Answer:
The equation of line parallel to given line passing through (8,-7) is:
[tex]y = -\frac{5}{4}x+3[/tex]
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
[tex]4y = -5x+16[/tex]
Dividing both sides by 4
[tex]\frac{4y}{4} = -\frac{5x}{4} + \frac{16}{4}\\y = -\frac{5}{4}x+4[/tex]
Slope intercept form is:
[tex]y=mx+b[/tex]
The slope of given line is:
[tex]m = -\frac{5}{4}[/tex]
Let m1 be the slope of line parallel to given line
"The slopes of two parallel lines are equal"
[tex]m = m_1\\m_1 = -\frac{5}{4}[/tex]
The equation of line parallel to given line will be:
[tex]y = m_1x+b[/tex]
Putting the value of slope
[tex]y = -\frac{5}{4}x+b[/tex]
Putting the point (8,-7) in the equation
[tex]-7 = -\frac{5}{4}(8)+b\\-7 = -(5)(2) + b\\-7 = -10+b\\b = -7 +10\\b = 3[/tex]
Putting the value of b
[tex]y = -\frac{5}{4}x+3[/tex]
Hence,
The equation of line parallel to given line passing through (8,-7) is:
[tex]y = -\frac{5}{4}x+3[/tex]
