Question

Write the equation of a line perpendicular to y=0.25x-7 and passes through the point -6,8

Answer 1

the equation of a line perpendicular to $y=0.25x-7$ and passes through the point (-6,8) is $y=-4x+30$

Step-by-step explanation:

We need to write equation of line perpendicular to $y=0.25x-7$ and passes through the point (-6,8)

Since the line $y=0.25x-7$ is in slope-inetercept form $y=mx+b$

the slope of line m = 0.25

As, the new line is perpendicular to the given line so, slopes of lines that are perpendicular are: slope = -1/slope1

So, slope will be -1/0.25 = -4

Now, finding y-intercept.

$y=mx+b8=4(-6)+b\\8=-24+b\\b=8+24\\b=30$

So, value of y-intercept b = 30

Now the equation of the required line having slope m= -4 and b =30 will be:

$y=mx+b\\y=-4x+30$

So, the equation of a line perpendicular to $y=0.25x-7$ and passes through the point (-6,8) is $y=-4x+30$

Keywords: Equation of line, slope-intercept form

Learn more about slope-intercept form at:

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