Question

Question-3 Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Two vertices of triangle are (-3, 1) and (0, -2).

• The centroid is at the origin.

Let assume that

• The triangle is ABC having (- 3, 1) and (0, - 2) as Coordinates of A and B respectively.

• Coordinates of Centroid (0, 0) is represented as G.

• Let third vertex be C (a, b).

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies (x, \: y)= \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$

We have

• • x₁ = - 3

• • x₂ = 0

• • x₃ = a

• • y₁ = 1

• • y₂ = - 2

• • y₃ = b

• • x = 0

• • y = 0

So, on substituting the values, we get

$\rm :\longmapsto\:(0,0) = \bigg(\dfrac{ - 3 + 0 + a}{3}, \: \dfrac{1 - 2 + b}{3} \bigg)$

$\rm :\longmapsto\:(0,0) = \bigg(\dfrac{ - 3+ a}{3}, \: \dfrac{ - 1+ b}{3} \bigg)$

So, on comparing we get

$\rm :\longmapsto\:a - 3 = 0 \: \: \: and \: \: \: b - 1 = 0$

$\rm :\longmapsto\:a = 3 \: \: \: and \: \: \: b = 1$

Hence, Coordinates of C is (3, 1)

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Learn More :-

1. Section formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$