Question

The centroid of a triangle is the point (6,-1) if two vertices are( 3 ,4)and (-2,5) find the third vertices

Answer 1

Given:-

• First vertices of triangle = (3,4)
• Second vertices of traingle = (-2,5)
• Centroid of triangle = (6,-1)

To find:-

• Third vertices of triangle.

Let the vertex is (x,y).

Given centroid = (6,-1).

We know that,

$\sf Centroid \: of \: triangle = \big( \dfrac{x1 + x2 + x3}{3} , \dfrac{y1 + y2 + y3}{3} \big)$

$\sf \dfrac{x + 3 + ( - 2)}{3} = 6\: , \: \dfrac{y + 4 + 5}{3} = - 1 \\ \sf x + 1 = 18 \: , \: y + 9 = - 3 \\ \sf \: \boxed{ \bf \: x = 17 \: , y = - 12}$

Hence, required third vertex of triangle is (17,-12).

More to know:-

Midpoint formula = (x, y) = [½(x1 + x2), ½(y1 + y2)]

Area of triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

Distance formula = √[(x2-x1) + (y2-y1)]

Answer 2

Answer:

$\huge \fbox {given}$

• Centroid of triangle = (6,-1)
• First vertices = (3,4)
• Second vertices = (-2,5)

$\huge \bf \: to \: find$

Third vertices of the triangle

$\huge \bf \: solution$

Let the vertices be x and y

We know that centroid of triangle

$( \frac{x1 + x2 + x3}{3} )( \frac{y1 + y2 + y3}{3} )$

$\frac{x + 3 + - 2}{3} = 6$

$\frac{y + 4 + 5}{3} = 1$

$x + 1 = 18$

$y + 9 = - 3$

$x = 18 - 1$

$x = 17$

$y = - 12$

Hence required answer is (17,-12)