Question

Find the length of the hypotenuse of the right triangle whose vertices are given by the points $(-2,1),(1,1)$ and $(1,2)$

• A. $\sqrt{10}$
• B. $\sqrt{5}$
• C. $10$
• D. $2\sqrt{5}$

Answer 1

Answer: (A)

$\sqrt{10}$

Let the points be $A(-2,1)$, $B(1,1)$ and $C(1,2)$

Given: $△ABC$ is right-angled.

By distance formula,,

$d(A,B)=\sqrt{(1+2{)}^2+(1-1{)}^2}=\sqrt{9}=3$

$d(A,C)=\sqrt{(1+2{)}^2+(2-1{)}^2}=\sqrt{10}$

$d(B,C)=\sqrt{(1-1{)}^2+(2-1{)}^2}=\sqrt{1}=1$

Hence, the longest side i.e. hypotenuse of the right angled $△ABC$ is $AC=\sqrt{10}$ units.