Question

two vertices of a triangle are (-4,5) and (-5,-2) if the centroit triangle is the origin determine 3 rd vertex of a triangle
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Answer 1

↝ Let us assume a triangle ABC having coordinates of vertex A (- 4, 5) and coordinates of vertex B be (- 5, - 2).

↝ The coordinates of Centroid G is (0, 0).

↝ We have to find the coordinates of vertex C.

↝ Let coordinates of vertex C be (e, f).

↝ We know,

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

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

Here,

• a = - 4

• c = - 5

• e = e

• x = 0

• y = 0

• b = 5

• d = - 2

• f = f

↝ On substituting all these values, we get

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

$\rm :\longmapsto\:(0,0) = \bigg(\dfrac{ - 9+ e}{3}, \dfrac{3 + f}{3} \bigg)$

↝ On comparing, we get

$\rm :\longmapsto\:\dfrac{ - 9 + e}{3} = 0 \: \: \: and \: \: \: \dfrac{3 + f}{3} = 0$

$\rm :\implies\:e = 9 \: \: \: and \: \: \: f = - 3$

$\boxed{ \bf \: Hence, \: Coordinates \: of \: vertex \: C \: is \: (9, - 3)}$

Additional Information :-

Distance Formula :-

Distance Formula is used to find the distance between 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 internally, then

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

3. Midpoint Formula :-

Midpoint Formula is used to find the midpoint of line segment joining the points

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