Question

Using slope, show that a(1,1), b(10,4), and c(7,7) are the vertices of a right triangle

Answer 1

• Slope of AC×Slope of BC=-1

`\\ \rm\hookrightarrow (7-1)/(7-1)* (7-4)/(7-10)=-1`

`\\ \rm\hookrightarrow (6)/(6)* (3)/(-3)=-1`

`\\ \rm\hookrightarrow 1(-1)=-1`

Hence they are vertices of right angled triangle

C is the right angle

Answer 2

Determine the slopes (gradients) of each line.

If the slopes of two of the lines are negative reciprocals (if their product is -1), the lines are perpendicular (the vertex is 90°).

Using slope formula:
`m=(y_2-y_1)/(x_2-x_1)`

Slope of AC:

A = (1, 1) =
`(x_1,y_1)`

C = (7, 7) =
`(x_2,y_2)`

`m=(y_2-y_1)/(x_2-x_1)=(7-1)/(7-1)=1`

Slope of AB:

A = (1, 1) =
`(x_1,y_1)`

B = (10, 4) =
`(x_2,y_2)`

`m=(y_2-y_1)/(x_2-x_1)=(4-1)/(10-1)=\frac13`

Slope of BC:

B = (10, 4) =
`(x_1,y_1)`

C = (7, 7) =
`(x_2,y_2)`

`m=(y_2-y_1)/(x_2-x_1)=(7-4)/(7-10)=-1`

Therefore, as AC and BC are negative reciprocals of each other (1 x -1 = -1), m∠C = 90° which proves that the triangle is a right triangle.