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The points $A( 4,\ 7) ,\ B( p,\ 3)$ and $C( 7,\ 3)$ are the vertices of a right triangle, right-angled at B, Find the values of $p$.
Given: The vertices of a right triangle, right-angled at B,The points A(4, 7), B(p, 3) and C(7, 3).
To do: To find the values of P.
Solution:
$\vartriangle ABC$ is right angled at B.
$AC^{2} =AB^{2} +BC^{2} .............( 1)$
Also, $A=( 4,\ 7),\ B=( p,\ 3)$ and $C=( 7,\ 3)$
On using the distance formula,
We have,
$AB=\sqrt{\left( p-4\right)^{2} +\left( 3-7\right)^{2}}$
$\Rightarrow AB=\sqrt{\left( p-4\right)^{2} +\left( -4\right)^{2}}$
$\Rightarrow AB=\sqrt{\left( p-4\right)^{2} +16}$
Similarly, $BC=\sqrt{\left( 7-p\right)^{2} +\left( 3-3\right)^{2}}$
$\Rightarrow BC=\sqrt{\left( 7-p\right)^{2}}$
$\Rightarrow BC=7-p$
And $AC=\sqrt{\left( 7-4\right)^{2} +\left( 3-7\right)^{2}}$
$\Rightarrow AC=\sqrt{\left( 3\right)^{2} +\left( -4\right)^{2}}$
$\Rightarrow AC=\sqrt{9+16}$
$\Rightarrow AC=\sqrt{25}$
$\Rightarrow AC=5$
On subtittuting these obtained values of AB, BC and AC in $\left( 1\right)$ ,
$5^{2} =\left( p-4\right)^{2} +16+\left( 7-p\right)^{2}$
$\Rightarrow p^{2} +16-8p+16+49+p^{2} -14p=25$
$\Rightarrow 2p^{2} -22p+56=0$
$\Rightarrow 2\left( p^{2} -11p+28\right) =0$
$\Rightarrow \left( p^{2} -11p+28\right) =0$
$\Rightarrow p^{2} -7p-4p+28=0$
$\Rightarrow p\left( p-7\right) -4\left( p-7\right) =0$
$\Rightarrow \left( p-4\right)\left( p-7\right) =0$
If $p-4=0$
$\Rightarrow p=4$
If $p-7=0$
$\Rightarrow p=7$
Therefore, The value of $p=4,\ 7$.
Updated on: 10-Oct-2022
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