Question

Question 8
(a) A(5,q), B(-4, 3) and C(p, -2) are the vertices of the triangle ABC whose centroid
is the origin. Calculate the values of p and q.
​

Answer 1

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

Given that

↝ In triangle ABC,

↝ Coordinates of vertex A is (5, q),

↝ Coordinates of vertex B is (- 4, 3),

↝ Coordinates of vertex C is (p, - 2)

↝ Coordinates of Centroid G is (0, 0).

↝ We have to find the value of p and q.

We know that,

↝ Centroid is a point where the medians of a triangle meet and is represented by the symbol G. If the coordinates of triangle are (a, b), (c, d) and (e, f) and coordinates of Centroid G be (x, y) then

$\rm :\longmapsto\:(x,y) = \bigg(\dfrac{a + c + e}{3}, \dfrac{b + d + f}{3} \bigg)$

Here,

• a = 5

• c = - 4

• e = p

• b = q

• d = 3

• f = - 2

• x = 0

• y = 0

↝ On substituting all these values, we get

$\rm :\longmapsto\:(0,0) = \bigg(\dfrac{5 - 4 + p}{3}, \dfrac{q + 3 - 2}{3} \bigg)$

$\rm :\longmapsto\:(0,0) = \bigg(\dfrac{1+ p}{3}, \dfrac{q + 1}{3} \bigg)$

↝ On comparing, we get

$\rm :\longmapsto\:\dfrac{p + 1}{3} = 0 \: \: \: and \: \: \: \dfrac{q + 1}{3} = 0$

$\rm :\implies\:p = - 1 \: \: \: and \: q = - 1$

Additional Information :-

1. Distance Formula:-

Distance Formula is used to find the distance between the two points A and B.

$\sf\:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

2. Section Formula :-

Section Formula is used to find the coordinates of the point which divides the line segment joining the points in the ratio m : n,

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n},  \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

3. Midpoint Formula :-

Midpoint Formula is used to find the midpoint of line segment.

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2},  \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$