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  • If A(2, 2), B(5, 2) and C(k, 8) are the vertices of a right-angled triangle ABC with angle B =90 then find the value of k

If A(– 2, 2), B(5, 2) and C(k, 8) are the vertices of a right-angled triangle ABC with \angle B =90^{\circ}, then find the value of k.

 

 

 

 
 
 
 
 

Answers (1)

Given vertices of right angled triangle \bigtriangleup ABC are A(-2,2),B(5,2),\;\text{and}\;C(k,8)

Using distance formula :

D=\sqrt{(c-a)^2+(d-b)^2}

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

AC=\sqrt{(k+2)^2+(8-2)^2}=\sqrt{(k+2)^2+36}

BC=\sqrt{(k-5)^2+(8-2)^2}=\sqrt{(k-5)^2+36}

Using Pythagoras theorem,

AC^2=AB^2+BC^2

\left ( \sqrt{(k+2)^2+36} \right )^2=7^2+\left ( \sqrt{(k-5)^2+36} \right )^2

(k+2)^2+36 =7^2+(k-5)^2+36

k^2+4+4k+36 =49+k^2+25-10k+36

4k+40 =110-10k

14k=70

k=5

Posted by

Ravindra Pindel

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