Solution
FOR THE FIRST LENGTH
For triangle AB, |AB|, ( where A is (0,4); and B is 4,10)
X1 = 0 ; y1 = 4
X2 = 4 ; y2 = 10
Find the length of the sides of triangle

FOR THE SECOND LENGTH
For triangle BC, |BC|, ( where B is (4,10); and C is( 7,8)
X1 = 4 ; y1 = 10
X2 = 7 ; y2 = 8
Find the length of the sides of triangle

FOR THE THIRD LENGTH
For triangle AC, |AC|, ( where A is (0,4); and C is 4,10)
X1 = 0 ; y1 = 4
X2 = 7 ; y2 = 8
Find the length of the sides of triangle

|AC| = 8.1
Hence, show that AB is perpendicular to BC
Note: Find two of any of the gradient of the sides of the triangle ( let’s go with AB and BC)
Gradients AB

A (0,4) ; B (4,10)
X1 = 0 ; y1 = 4
X2 = 4 ; y2 = 10
Gradient AB

Gradient BC

A (-4,10) ; B (7,8)
X1 = -4 ; y1 = 10
X2 = 7 ; y2 = 8
Gradient AB

The product of gradient of AB and BC
× = -1
Since that ABC gradient is -1 that shows that AB Is perpendicular to BC