#### Question

If A(5, 2), B(2, −2) and C(−2, *t*) are the vertices of a right angled triangle with ∠B = 90°, then find the value of *t*.

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#### Solution

Using distance formula, we have

AB = `sqrt((2-5)^2+(-2-2)^2)=sqrt(9+16)=5`

BC = `sqrt((-2-2)^2+(t+2)^2)=sqrt(t^2+4t+20)`

AC = `sqrt((-2-5)^2+(t-2)^2)=sqrt(t^2-4t+53) `

Now, it is given that △ABC is right angled at B.

Using the Pythagorean theorem, we have

AB^{2} + BC^{2} = AC^{2}

∴25+t^{2}+4t+20=t^{2}−4t+53 [From (1), (2) and (3)]

⇒45+4t=−4t+53

⇒8t=8

⇒t=1

Hence, the value of *t* is 1.

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#### Reference Material

Solution for question: If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t. concept: null - Distance Formula. For the course CBSE