The equation of line that passes through(5,0) and is perpendicular to the graph of y+ 1=2(x-3) is:
[tex]y = -\frac{1}{2}x+\frac{5}{2}[/tex]
Step-by-step explanation:
Given equation of line is:
[tex]y+1 = 2(x-3)[/tex]
The given equation is in point-slope form
So the factor with x will be the slope of the line
Let m1 be the slope of given line
m1 = 2
Let m2 be the slope of second line
As we know that the product of slopes of two perpendicular lines is: -1
So,
[tex]m_1.m_2 = -1\\2 . m_2 = -1\\m_2 = -\frac{1}{2}[/tex]
The slope intercept form is:
[tex]y=mx+b[/tex]
Putting the value of slope
[tex]y = -\frac{1}{2}x+b[/tex]
Putting the point in the equation (5,0)
[tex]0 = -\frac{1}{2}(5) +b\\0 = -\frac{5}{2}+b\\b = \frac{5}{2}[/tex]
Putting the value of b in the equation
[tex]y = -\frac{1}{2}x+\frac{5}{2}[/tex]
Hence,
The equation of line that passes through(5,0) and is perpendicular to the graph of y+ 1=2(x-3) is:
[tex]y = -\frac{1}{2}x+\frac{5}{2}[/tex]
Keywords: Equation of line, slope
Learn more about equation of line at:
- brainly.com/question/702593
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