Question

Given that (1, -2), (0, 5) and (-3, 1) are the vertices of a right triangle, how do you determine an equation of the line that passes through the endpoints of the hypotenuse?

Answer 1

Hint: To determine the equation of the line that passes through the endpoints of the hypotenuse, we have to determine the hypotenuse among the given points. We will consider the given points to be \[A\left( 1,\text{ }-2 \right),B\left( 0,\text{ }5 \right),C\left( -3,\text{ }1 \right)\] . We have to find the slopes of AB, BC and CA. We know that slope of a line AB with points $A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}})$ is given as $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Then we will find the sides which are perpendicular. We know that if a line is perpendicular to another, the product of their slopes will be equal to -1. So will multiply ${{m}_{1}}{{m}_{2}},{{m}_{2}}{{m}_{3}}\text{ and }{{m}_{3}}{{m}_{1}}$ and check which one equals to -1. The remaining side will be the hypotenuse. By using the point-slope formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ , we will get the required solution.

Complete step by step solution:
We have to determine an equation of the line that passes through the endpoints of the hypotenuse. So first, we have determine the hypotenuse among the given points.
Let us consider the given points to be \[A\left( 1,\text{ }-2 \right),B\left( 0,\text{ }5 \right),C\left( -3,\text{ }1 \right)\] . Now, we have to find the slopes of AB, BC and CA. We know that slope of a line AB with points $A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}})$ is given as $m=\dfrac{\text{Change in y-coordinates}}{\text{Change in x-coordinates}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Now, let us find the slopes of AB, BC and CA.
Slope of AB, ${{m}_{1}}=\dfrac{5-\left( -2 \right)}{0-1}=\dfrac{5+2}{-1}=-7$
Slope of BC, ${{m}_{2}}=\dfrac{1-5}{-3-0}=\dfrac{-4}{-3}=\dfrac{4}{3}$
Slope of CA, ${{m}_{2}}=\dfrac{1-\left( -2 \right)}{-3-1}=\dfrac{1+2}{-4}=\dfrac{-3}{4}$
We are given that the triangle is a right-angled triangle. So we have to find the sides which are perpendicular. For this we will multiply the slopes of two lines. We know that if a line is perpendicular to another, the product of their slopes will be equal to -1. So let us multiply the slopes as follows.
$\begin{align}
  & {{m}_{1}}\times {{m}_{2}}=-7\times \dfrac{4}{3}=\dfrac{28}{3}\ne 1 \\
 & {{m}_{2}}\times {{m}_{3}}=\dfrac{4}{3}\times \dfrac{-3}{4}=-1 \\
 & {{m}_{3}}\times {{m}_{1}}=\dfrac{-3}{4}\times -7=\dfrac{21}{4}\ne 1 \\
\end{align}$
We can see that product of slopes of BC and CA is equal to -1. Therefore, BC and CA are perpendicular with $\angle C={{90}^{{}^\circ }}$ .
Hence, AB will be hypotenuse.

We know that equation of a line passing through points $\left( {{x}_{1}},{{y}_{1}} \right)$ with slope m is given by the point-slope formula, that is,
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ .
Hence, equation of the line that passes through the endpoints of the hypotenuse, that is, \[A\left( 1,\text{ }-2 \right)\]is
$\begin{align}
  & y-\left( -2 \right)=-7\left( x-1 \right) \\
 & \Rightarrow y+2=-7\left( x-1 \right) \\
\end{align}$
Let us solve the above equation.
$\begin{align}
  & \Rightarrow y+2=-7x+7 \\
 & \Rightarrow y=-7x+7-2 \\
 & \Rightarrow y=-7x+5 \\
\end{align}$
Hence, the required equation of line is $y=-7x+5$ .

Note: We can also find the required equation of the line by using the point \[B\left( 0,\text{ }5 \right)\] as follows. Let us substitute this point in the point-slope formula.
$\begin{align}
  & y-5=-7\left( x-0 \right) \\
 & \Rightarrow y-5=-7x \\
 & \Rightarrow y=-7x+5 \\
\end{align}$