Chapter 1, Problem 126RE

To determine

To calculate:The equation for the line passing through the point $\left(1,7\right)$ and also seen to be perpendicular to a line with equation $x-3y+16=0$ .

Answer to Problem 126RE

The equation for the line passing through the point $\left(1,7\right)$ and perpendicular to a line $x-3y+16=0$ is $y=-3x+10$ .

Explanation of Solution

Given information:

A line passing through the point $P_1=\left(1,7\right)$ and perpendicular to a line $x-3y+16=0$

Formula used:

Slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Point-slope equation for a given line passing along point $P_1=\left(x_1,y_1\right)$ is:

  $y-y_1=m\left(x-x_1\right)$

Slope-intercept equation for a given line which has slope as $m$ and $y$ −intercept as $b$ is:

  $y=mx+b$

Two-intercept equation for a given line which has $x$ -intercept as $a$ and $y$ −intercept as $b$ is:

  $\frac{x}{a}+\frac{y}{b}=1$

When two lines are perpendicular then the product of their slopes is zero that is $m_1.m_2=-1$

Calculation:

Consider thegiven point $P_1=\left(1,7\right)$ and line $x-3y+16=0$

Recall the slope-intercept equation for a given line which has slope as $m$ and $y$ −intercept as $b$ is:

  $y=mx+b$

Let the line passing through point $P_1=\left(1,7\right)$ and perpendicular to line $x-3y+16=0$ has the general form:

  $y=m_{1}x+b_1$

Where $m_1$ is the slope and $b_1$ is the $y$ -intercept of the line.

Now, we try to express the line $x-3y+16=0$ in the same form:

Firstly let its general form be $y=m_{2}x+b_2$

Where $m_2$ is the slope and $b_2$ is the $y$ -intercept of the line.

  $\begin{array}{l}x-3y+16=0\\ -3y=-x-16\\ y=\frac{-x-16}{-3}\\ y=\left(\frac{1}{3}\right)x+\frac{16}{3}\end{array}$

Comparing with $y=m_{2}x+b_2$ to get:

  $\begin{array}{l}{m}_2=\frac{1}{3}\\ b_2=\frac{16}{3}\end{array}$

As it’s given that line $y=m_{1}x+b_1$ and line $y=m_{2}x+b_2$ are perpendicular.

Therefore, recall that when two lines are perpendicular then the product of their slopes is zero that is $m_1.m_2=-1$

That is:

  $\begin{array}{l}{m}_1.\frac{1}{3}=-1\\ m_1=-3\end{array}$

Therefore, equation of the required line reduces to:

  $y=-3x+b_1$ $\left(1\right)$

Also it is given that this line passes through $P_1=\left(1,7\right)$ .

Then in order to proceed following should be known:

  $\begin{array}{l}x=1\\ y=7\end{array}$

Putting this in equation $\left(1\right)$ to get:

  $\begin{array}{l}7=-3\left(1\right)+b_1\\ b_1=7+3\\ b_1=10\end{array}$

Put this value of $b_1$ in $\left(1\right)$ to get the required equation as:

  $y=-3x+10$

Hence, the equation for the line passing through the point $\left(1,7\right)$ and perpendicular to a line $x-3y+16=0$ is $y=-3x+10$ .