from a handpicked tutor in LIVE 1-to-1 classes

# Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC

(i) The median from A meets BC at D. Find the coordinates of the point D

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1

(iv) What do yo observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle

**Solution:**

A median of a triangle is a line segment joining the vertex to the midpoint of the opposite side, thus bisecting that side. Every triangle has exactly three medians, one from each vertex, and they all intersect each other at the triangle's centroid.

Let's draw the figure,

Let A(x₁, y₁) = (4, 2)

Let B(x₂, y₂ ) = (6, 5)

Let C(x₃, y₃ ) = (1, 4)

(i) Let AD be the median of the triangle ABC

Hence D is the midpoint of BC which can be found using the Mid Point formula.

M = [(x₁ + x₂) / 2, (y₁ + y₂) /2]

Coordinates of D = [(6 + 1)/2, (5 + 4)/2]

= [7/2, 9/2]

(ii) Let's draw the figure,

Point P divides the side AD in a ratio m: n = 2: 1.

P (x, y) = [mx₂ + nx₁/m + n , my₂ + ny₁/m + n]

Coordinates of P = [(2 × 7 / 2 + 1 × 4) / (2 + 1), (2 × 9 / 2 + 1 × 2) / (2 + 1)]

= [11 / 3, 11 / 3]

(iii) Let's draw the figure,

Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC. The co-ordinates of E can be calculated using the Mid-Point Formula.

M = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Coordinates of E = [(4 + 1)/2, (2 + 4)/2] = [5/2, 3]

Point Q divides the side BE in a ratio 2:1.

By Section formula

P (x, y) = [mx₂ + nx₁_{ }/ m + n , my₂ + ny₁_{ }/ m + n] .....Equation(1)

Coordinates of Q = [(2 × 5 / 2 + 1 × 6) / (2 + 1), (2 × 3 + 1 × 5) / (2 + 1)]

= [11/3, 11/3]

Median CF of the triangle will divide the side AB in two equal parts.

Therefore, F is the mid-point of side AB. The co-ordinates of F can be calculated using the Mid-Point Formula.

Coordinates of F = [(4 + 6) / 2, (2 + 5) / 2]

= [5, 7 / 2]

Point R divides the side CF in a ratio 2:1

By Section formula

P (x, y) = [mx₂ + nx₁_{ }/ m + n , my₂_{ }+ ny₁_{ }/ m + n] .....Equation(1)

Coordinates of R = [(2 × 5 + 1 × 1) / (2 + 1), (2 × 7/2 + 1 × 4) / (2 + 1)]

= [11/3, 11/3]

(iv) It can be observed that the coordinates of points P, Q, R are the same.

Therefore, all these are representing the same point on the plane that is, the centroid of the triangle.

(v) Consider a triangle, ∆ABC, having its vertices as A( x₁, y₁), B( x₂, y₂), and C (x₃, y₃)

The median AD of the triangle will divide the side BC into two equal parts.

Therefore, D is the mid-point of side BC.

The coordinates of D can be calculated using the Mid-Point Formula.

D = [(x₂ + x₃) / 2, (y₂ + y₃) / 2]

Let the centroid of this triangle be O.

Point O divides the side AD in a ratio of 2:1.

Coordinates of O = [(2 × (x₂ + x₃) / 2 + 1 × x₁)/ (2 + 1), (2 × (y₂ + y₃) / 2 + 1 × y₁)_{ }/ (2 + 1)]

= [(x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3]

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 7

**Video Solution:**

## Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC. (i) The median from A meets BC at D. Find the coordinates of the point D. (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1. (iv) What do yo observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.] (v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.

Maths NCERT Solutions Class 10 Chapter 7 Exercise 7.4 Question 7

**Summary:**

The median from A meets BC at D is (7/2, 9/2) and the coordinates of the point P on AD such that AP: PD = 2:1 is [11/3, 11/3] and the coordinates of point Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR: RF = 2:1 is (11/3, 11/3). It can be observed that the coordinates of point P, Q, R are the same. Therefore, all these are representing the same point on the plane i.e.,the centroid of the triangle. If A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle is Coordinates of O = [2 × (x₂ + x₃)/2 + 1 × x₁/(2 + 1), 2 × (y₂ + y₃)/2 + 1 × y₁_{ }/ (2 + 1)] = [(x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3]

**☛ Related Questions:**

- Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A (2, - 2) and B (3, 7).
- Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
- Find the centre of a circle passing through the points (6, - 6), (3, - 7) and (3, 3).
- The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.

visual curriculum