In $ \triangle \mathrm{ABC} \angle \mathrm{A} $ is a right angle and its vertices are $ \mathrm{A}(1,7), \mathrm{B}(2,4) $ and $ \mathrm{C}(k, 5) $. Then, find the value of $ k $.

Given:

In \( \triangle \mathrm{ABC} \angle \mathrm{A} \) is a right angle and its vertices are \( \mathrm{A}(1,7), \mathrm{B}(2,4) \) and \( \mathrm{C}(k, 5) \).

To do:

We have to find the value of \( k \).

Solution:

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)

\( =\sqrt{(2-1)^{2}+(4-7)^{2}} \)

\( =\sqrt{(1)^{2}+(-3)^{2}} \)

\( =\sqrt{1+9} \)

\( =\sqrt{10} \)

Similarly,

\( \mathrm{BC}=\sqrt{(k-2)^{2}+(5-4)^{2}} \)

\( =\sqrt{(k-2)^{2}+(1)^{2}} \)

\( =\sqrt{k^2+4-4k+1} \)

\( =\sqrt{k^2-4k+5} \)

\( \mathrm{CA}=\sqrt{(k-1)^{2}+(5-7)^{2}} \)

\( =\sqrt{(k-1)^{2}+(-2)^{2}} \)

\( =\sqrt{k^2+1-2k+4} \)

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

Here,

\( \angle \mathrm{A} \) is the right angle.

Therefore,

\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{10})^{2}+(\sqrt{k^2-2k+5})^{2} \)

\( =10+k^2-2k+5=k^2-2k+15 \)

\( \mathrm{BC}^{2}=(\sqrt{k^2-4k+5})^{2}=k^2-4k+5 \)

We know that,

\( \mathrm{AB}^{2}+\mathrm{CA}^{2}=\mathrm{BC}^{2} \)

This implies,

$k^2-2k+15=k^2-4k+5$

$4k-2k=5-15$

$2k=-10$

$k=-5$

Therefore, the value of $k$ is $-5$.

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Updated on: 10-Oct-2022

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