Question

Prove using the Pythagorean theorem that AC is perpendicular to AB given points ????(−2, −2), ????(5, −2), and

????(−2,22).

Answer 1

`(BC)^2=(AC)^2+(AB)^2`. It means AC is perpendicular to AB.

Explanation:

Vertices of given triangle are A(−2, −2), B(5, −2), and C(−2,22).

Distance formula:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Using distance formula we get

`AB=√((5-(-2))^2+(-2-(-2))^2)=√(7^2)=7`

`BC=√((-2-5)^2+(22-(-2))^2)=√(7^2+24^2)=√(625)=25`

`AC=√((-2-(-2))^2+(22-(-2))^2)=√(24^2)=24`

According to Pythagoras theorem

`hypotenuse^2=perpendicular^2+base^2`

`(BC)^2=(AC)^2+(AB)^2`

Perpendicular = AC

Base = AB

It means AC is perpendicular to AB.

Hence proved.