Question

What is the equation of a line that passes through the point (8, −2) and is parallel to the line whose equation is 3x + 4y = 15? Enter your answer in the box

Answer 1

Step 1

Find the slope of the given line

we have

$3x+4y=15$

Isolate the variable y

Subtract $3x$ both sides

$3x+4y-3x=15-3x$

$4y=-3x+15$

Divide by $4$ both sides

$y=-\frac{3}{4}x+ \frac{15}{4}$

the slope of the given line is

$m=-\frac{3}{4}$

Step 2

Find the equation of the line that passes through  the point $(8,-2)$ and is parallel to the given line

we know that

if two lines are parallel, then their slopes are equal

The equation of the line into point-slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{3}{4}$

$(x1,y1)=(8,-2)$

substitute in the equation

$y+2=-\frac{3}{4}(x-8)$

$y=-\frac{3}{4}x+6-2$

$y=-\frac{3}{4}x+4$  or  $3x+4y=16$

therefore

the answer is

$y=-\frac{3}{4}x+4$

or

$3x+4y=16$