Question

the centroid of a triangle whose two vertices are (3,-5) and (10,-5) is (2,-1) find the third vertex​

Answer 1

Answer:

Given,

( x1 , y1 ) = (3,-5)

( x2 , y2 ) = (10,-5)

Let the 3rd vertex be ( x3 , y3 )

since,

$centroid \: = (\frac{x1 + x2 + x3}{3} , \: \frac{y1 + y2 + y3}{3})$

$(2, - 1) = (\frac{3 + 10 + x3}{3} , \: \frac{ - 5 - 5 + y3}{3})$

$(2, - 1) = (\frac{13 + x3}{3},\frac{ - 10 + y3}{3})$

therefore,

$2 = \frac{13 + x3}{3} \: and \: - 1 = \frac{ - 10 + y3}{3}$

2×3 = 13+x3 and -1×3 = -10+y3

6-13 =x3 and -3+10 = y3

∴ x3 = -7 and 7 = y3

3rd vertex = ( -7 , 7)

hope it helped you :)