Question

Write an equation of the line through the point (5, 3) that is perpendicular to the line –3x+7y=5.

Answer 1

Answer:

7x + 3y = 44

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 = 5 into this form by adding 3x to both sides

7y = 3x + 5 ( divide all terms by 7 )

y = $\frac{3}{7}$ x + $\frac{5}{7}$ ← in slope- intercept form

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

Given a line with slope m then 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

To find c substitute (5, 3) into the partial equation

3 = - $\frac{35}{3}$ + c ⇒ c = 3 + $\frac{35}{3}$ = $\frac{44}{3}$

y = - $\frac{7}{3}$ x + $\frac{44}{3}$ ← in slope- intercept form

Multiply through by 3

3y = - 7x + 44 ( add 7x to both sides )

7x + 3y = 44 ← in standard form

Answer 2

Answer:

equation: y = -7/3x + 14 2/3

Step-by-step explanation:

–3x+7y=5

y = 3/7 x + 5

slope of perpendicular line: - 7/3

y = mx + b for (5 , 3)

b = y - mx = 3 - ((-7/3) * 5) = 3 + 35/3 = 44/3 = 14 2/3

equation: y = -7/3x + 14 2/3