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

Answer:

Step-by-step explanation:

Use the point-slope formula by the line

y - y_1 = m(x - x_1)      when  x_1 =8   and  y_1 = -2    m :  is the slope (same slope for the line whose equation is 3x + 4y = 15 because are  parallel)

calculate : m

3x + 4y = 15

4y = -3x +15

y = (-3/4)x +15/4     so   m = -3/4

an equation is : y +2 =(-3/4)(x-8)

Answer 2

Answer:

The required equation of line is $3x+4y=16$

Step-by-step explanation:

Given : A line that passes through the point (8, -2) and is parallel to the line whose equation is $3x + 4y = 15$

To find : What is the equation of a line ?

Solution :

We know that,

When two lines are parallel then their slopes are equal.

The equation of line is $3x + 4y = 15$

Convert into slope form $y=mx+c$,

$4y =-3x+ 15$

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

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

The slope of the line is $m=-\frac{3}{4}$

The line passes through (8,-2).

The general point slope form is $(y-y_1)=m(x-x_1)$

i.e. $(y-(-2))=-\frac{3}{4}(x-8)$

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

$4y+8=-3x+24$

$3x+4y=16$

Therefore, the required equation of line is $3x+4y=16$