If A(4,3), B(-1,y) and C(3,4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.

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Solution

Given the triangle ABC,right angled at A.

$N o w, A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$A B = \sqrt{(- 1 - 4)^{2} + (y - 3)^{2}}$

$A B = \sqrt{(- 5)^{2} + (y - 3)^{2}}$

$A B = \sqrt{25 + (y - 3)^{2}}$

$A B = \sqrt{25 + y^{2} + 9 - 6 y}$

$A B = \sqrt{34 + y^{2} - 6 y}$

$B C = \sqrt{(3 - (- 1))^{2} + (4 - y)^{2}}$

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

$B C = \sqrt{16 + 16 + y^{2} - 8 y}$

$B C = \sqrt{32 + y^{2} - 8 y}$

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

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

$A C = \sqrt{1 + 1}$

$A C = \sqrt{2} u n i t s$

Given, $△ A B C$ is a right angled triangle, right angled at A

So,by Pythagoras theorem $B C^{2} = A C^{2} + A B^{2}$

${(\sqrt{32 + y^{2} - 8 y})}^{2} = (\sqrt{2})^{2} + {(\sqrt{32 + y^{2} - 6 y})}^{2}$

$32 + y^{2} - 8 y = 2 + 34 + y^{2} - 6 y$

-2y=4

y=-2

Hence,the value of y is -2.

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