LIWA

Section and Mid-Point Formula

ICSE Class 10 · Mathematics · Chapter 13

What this chapter covers

Key definitions

Internal division of a line segment is the division of the segment by a point that lies on the segment itself, between its two end points.

The section formula is the rule that gives the coordinates of the point which divides the line segment joining two given points in a given ratio.

The mid-point of a line segment is the point on the segment that divides it internally in the ratio 1 : 1.

The mid-point formula is the rule that the coordinates of the mid-point of a segment are the averages of the corresponding coordinates of its two end points.

A median of a triangle is the line segment joining a vertex of the triangle to the mid-point of the opposite side.

The centroid of a triangle is the point at which the three medians of the triangle intersect.

The centroid formula is the rule that each coordinate of the centroid of a triangle is the average of the corresponding coordinates of its three vertices.

Key points

Questions answered

Define internal division of a line segment.
Internal division of a line segment is the division of the segment by a point that lies on the segment, between its two end points.
Find the coordinates of the point which divides the line segment joining (1, -2) and (10, 4) internally in the ratio 2 : 1.
x = (2 × 10 + 1 × 1)/(3) = 7, y = (2 × 4 + 1 × (-2))/(3) = 2. The point is (7, 2).
Find the mid-point of the line segment joining (-6, 4) and (2, -8).
((-6 + 2)/(2), (4 - 8)/(2)) = (-2, -2).
M(-1, 2) is the mid-point of AB, where A is (3, -4). Find B.
B = (2 × (-1) - 3, 2 × 2 - (-4)) = (-5, 8).
In what ratio does the x-axis divide the line segment joining (2, -3) and (5, 6)? Find the point of division.
With ratio k : 1, 6k - 3 = 0, so k = (1)/(2) and the ratio is 1 : 2. The point is ((1 × 5 + 2 × 2)/(3), 0) = (3, 0).
In what ratio does the y-axis divide the line segment joining (-4, 7) and (8, 1)? Find the point of division.
8k - 4 = 0 gives k = (1)/(2), so the ratio is 1 : 2. The point is (0, (1 × 1 + 2 × 7)/(3)) = (0, 5).
Find the centroid of the triangle with vertices (-2, 3), (5, -1) and (3, 7).
((-2 + 5 + 3)/(3), (3 - 1 + 7)/(3)) = (2, 3).
The centroid of triangle ABC is (2, -1). If A is (5, 2) and B is (-3, 1), find C.
C = (3 × 2 - 5 + 3, 3 × (-1) - 2 - 1) = (4, -6).

Past board exam questions

ICSE 2027 sample paper · 4 marks

A line is formed joining the two points P(−4, 1) and Q(11, −2). Straight line PQ intersects the x-axis at R. Find the: (a) ratio PR : RQ. (b) coordinates of the point R. (c) equation of the line perpendicular to PQ at R.

Show answer
(a) Let R(x, 0) divide PQ in the ratio m : n. y-coordinate: (m(-2) + n(1))/(m + n) = 0 ⇒ n = 2m ⇒ m : n = 1 : 2. So PR : RQ = 1 : 2. (b) x = (1(11) + 2(-4))/(1 + 2) = (3)/(3) = 1. R = (1, 0). (c) Slope of PQ = (-2 - 1)/(11 - (-4)) = (-3)/(15) = -(1)/(5). Slope of the perpendicular = 5. Equation: y - 0 = 5(x - 1) ⇒ 5x - y = 5.

ICSE 2026 sample paper · 4 marks

P is a point on the x- axis which divides the line joining A (- 6, 2) and B (9, - 4). Find: (a) the ratio in which P divides the line segment AB. (b) the coordinates of the point P. (c) equation of a line parallel to AB and passing through (-3, -2).

Show answer
(a) Let P divide AB in the ratio m : n. P is on the x-axis, so its y-coordinate is 0: (m(-4) + n(2))/(m + n) = 0 ⇒ 4m = 2n ⇒ m : n = 1 : 2. (b) x = (1(9) + 2(-6))/(1 + 2) = (-3)/(3) = -1. P = (−1, 0). (c) Slope of AB = (-4 - 2)/(9 - (-6)) = (-6)/(15) = -(2)/(5). Line through (−3, −2): y + 2 = -(2)/(5)(x + 3) ⇒ 5y + 10 = -2x - 6 ⇒ 2x + 5y + 16 = 0.

ICSE 2026 board paper · 1 mark

In the triangle with vertices A(0, 2y), O(0, 0) and B(2x, 0), ΔAOB is a right-angled triangle and C is the mid-point of AB. The coordinates of the point which is equidistant from the three vertices of ΔAOB is:

  1. (x, y)
  2. (y, x)
  3. ((x)/(2), (y)/(2))
  4. ((2x)/(3), (2y)/(3))
Show answer
(x, y)

ICSE 2026 board paper · 3 marks

A(−10, −2) and B(2, 10) are two end points of a line segment. If AB intersects the x-axis at P, find the: (a) ratio in which 'P' divides AB. (b) coordinates of point P.

Show answer
(a) Let P divide AB in the ratio k : 1. P lies on the x-axis, so its y-coordinate is 0: (10k + (-2))/(k + 1) = 0 ⇒ k = (1)/(5). Ratio = 1 : 5. (b) x = (1(2) + 5(-10))/(1 + 5) = (-48)/(6) = -8. P = (−8, 0).

ICSE 2025 compartment paper · 1 mark

The line segment joining A(–7, 2) and B(3, –8) is divided by the x-axis in the ratio:

  1. 1 : 4
  2. 3 : 7
  3. 4 : 1
  4. 7 : 3
Show answer
1 : 4

ICSE 2025 sample paper · 3 marks

Find the coordinates of the centroid P of the ΔABC, whose vertices are A(–1, 3), B(3, –1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.

Show answer
Centroid P = ((-1 + 3 + 0)/(3), (3 + (-1) + 0)/(3)) = ((2)/(3), (2)/(3)). Slope of AB = (-1 - 3)/(3 - (-1)) = (-4)/(4) = -1. The required line has slope −1: y - (2)/(3) = -1(x - (2)/(3)) ⇒ x + y = (4)/(3) ⇒ 3x + 3y = 4.

ICSE 2024 sample paper · 4 marks

A (a, b), B(-4, 3) and C(8,-6)are the vertices of a ∆ABC. Point D is on BC such that BD : DC is 2 : 1 and M (6, 0) is mid point of AD. Find: (a) coordinates of point D. (b) coordinates of point A. (c) equation of a line passing through M and parallel to line BC.

Show answer
(a) D divides BC in the ratio 2 : 1: D = ((2 × 8 + 1 × (-4))/(3), (2 × (-6) + 1 × 3)/(3)) = ((12)/(3), (-9)/(3)) = (4, -3). (b) M(6, 0) is the mid-point of AD: (a + 4)/(2) = 6 ⇒ a = 8; (b - 3)/(2) = 0 ⇒ b = 3. A = (8, 3). (c) Slope of BC = (-6 - 3)/(8 - (-4)) = (-9)/(12) = -(3)/(4). Line through M(6, 0) with slope -(3)/(4): y - 0 = -(3)/(4)(x - 6) ⇒ 4y = -3x + 18 ⇒ 3x + 4y - 18 = 0.

ICSE 2023 board paper · 1 mark

The coordinates of the vertices of ΔABC are respectively (–4, –2), (6, 2) and (4, 6). The centroid G of ΔABC is:

  1. (2, 2)
  2. (2, 3)
  3. (3, 3)
  4. (0, –1)
Show answer
(2, 2)