Answer:
The equation of line that passes through the point (8,-7) and is parallel to the line 5x+4y=16 is [tex]\mathbf{y=-\frac{5}{4}x+3}[/tex]
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 [tex]y=mx+b[/tex] 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 [tex]5x+4y=16[/tex] can be found by writing in slope-intercept form [tex]y=mx+b[/tex]
[tex]5x+4y=16\\4y=-5x+16\\y=-\frac{5}{4}x+16[/tex]
Comparing with [tex]y=mx+b[/tex] the slope m is: [tex]m=-\frac{5}{4}[/tex]
So, the slope of required line is: [tex]m=-\frac{5}{4}[/tex]
Now, finding y-intercept b
y-intercept can be found using slope [tex]m=-\frac{5}{4}[/tex] and point (8,-7)
[tex]y=mx+b\\-7=-\frac{5}{4}(8)+b\\-7=-5(2)+b\\-7=-10+b\\b=-7+10\\b=3[/tex]
So, we get y-intercept: b=3
Now, the equation of required line having slope [tex]m=-\frac{5}{4}[/tex] and y-intercept b=3 is:
[tex]y=mx+b\\y=-\frac{5}{4}x+3[/tex]
So, the equation of line that passes through the point (8,-7) and is parallel to the line 5x+4y=16 [tex]\mathbf{y=-\frac{5}{4}x+3}[/tex]
