The equation of line perpendicular to given line through (-6,7) is:
[tex]y=\frac{1}{6}x+8[/tex]
Further explanation:
Given equation of line is:
[tex]6x+y=3\\y= -6x+3[/tex]
The co-efficient of x is the slope of given line.
Let m1 be the slope of given line
and
m2 be the slope of line perpendicular to given line
Then
[tex]m_1=-6[/tex]
Product of slopes of perpendicular lines is -1
[tex]m_1*m_2 = -1\\-6*m_2=-1\\m_2=\frac{-1}{-6}\\m_2=\frac{1}6}[/tex]
The equation of new line can be written as:
[tex]y=m_2x+b\\[/tex]
Putting m2
[tex]y=\frac{1}{6}x+b[/tex]
To find the value of b, we will put (-6,7) in equation
[tex]7=\frac{1}{6}(-6)+b\\7=-1+b\\b=7+1\\b=8[/tex]
Putting the values of b and m in general equation
[tex]y=\frac{1}{6}x+8[/tex]
Keywords: Slope, Point-slope form, perpendicular lines
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