Question

What is an equation of the line that passes through the point (8,-7)(8,−7) and is parallel to the line 5x+4y=165x+4y=16?

Answer 1

Answer:

The equation of line that passes through the point (8,-7) and is parallel to the line 5x+4y=16 is  $\mathbf{y=-\frac{5}{4}x+3}$

Step-by-step explanation:

We need to write an equation of the line that passes through the point (8,-7) and is parallel to the line 5x+4y=16.

The equation will be of form $y=mx+b$ where m is slope and b is y-intercept.

Finding slope of the line:

Since both the lines are parallel, and we know that parallel lines have same slope.

The slope of given line $5x+4y=16$ can be found by writing in slope-intercept form $y=mx+b$

$5x+4y=16\\4y=-5x+16\\y=-\frac{5}{4}x+16$

Comparing with $y=mx+b$ the slope m is: $m=-\frac{5}{4}$

So, the slope of required line is: $m=-\frac{5}{4}$

Now, finding y-intercept b

y-intercept can be found using slope $m=-\frac{5}{4}$ and point (8,-7)

$y=mx+b\\-7=-\frac{5}{4}(8)+b\\-7=-5(2)+b\\-7=-10+b\\b=-7+10\\b=3$

So, we get y-intercept: b=3

Now, the equation of required line having slope $m=-\frac{5}{4}$ and y-intercept b=3 is:

$y=mx+b\\y=-\frac{5}{4}x+3$

So, the equation of line that passes through the point (8,-7) and is parallel to the line 5x+4y=16 $\mathbf{y=-\frac{5}{4}x+3}$