The equation of line $c$ is $y = -7x + 10$ . Line $d$ is perpendicular to line $c$ and passes through $(7,4)$ . What is the equation of line $d$ ? $\newline$ Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

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Algebra 1

Write an equation for a parallel or perpendicular line

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Q. The equation of line

c

y = -7x + 10

. Line

d

is perpendicular to line

c

and passes through

(7,4)

. What is the equation of line

d

\newline

Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

Determine slope of line c: Determine the slope of line c. The equation of line c is given as $y = -7x + 10$ . The slope ( $m$ ) of a line in the slope-intercept form $y = mx + b$ is the coefficient of $x$ . Therefore, the slope of line c is $-7$ .
Find slope of line $d$ : Find the slope of line $d$ . Since line $d$ is perpendicular to line $c$ , its slope will be the negative reciprocal of the slope of line $c$ . The negative reciprocal of $-7$ is $\frac{1}{7}$ . Therefore, the slope of line $d$ is $\frac{1}{7}$ .
Use point-slope form: Use the point-slope form to find the equation of line $d$ . We have the slope of line $d$ ( $\frac{1}{7}$ ) and a point through which it passes $(7,4)$ . The point-slope form of a line is $(y - y_1) = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the line. Plugging in the values, we get $(y - 4) = \frac{1}{7}(x - 7)$ .
Simplify to slope-intercept form: Simplify the equation to slope-intercept form. $\newline$ To get the slope-intercept form $y = mx + b$ , we distribute the slope on the right side of the equation and then isolate $y$ : $\newline$ $y - 4 = \frac{1}{7} \times x - \frac{1}{7} \times 7$ $\newline$ $y - 4 = \frac{1}{7} \times x - 1$ $\newline$ $y = \frac{1}{7} \times x - 1 + 4$ $\newline$ $y = \frac{1}{7} \times x + 3$
Verify the equation: Verify the equation. $\newline$ The equation $y = \frac{1}{7} \times x + 3$ is in slope-intercept form, with a slope of $\frac{1}{7}$ and a $y$ -intercept of $3$ . This line is perpendicular to line $c$ , which has a slope of $-7$ , and passes through the point $(7,4)$ , as required.