Find the equation of the median from b in ABC whose vertices are (1,5), B(5,3) and C(-3, -2)

y = x + 6

x = 1

y = ¼(x - 5) + 3

Step-by-step explanation:

Vetices are;

A(1,5), B(5,3) and C(-3, -2)

Thus;

Median of AB is; D = (1 + 5)/2, (5 + 3)/2

D = (3, 4)

Median of BC is; E = (5 + (-3))/2, (3 + (-2))/2

E = (1, 0.5)

Median of AC is; F ; (-3 + 1)/2, (-2 + 5)/2

F = (-1, 1.5)

Thus, the median lines will be;

CD, AE & BF.

Thus;

Equation of CD is;

(y - (-3))/(x - (-2)) = (-2 - 4))/(-3 - 3)

(y + 4)/(x + 2) = -6/-6

y - 4 = 1(x + 2)

y = 4 + x + 2

y = x + 6

Equation of AE;

(y - 5)/(x - 1) = (0.5 - 5)/(1 - 1)

(y - 5)/(x - 1) = -4.5/0

Cross multiply to get;

0(y - 5) = -4.5(x - 1)

-4.5x = -4.5

x = 1

Equation of BF;

(y - 3)/(x - 5) = (1.5 - 3)/(-1 - 5)

(y - 3)/(x - 5) = -1.5/-6

(y - 3)/(x - 5) = 1/4

y - 3 = ¼(x - 5)

y = ¼(x - 5) + 3