Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula we get

$AB=\sqrt{(5-(-2))^2+(-2-(-2))^2}=\sqrt{7^2}=7$

$BC=\sqrt{(-2-5)^2+(22-(-2))^2}=\sqrt{7^2+24^2}=\sqrt{625}=25$

$AC=\sqrt{(-2-(-2))^2+(22-(-2))^2}=\sqrt{24^2}=24$

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.