The equation of line perpendicular to given line passing through (8,-5) is:
[tex]y = -\frac{3}{4}x +1[/tex]
Step-by-step explanation:
Given equation of line is:
[tex]4x-3y = 15\\3y = 4x -15\\\frac{3y}{3} = \frac{4x-15}[3}\\y = \frac{4}{3}x - \frac{15}{3}\\ y = \frac{4}{3}x - 5[/tex]
Let m1 be the slope of given line
then
m1 = 4/3
Let m2 be the slope of line perpendicular to given line
As we know that the product of slopes of two perpendicular line is -1
[tex]m_1.m_2 = -1\\\frac{4}{3} . m_2 = -1\\m_2= -\frac{3}{4}[/tex]
Equation of line is given by:
[tex]y = m_2x+b[/tex]
Putting the value of the slope
[tex]y = -\frac{3}{4}x +b[/tex]
Putting the point (8,-5) in equation
[tex]-5 = -\frac{3}{4} (8) +b\\-5 = -3 * 2 +b\\-5 = -6 +b\\b = -5+6\\b = 1[/tex]
Putting the value of b in equation
[tex]y = -\frac{3}{4}x +1[/tex]
Hence,
The equation of line perpendicular to given line passing through (8,-5) is:
[tex]y = -\frac{3}{4}x +1[/tex]
Keywords: Equation of line, slope
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