Q.Using the distance formula, show that the points $A(0, 0)$, $B(5, 0)$ and $C(5, 12)$ form a right-angled triangle.

Compute all three side lengths with the distance formula, then check whether the Pythagorean relation holds among them — if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled.

$AB=\sqrt{(5-0)^2+0^2}=5$; $BC=\sqrt{0+(12-0)^2}=12$; $AC=\sqrt{5^2+12^2}=\sqrt{169}=13$.

$AB^2+BC^2=25+144=169=AC^2$.

Compute all three sides using the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

$AB$ (from $A(0,0)$ to $B(5,0)$):

$BC$ (from $B(5,0)$ to $C(5,12)$):

$AC$ (from $A(0,0)$ to $C(5,12)$):

Now check the Pythagorean relation among the three sides:

Since $AB^2+BC^2=AC^2$, with $AC$ the longest side, the angle opposite $AC$ — which is the angle at vertex $B$ — is a right angle. So $\triangle ABC$ is right-angled at $B$.

A common error is checking the Pythagorean relation with the wrong side singled out as the "hypotenuse" — the longest side must always be the one squared alone on one side of the equation; testing $AB^2+AC^2=BC^2$ here (with the shorter side $BC$ on the right) would wrongly fail even though the triangle is genuinely right-angled.