | 31 | scalene | 41 | p=2, q=4, sum=6 |
| 2 | $2\sqrt{a^2+b^2}$ | 12 | 4 | 22 | 9 | 32 | a=1, b=-3 | 42 | (2, 2) |
| 3 | $6\sqrt{2}$ | 13 | $\pm 4$ | 23 | 3 units | 33 | 2a | 43 | 2:5 approx (check coords) |
| 4 | 7 or -5 | 14 | Yes | 24 | 5 units | 34 | a=1, b=1 | 44 | 2 or 8 |
| 5 | (4, 3) | 15 | (-7, 0) | 25 | a | 35 | (x, y) | 45 | 72 sq units |
| 6 | a=2, b=2 | 16 | 12 units | 26 | (0, 3) | 36 | 6 sq units | 46 | Yes |
| 7 | (1, 0) | 17 | (2, 3) | 27 | x=-2, y=-1 | 37 | False | 47 | (3, 4) |
| 8 | (1.14, -1.28) | 18 | (-6, 7) | 28 | 1:2 | 38 | 13 | 48 | x - y = 2 |
| 9 | 1:1 | 19 | 3 or -9 | 29 | (2, 0) approx | 39 | -1 | 49 | (4, 0) |
| 10 | (-1, 0) & (-4, 2) | 20 | Proof | 30 | 1/a + 1/b = 1 | 40 | 6 | 50 | (-7, 0) |