Chapter B, Problem 42E

Show that the lines 3x − 5y + 19 = 0 and 10x + 6y − 50 = 0 are perpendicular and find their point of intersection.

To determine

To show: The lines $3x-5y+19=0$ and $10x+6y-50=0$ are perpendicular and also find their point of intersection.

Answer to Problem 42E

The point of the intersection is $\left(2,5\right)$.

Explanation of Solution

Formula used:

The two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1{m}_2=-1$.

Calculation:

It is enough to show that $m_1{m}_2=-1$, in order to say the two lines are perpendicular.

Find the slope of the line $3x-5y+19=0$ as follows.

$\begin{array}{l}3x-5y+19=0\\ -5y=19-3x\\ -y=\frac{19}{5}-\frac{3}{5}x\end{array}$

$y=\frac{3}{5}x-\frac{19}{5}$. (1)

Therefore, the slope is $m_1=\frac{3}{5}$.

Find the slope of the line $10x+6y-50=0$ as follows.

$\begin{array}{l}10x+6y-50=0\\ 6y=50-10x\\ y=\frac{50}{6}-\frac{10}{6}x\end{array}$

$y=-\frac{5}{3}x+\frac{25}{3}$. (2)

Therefore, the slope is $m_2=-\frac{5}{3}$.

Here,

$\begin{array}{c}{m}_1{m}_2=\frac{3}{5}\times -\frac{5}{3}\\ m_1{m}_2=-1\end{array}$.

From the given formula, the given two lines are perpendicular.

Hence, it is proved that the lines $3x-5y+19=0$ and $10x+6y-50=0$ are perpendicular.

In order to find the intersection of their points, equate the equations (1) and (2) as follows.

$\begin{array}{l}\frac{3}{5}x+\frac{19}{5}=-\frac{5}{3}x+\frac{25}{3}\\ \frac{3}{5}x+\frac{5}{3}x=-\frac{19}{5}+\frac{25}{3}\\ \frac{9x+25x}{15}=\frac{-57+125}{15}\\ 9x+25x=-57+125\end{array}$

On further simplification,

$\begin{array}{c}34x=68\\ x=\frac{68}{34}\\ x=2\end{array}$

Now substitute $x=2$ in either $3x-5y+19=0$ or $10x+6y-50=0$.

$\begin{array}{l}3\left(2\right)-5y+19=0\\ 6-5y=-19\\ -5y=-19-6\\ y=\frac{25}{5}\\ y=5\end{array}$

Therefore, the point of the intersection is $\left(2,5\right)$.