Chapter 1.3, Problem 73E

i.

To determine

ToFind:The equation of the line that passes through the point $\left(2,1\right)$ and parallel to $4x-2y=3$

Answer to Problem 73E

The equation of the line that passes through the point $\left(2,1\right)$ and parallel to $4x-2y=3$ is $y=2x-3$

Explanation of Solution

Given:

A point $\left(2,1\right)$ and the parallel line $4x-2y=3$

Concept Used:

Two distinct non vertical lines are parallel if and only of their slopes are equal.

That is $m_1=m_2$

Calculation:

Given a point $\left(2,1\right)$ and the parallel line $4x-2y=3$

Finding the slope of the parallel line $4x-2y=3$

That is $4x-2y=3$

  $\begin{array}{c}2y=4x-3\\ y=2x-\frac{3}{2}\end{array}$

Comparing with the general slope intercept form of the equation of a line

  $y=mx+c$ where is $m$ slope

Hence, slope of the parallel line is $m_1=2$

Since, the lines are parallel.

So, the slope of required line is $m_2=2$

Since, the required line passes through the point $\left(2,1\right)$

The equation of the line with point $\left(x_1,y_1\right)$ and slope $m$ is given by

  $\left(y-y_1\right)=m\left(x-x_1\right)$

  $\begin{array}{c}\left(y-1\right)=2\left(x-2\right)\\ y-1=2x-4\\ y=2x-4+1\\ y=2x-3\end{array}$

Therefore,the equation of the line that passes through the point $\left(2,1\right)$ and parallel to $4x-2y=3$ is $y=2x-3$

Conclusion:

The equation of the line that passes through the point $\left(2,1\right)$ and parallel to $4x-2y=3$ is $y=2x-3$

b.

To determine

ToFind:The equation of the line that passes through the point $\left(2,1\right)$ and perpendicular to $4x-2y=3$

Answer to Problem 73E

The equation of the line that passes through the point $\left(2,1\right)$ and perpendicular to $4x-2y=3$ is $2y=x$

Explanation of Solution

Given:

A point $\left(2,1\right)$ and the perpendicular line $4x-2y=3$

Concept Used:

Two distinct non vertical lines are perpendicular if and only of their slopes are negative reciprocals of each other.

That is $m_1=\frac{1}{m_2}$

Calculation:

Given a point $\left(2,1\right)$ and the perpendicular line $4x-2y=3$

Finding the slope of the perpendicular line $4x-2y=3$

That is $4x-2y=3$

  $\begin{array}{c}2y=4x-3\\ y=2x-\frac{3}{2}\end{array}$

Comparing with the general slope intercept form of the equation of a line

  $y=mx+c$ where is $m$ slope

Hence, slope of the perpendicular line is $m_1=2$

Since, the lines are perpendicular.

So, the slope of required line is

  $\begin{array}{c}{m}_2=\frac{1}{m_1}\\ =\frac{1}{2}\end{array}$

Since, the required line passes through the point $\left(2,1\right)$

The equation of the line with point $\left(x_1,y_1\right)$ and slope $m$ is given by

  $\left(y-y_1\right)=m\left(x-x_1\right)$

  $\begin{array}{c}\left(y-1\right)=\frac{1}{2}\left(x-2\right)\\ 2\left(y-1\right)=x-2\\ 2y-2=x-2\\ 2y=x\end{array}$

Therefore, the equation of the line that passes through the point $\left(2,1\right)$ and perpendicular to $4x-2y=3$ is $2y=x$

Conclusion:

The equation of the line that passes through the point $\left(2,1\right)$ and perpendicular to $4x-2y=3$ is $2y=x$