Question

Show that point A(7,5), B(2,3) and C(6,-7) are the vertices of a right triangle. Also find its area.

Answer 1

Consider the problem

Given points are

$A(7,5),B(2,3),C(6,-7)$

Let the given lines of triangle be $AB,BC,AC$

Let , By distance formula

Distance of $AB$

$AB=\sqrt{(3-5{)}^2+(2-7{)}^2}=\sqrt{(-2{)}^2+(-5{)}^2}=\sqrt{4+25}=\sqrt{29}$

Similarly,

$BC=\sqrt{(-7-3{)}^2+(6-2{)}^2}=\sqrt{100+16}=\sqrt{116}$

$AC=\sqrt{(-7-5{)}^2+(6-7{)}^2}=\sqrt{145}$

Now, Pythagoras theorem

$A{C}^2=B{C}^2+A{B}^2$

$(\sqrt{145}{)}^2=(\sqrt{116}{)}^2+(\sqrt{29}{)}^2$

$145=116+29$

$145=145$

L.H.S.=R.H.S

Hence, prove that they are at right angles.