Question

The vertices of a triangle are (1, 2), (h, −3) and (−4, k). If the centroid of the triangle is at the point (5, −1) then find the value of `sqrt(("h" + "k")^2 + ("h" + 3"k")^2`

Answer 1

Let the vertices A(1, 2), B(h, −3) and C(−4, k)

Centroid of a ΔABC = `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

(5, −1) = `((1 + "h" - 4)/3, (2 - 3 + "k")/3)`

= `((-3 + "h")/3, (-1 + "k")/3)`

`(-3 + "h")/3` = 5

−3 + h = 15

h = 15 + 3 = 18

and 

`(-1 + "k")/3` = −1

−1 + k = −3

k = −3 + 1

k = −2

The value of `sqrt(("h" + "k")^2 + ("h" + 3"k")^2`

= `sqrt((18 - 2)^2 + (18 - 6)^2`

= `sqrt(16^2 + 12^2)`

= `sqrt(256 + 144)`

= `sqrt(400)`

= 20

The value of `sqrt(("h" + "k")^2 + ("h" + 3"k")^2` = 20 units