Question

Write the equation of the line that is perpendicular to 3x + 7y = 15 and passes through (4, 9)

Answer 1

Answer:

7x - 3y = 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange 3x + 7y = 15 into this form

Subtract 3x from both sides

7y = - 3x + 15 ( divide all terms by 7 )

y = - $\frac{3}{7}$ x + $\frac{15}{7}$ ← in slope- intercept form

with slope m = - $\frac{3}{7}$

Given a line with slope m the the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{-\frac{3}{7} }$ = $\frac{7}{3}$, thus

y = $\frac{7}{3}$ x + c ← is the partial equation of the line

To find c substitute (4, 9) into the partial equation

9 = $\frac{28}{3}$ + c ⇒ c - $\frac{1}{3}$

y = $\frac{7}{3}$ x - $\frac{1}{3}$ ← in slope- intercept form

Multiply through by 3

3y = 7x - 1 ( subtract 3y from both sides )

0 = 7x - 3y - 1 ( add 1 to both sides )

1 = 7x - 3y, that is

7x - 3y = 1 ← in standard form