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