The equation of line perpendicular to 4y = x-8 passing through (3,-3) is:
[tex]y = -4x+9[/tex]
Step-by-step explanation:
Given equation of line is:
[tex]4y=x-8[/tex]
We have to convert the given line in slope-intercept form to find the slope of the line
So,
Dividing both sides by 4
[tex]\frac{4y}{4} =\frac{x-8}{4}\\y = \frac{1}{4}x-\frac{8}{4}\\y = \frac{1}{4}x-2[/tex]
Let m1 be the slope of given line
Then
[tex]m_1 = \frac{1}{4}\\[/tex]
Let m2 be the slope of line perpendicular to given line
As we know that produt of slopes of two perpendicular lines is -1
[tex]m_1.m_2 = -1\\\frac{1}{4} . m_2 = -1\\m_2 = -1*4\\m_2 = -4[/tex]
The slope intercept form of line is given by:
[tex]y = m_2x+b[/tex]
Putting the value of slope
[tex]y = -4x+b[/tex]
to find the value of b, putting (3,-3) in equation
[tex]-3 = (-4)(3) + b\\-3 = -12 +b\\b = -3+12\\b = 9[/tex]
Putting the value of b in the equation
[tex]y = -4x+9[/tex]
Hence,
The equation of line perpendicular to 4y = x-8 passing through (3,-3) is:
[tex]y = -4x+9[/tex]
Keywords: Equation of line, slope
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