Introduction
In Class 9 you learned how to plot points on a coordinate plane and how to locate a point given its coordinates. Now we use coordinates to find the distance between two points and the point that divides a line segment internally in a given ratio. These are the two topics listed for CBSE 2026-27.
Distance Formula
Given two points on the plane, we want the straight-line distance between them, the length of the line segment joining them.
The distance between two points P(x1, y1) and Q(x2, y2) is
PQ = √[(x2 − x1)2 + (y2 − y1)2]
It comes from applying the Pythagoras Theorem to the right triangle formed by the horizontal distance |x2 − x1| and the vertical distance |y2 − y1| between the points.
A few points to keep in mind:
- The formula works whichever point you call (x1, y1) and whichever you call (x2, y2), swapping them gives the same answer because we square the differences.
- The distance is always non-negative. Take the positive square root.
- The distance of a point P(x, y) from the origin O(0, 0) is a special case:
OP = √(x2 + y2)
Find the distance between A(2, 3) and B(6, 6).
AB = √[(6 − 2)2 + (6 − 3)2] = √[42 + 32] = √[16 + 9] = √25 = 5
This is another appearance of the 3-4-5 right triangle, the differences in x and y (4 and 3) are the legs of the right triangle.
Check whether the points A(1, 2), B(4, 6) and C(7, 2) form an isosceles triangle.
Find all three side lengths.
AB = √[(4−1)2 + (6−2)2] = √[9 + 16] = 5
BC = √[(7−4)2 + (2−6)2] = √[9 + 16] = 5
CA = √[(7−1)2 + (2−2)2] = √[36 + 0] = 6
Since AB = BC = 5, the triangle is isosceles with AB = BC.
Find the point on the x-axis that is equidistant from A(2, −5) and B(−2, 9).
Any point on the x-axis has the form P(x, 0). Equidistant means PA = PB, so PA2 = PB2:
PA2 = (x − 2)2 + (0 + 5)2 = (x − 2)2 + 25
PB2 = (x + 2)2 + (0 − 9)2 = (x + 2)2 + 81
Setting them equal:
(x − 2)2 + 25 = (x + 2)2 + 81
x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81
−4x + 29 = 4x + 85 ⟹ −8x = 56 ⟹ x = −7
So the required point is (−7, 0).
The distance formula is also the tool for proving that a given set of points forms a particular type of triangle or quadrilateral, just calculate the side lengths (and diagonals for quadrilaterals) and compare them.
Section Formula
Suppose you have two points A and B, and a point P on the line segment AB that divides it in a given ratio. The section formula tells you the coordinates of P without drawing a graph.
The coordinates of the point P(x, y) which divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are
P = ( (m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2) )
Here, m1 is the part next to B and m2 is the part next to A, think of m1 as multiplying the coordinates of B and m2 as multiplying the coordinates of A.
An important special case is the midpoint, where m1 = m2 = 1:
The midpoint of the segment joining A(x1, y1) and B(x2, y2) is
M = ( (x1 + x2)/2, (y1 + y2)/2 )
Note that the section formula as given in Class 10 deals with internal division, the point P lies between A and B. The formula works for the ratio counted as AP : PB = m1 : m2.
Find the coordinates of the point that divides the segment joining A(2, −3) and B(5, 6) internally in the ratio 1 : 2.
Here m1 = 1, m2 = 2, A = (2, −3), B = (5, 6).
x = (1·5 + 2·2)/(1+2) = (5 + 4)/3 = 9/3 = 3
y = (1·6 + 2·(−3))/(1+2) = (6 − 6)/3 = 0/3 = 0
So the point is (3, 0).
Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Any point on the x-axis has y-coordinate 0. Let the point be P and let AP : PB = k : 1. Using the section formula for the y-coordinate:
y = (k·5 + 1·(−5))/(k+1) = (5k − 5)/(k+1)
Set y = 0: 5k − 5 = 0 ⟹ k = 1. So the x-axis divides the segment in the ratio 1 : 1 (it bisects it).
The x-coordinate is then:
x = (1·(−4) + 1·1)/(1+1) = (−3)/2 = −1.5
So the point of division is (−3/2, 0).
A useful tip: when you need to find the ratio in which a point divides a segment, assume the ratio is k : 1, write the section formula, and solve for k using the known coordinate.
Summary
- Distance formula: distance between P(x1, y1) and Q(x2, y2) is √[(x2 − x1)2 + (y2 − y1)2]. Distance from the origin: √(x2 + y2).
- Section formula (internal): point dividing A(x1, y1)B(x2, y2) in the ratio m1 : m2 is ((m1x2 + m2x1)/(m1+m2), (m1y2 + m2y1)/(m1+m2)). Midpoint = ((x1+x2)/2, (y1+y2)/2).
Practice Questions
- Find the distance between the points P(−5, 7) and Q(−1, 3). Also find the distance of the point (4, 3) from the origin.
- Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
- Find the coordinates of the point which divides the line segment joining A(−2, 2) and B(2, 8) in the ratio 3 : 1.
- Find the ratio in which the point (−3, p) divides the line segment joining the points (−5, −4) and (−2, 3). Also find the value of p.