Chapter 1, Problem 127RE

To determine

To calculate:The equation for the line passing through the origin and is also parallel to line $3x+15y=22$

Answer to Problem 127RE

The equation for the line passing through the origin and is also parallel to line $3x+15y=22$ is $y=\frac{-1}{5}x$

Explanation of Solution

Given information:

A line passing through origin $\left(0,0\right)$ and parallel to line $3x+15y=22$

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:

  $m=\frac{y_2-y_1}{x_2-x_1}$

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 $m_1.m_2=-1$

When two lines are parallel then their slope are equal that is $m_1=m_2$

Calculation:

Consider theline $3x+15y=22$

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 $3x+15y=22$ 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 $3x+15y=22$ in the same form:

  $\begin{array}{l}3x+15y=22\\ 15y=-3x+22\\ y=\frac{-3}{15}x+\frac{22}{15}\\ y=\frac{-1}{5}x+\frac{22}{15}\end{array}$

Therefore,

  $m_1=-\frac{1}{5}$

Now let the line given parallel to $3x+15y=22$ has its general form to be $y=m_{2}x+b_2$

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

Recall when two lines are parallel then their slope are equal that is $m_1=m_2$

Hence, $m_2=-\frac{1}{5}$

Therefore, equation of the required line reduces to:

  $y=\frac{-1}{5}x+b_2$ $\left(1\right)$

Also it is given that this line passes through origin, therefore, it must satisfy $\left(0,0\right)$

That is:

  $\begin{array}{l}0=\frac{-3}{15}\left(0\right)+b_2\\ b_2=0\end{array}$

Hence, $\left(1\right)$ reduces to:

  $\begin{array}{l}y=\frac{-1}{5}x+0\\ y=\frac{-1}{5}x\end{array}$

Hence, the equation for the line passing through the origin and is also parallel to line $3x+15y=22$ is $y=\frac{-1}{5}x$