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 parallel to given line passing through (8,-7) is:

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

Step-by-step explanation:

Given line is:

5x+4y=16

first of all, we have to convert the equation of given line in slope-intercept form

$4y = -5x+16$

Dividing both sides by 4

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

Slope intercept form is:

$y=mx+b$

The slope of given line is:

$m = -\frac{5}{4}$

Let m1 be the slope of line parallel to given line

"The slopes of two parallel lines are equal"

$m = m_1\\m_1 = -\frac{5}{4}$

The equation of line parallel to given line will be:

$y = m_1x+b$

Putting the value of slope

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

Putting the point (8,-7) in the equation

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

Putting the value of b

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

Hence,

The equation of line parallel to given line passing through (8,-7) is:

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