Chapter 1, Problem 128RE

To determine

To calculate:The equation for the line passing through $\left(5,2\right)$ and is also parallel to the line which passes through the points $\left(-1,-3\right)$ and $\left(3,2\right)$ .

Answer to Problem 128RE

The equation for the line passing through $\left(5,2\right)$ and is also parallel to the line which passes through the points $\left(-1,-3\right)$ and $\left(3,2\right)$ is $5x-4y=17$

Explanation of Solution

Given information:

A line passing through $\left(5,2\right)$ and also parallel to the line which passes through the points $\left(-1,-3\right)$ and $\left(3,2\right)$

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:

  $y=mx+b$

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

Calculation:

Consider thepoints $\left(-1,-3\right)$ and $\left(3,2\right)$

Let the slope of the line passing through these points be $m_1$

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:

  $m=\frac{y_2-y_1}{x_2-x_1}$

To proceed further the following must be known:

  $\begin{array}{l}{x}_1=-1\\ x_2=3\\ y_1=-3\\ y_2=2\end{array}$

Therefore,

  $\begin{array}{l}{m}_1=\frac{2-\left(-3\right)}{3-\left(-1\right)}\\ m_1=\frac{5}{4}\end{array}$

Now let the line given parallel tothe line which passes through the points $\left(-1,-3\right)$ and $\left(3,2\right)$ has slope $m_2$

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

Hence, $m_2=\frac{5}{4}$

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

  $y-y_1=m\left(x-x_1\right)$

We are given that the required line passes through the point $\left(5,2\right)$

Therefore,

  $\begin{array}{l}{x}_1=5\\ y_1=2\end{array}$

And

  $m_2=m=\frac{5}{4}$

Pitting all the values, the required equation for the line is:

  $\begin{array}{l}\frac{5}{4}=\frac{y-2}{x-5}\\ 5\left(x-5\right)=4\left(y-2\right)\\ 5x-25=4y-8\\ 5x-4y=-8+25\\ 5x-4y=17\end{array}$

Hence, theequation for the line passing through $\left(5,2\right)$ and is also parallel to the line which passes through the points $\left(-1,-3\right)$ and $\left(3,2\right)$ is $5x-4y=17$