Chapter 4, Problem 6CA

(a)

To determine

To create: The equation of a line in slope-intercept

Answer to Problem 6CA

The equation of line passes through $(4,-5)$ and parallel to $x+2y=10$ is $y=-\frac{1}{2}x-3$

Explanation of Solution

Given:

Line passes through $(4,-5)$ and parallel to the line $x+2y=10$

Calculation:

Now the slope of line $x+2y=10$

Rewrite the equation

  $x+2y=10$

Take the $x$ variable on right side

  $2y=-x+10$

Divide by $2$ on both sides

  $y=-\frac{1}{2}x+5$

Compare the equation of given line with the slope-intercept form

So the slope is $m=-\frac{1}{2}$

Now using the slope $m=-\frac{1}{2}$ and the point $(4,-5)$ with the slope point form of line

  $y-y1=m(x-x1)$

Substitute the values

  $\begin{array}{l}y-(-5)=-\frac{1}{2}(x-4)\\ y+5=-\frac{1}{2}x+2\end{array}$

Subtract $5$ on both sides

  $y=-\frac{1}{2}x-3$

Therefore, The equation of line passes through $(4,-5)$ and parallel to $x+2y=10$ is $y=-\frac{1}{2}x-3$

Conclusion:

Therefore, The equation of line passes through $(4,-5)$ and parallel to $x+2y=10$ is $y=-\frac{1}{2}x-3$

(b)

To determine

To create: The equation of a line in slope-intercept

Answer to Problem 6CA

The equation of line passes through $(2,-1)$ and perpendicular to $x+2y=10$ is $y=2x-5$

Explanation of Solution

Given:

Line pass through $(2,-1)$ and perpendicular to the line $x+2y=10$

Calculation:

As the slope of the given line is $m=-\frac{1}{2}$

So, the slope of the line will be negative reciprocal $m=2$

Now using the slope $m=2$ and the point $(2,-1)$ with the slope point form of line

  $y-y1=m(x-x1)$

Substitute the values

  $\begin{array}{l}y-(-1)=2(x-2)\\ y+1=2x-4\end{array}$

Subtract $1$ on both sides

  $y=2x-5$

Therefore, The equation of line passes through $(2,-1)$ and perpendicular to $x+2y=10$ is $y=2x-5$

Conclusion:

Therefore, The equation of line passes through $(2,-1)$ and perpendicular to $x+2y=10$ is $y=2x-5$