Question

find the slope of the line that is a) parallel and b) perpendicular to the given line
1a) 5x+2y=10

1b) y=-7

1c) x=10

write an equation for the line in point/slope form and slope/intercept form that has the given condition.
2a) passes through (-7,2) and is parallel to 7x+2y=0

2b) passes through (3,-1) and is perpendicular to y=2x-3

Answer 1

I. Find the slope of the line that is a) parallel and b) perpendicular to the given line

1a) $5x + 2y = 10$

1b) $y = -7$

1c) $x = 10$

II. Write an equation for the line in point/slope form and slope/intercept form that has the given condition.

2a) passes through $(-7, 2)$ and is parallel to $7x + 2y = 0$

2b) passes through (3,-1) and is perpendicular to $y = 2x - 3$

Solution

Parallel lines and their slopes $m_{1} = m_{2}$ . Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel. Find the slope of the line that is parallel to the given line $5x + 2y = 10$

$5x + 2y = 10$

$2y = 10 - 5x$

$y = 5 - \frac{5}{2} x$

$y = -\frac{5}{2} x + 5$ is the formula for a line with as slope of $(- \frac{5}{2})$ , the reference slope from the reference line is $m_{1} = -\frac{5}{2}$ . Any line with a slope of $(- \frac{5}{2})$ is parallel to this line. Graphing equation is as follows:

The slope of the line that parallel to the given line $y = -\frac{5}{2} x + 5$ .

Example of the parallel line equation

Find the slope of the line that perpendicular to the given line $y = -\frac{5}{2} x + 5$ . If we visualize a line with positive slope (so it's an increasing line), then the perpendicular line must have negative slope (because it will be a decreasing line). Perpendicular slope have opposite sign. Any line perpendicular to this line has a slope of the negative reciprocal of this slope.

$$m _ {1} = - \frac {1}{m _ {2}}; m _ {1} = - \frac {1}{- \frac {5}{2}} = \frac {2}{5}$$

Graphing equations is as follows:

The slope of the line that perpendicular to the given line $y = -\frac{5}{2} x + 5$ .

1b) $y = -7$

The equation of a line: $y = mx + b, y = 0x - 7, m = 0$. A parallel line has slope of 0.

The perpendicular has a slope of $-\frac{1}{0}$ which is undefined slope.

1c) $x = 10$

A parallel line has an undefined slope; the perpendicular has a slope of 0.

II. Write an equation for the line in point/slope form and slope/intercept form that has the given condition.

2a) passes through $(-7,2)$ and is parallel to $7x + 2y = 0$

Firstly find the slope of the line that is parallel to the given line $7x + 2y = 0$.

$$\begin{array}{l} 7x + 2y = 0 \\ 2y = -7x \\ y = -\frac{7}{2}x \\ \end{array}$$

The reference slope from the reference line is $m_1 = -\frac{7}{2}$. Since a parallel line has an identical slope, then the parallel line through (-7,2) will have slope $m_1 = -\frac{7}{2}$. Use the point-slope form to find the line:

$$\begin{array}{l} y - 2 = -\frac{7}{2}(x + 7) \\ y - 2 = -\frac{7}{2}x - \frac{49}{2} \\ y = -\frac{7}{2}x - \frac{49}{2} + 2 \\ y = -\frac{7}{2}x - \frac{45}{2} \\ \end{array}$$

This is the parallel line that they asked for.

Slope/intercept equation of a line: $y = mx + b$. To find an equation for the line in point substitute the coordinate $(-7,2)$ into $y = -\frac{7}{2}x + b \Rightarrow 2 = -\frac{7}{2}(-7) + b \Rightarrow b = -\frac{45}{2}$. Slope/intercept form is $y = -\frac{7}{2}x - \frac{45}{2}$. Point/slope form is $y - 2 = -\frac{7}{2}(x + 7)$. Graphing equations is as follows:

II. Write an equation for the line in point/slope form and slope/intercept form that has the given condition.

2b) passes through $(3, -1)$ and is perpendicular to $y = 2x - 3$

Find the slope of the line $y = 2x - 3$ , the reference slope is $m_{1} = 2$ , and, for the perpendicular slope, we will this slope and change the sign. Then the perpendicular slope is $m_{1} = -\frac{1}{2}$ to our line. So now we can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now. To find an equation for the line in point substitute the coordinate $(3, -1)$ into $-1 = -\frac{1}{2} \cdot 3 + b = > b = -\frac{1}{2}$ . Slope/intercept form is $y = -\frac{1}{2} x + \frac{1}{2}$ . Point/slope form is $y - (-1) = -\frac{1}{2}(x - 3)$ or $y + 1 = -\frac{1}{2}(x - 3)$ . Graphing equations is as follows: