Question

Are line passes through(5,0) and is perpendicular to the graph of y+ 1=2(x-3).What equation represents the line in slope-intersept form?

Answer 1

The equation of line that  passes through(5,0) and is perpendicular to the graph of y+ 1=2(x-3) is:

$y = -\frac{1}{2}x+\frac{5}{2}$

Step-by-step explanation:

Given equation of line is:

$y+1 = 2(x-3)$

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,

$m_1.m_2 = -1\\2 . m_2 = -1\\m_2 = -\frac{1}{2}$

The slope intercept form is:

$y=mx+b$

Putting the value of slope

$y = -\frac{1}{2}x+b$

Putting the point in the equation (5,0)

$0 = -\frac{1}{2}(5) +b\\0 = -\frac{5}{2}+b\\b = \frac{5}{2}$

Putting the value of b in the equation

$y = -\frac{1}{2}x+\frac{5}{2}$

Hence,

The equation of line that  passes through(5,0) and is perpendicular to the graph of y+ 1=2(x-3) is:

$y = -\frac{1}{2}x+\frac{5}{2}$

Keywords: Equation of line, slope

Learn more about equation of line at:

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