Show that the points A


Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right-angled triangle.



Given: The 3 points are $A(7,10), B(-2,5)$ and $C(3,-4)$

$A B=\sqrt{(-2-7)^{2}+(5-10)^{2}}$


$=\sqrt{106}$ units …..(1)

$B C=\sqrt{(3+2)^{2}+(-4-5)^{2}}$


$=\sqrt{106}$ units …..(2)

$A C=\sqrt{(3-7)^{2}+(-4-10)^{2}}$


$=\sqrt{2} 12$ units

From equations 1 and 2, we have

⇒ AB = BC

Therefore, Δ ABC is an isosceles triangle …..(3)

Also, $\mathrm{AB}^{2}=106$ units  …..(4)

$\mathrm{BC}^{2}=106$ units ...........(5)

$\mathrm{AC}^{2}=212$ units .............(6)

From equations 4, 5 and 6, we have


So, it satisfies the Pythagoras theorem.

Δ ABC is right angled triangle …..(7)

From 3 and 7 , we have

$\triangle \mathrm{ABC}$ is an isosceles right angled triangle.

Hence, proved.


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