Question

13. If the vertices of a triangle are (1, 2) (P,
-3) and (-4, q), centrol be at the point (5,
-1), then find the values of p and q.​

Answer 1

$\maltese$Given :-

Vertices of a triangle are (1, 2) , (p, -3) , (-4 , q)

Centroid of a triangle is (5 , -1)

$\maltese$To find :-

Value of p, q

$\maltese$SOLUTION :-

As we know the centroid formula that is

$Centroid =\bigg( \dfrac{x_1+x_2+x_3}{3} , \dfrac{y_1+y_2+y_3}{3} \bigg)$

$x_1 = 1 \\x_2 = p \\x_3= -4$

$y_1 = 2\\y_2 = -3 \\y_3 = q$

Centroid is (5 , -1 ){ATQ}

According to vertices of triangle, Centroid is

$\red{\boxed{Centroid =\bigg( \dfrac{x_1+x_2+x_3}{3} , \dfrac{y_1+y_2+y_3}{3} \bigg)}}$

Substituting the values ,

$Centroid =\bigg( \dfrac{1+p- 4}{3} , \dfrac{2-3+q}{3} \bigg)$

$Centroid =\bigg(\dfrac{p-3}{3} ,\dfrac{q-1}{3} \bigg)$

But According to question it is (5, -1) So,

$(5, -1) =\bigg(\dfrac{p-3}{3} ,\dfrac{q-1}{3} \bigg)$

$5 =\dfrac{p-3}{3}$

$-1 = \dfrac{q-1}{3}$

$5(3) = p-3$

$15 = p-3$

$15+3 =p$

$\red{\boxed{p = 18}}$

$q- 1= -1(3)$

$q-1 = -3$

$q = -3+1$

$\purple{\boxed{q= -2}}$

So, the value of p, q are 18, -2

Know more formulae:-

Distance formula:-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Section formula Internal division:-

$\bigg(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\bigg)$

Section formula External division:-

$\bigg(\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}\bigg)$

Mid point formula:-

$\bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2}\bigg)$