Answer:
Equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1 is [tex]\mathbf{y=3x-21}[/tex]
Step-by-step explanation:
We need to write equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1
The equation will be in slope-intercept form i.e [tex]y=mx+b[/tex] where m is slope and b is y-intercept.
Finding Slope
Since the lines are parallel to each other, their slopes will be equal.
Slope of given line: y = 3x + 1 is 3 (Compare it with general equation [tex]y=mx+b[/tex] we get m = 3)
So, slope of required line is: m=3
Finding y-intercept
Using the point (6,-3) and slope m = 3 we can find y-intercept by using the formula:
[tex]y=mx+b\\-3=3(6)+b\\-3=18+b\\b=-3-18\\b=-21[/tex]
So, we get y-intercept: b= -21
Equation of required line
The equation of required line having slope m=3 and y-intercept b = -21 is:[tex]y=mx+b\\y=3x-21[/tex]
So, equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1 is [tex]\mathbf{y=3x-21}[/tex]
