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Write the equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point. y = -5x + 1; through (2, -1)

Respuesta :

ShyzaSling

Answer:

The equation is y = 1/5x -7/5

Step-by-step explanation:

The given equation is in the form of y = mx + b

⇒ m = -5

since the line is perpendicular, therefore perpendicular line has a slope reciprocal of the given slope, that is m = 1/5

By using point slope equation  (y-y1) = m(x-x1)

here (y1,x1) = (2,-1) and m = 1/5

(y-(-1)) = (1/5)(x-2)

y+1 = 1/5x -2/5

y+1-1 = 1/5x - 2/5 -1

y = 1/5x -7/5 is the desired equation.  

gmany

Answer:

[tex]\large\boxed{y=\dfrac{1}{5}x-1\dfrac{2}{5}}[/tex]

Step-by-step explanation:

[tex]\text{The slope-intercept form:}\\\\y=mx+b\\\\m-slope\\b-\ y-intercept[/tex]

[tex]\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ ||\ k\iff m_2=m_1\\------------------------\\\\\text{We have}\ k:y=-5x+1\to m_1=-5.\\\text{Therefore}\ m_2=-\dfrac{1}{-5}=\dfrac{1}{5}\\\\l:y=\dfrac{1}{5}x+b\\\\\text{The line}\ l\ \text{passes through point (2. -1). Substitute x = 2 and y = -1}\\\text{to the equation of the line}\ l:[/tex]

[tex]-1=\dfrac{1}{5}(2)+b\\\\-1=\dfrac{2}{5}+b\qquad\text{subtract}\ \dfrac{2}{5}\ \text{from both sides}\\\\-1-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{2}{5}+b\\\\-1\dfrac{2}{5}=b\to b=-1\dfrac{2}{5}\\\\\text{Finally we have:}\\\\l:y=\dfrac{1}{5}x-1\dfrac{2}{5}[/tex]