hunterfatdog
hunterfatdog hunterfatdog
  • 15-01-2019
  • Mathematics
contestada

What is the equation of a line that is perpendicular to y-3=-4(x+2) and passes through the point (-5,7)

Respuesta :

gmany
gmany gmany
  • 17-01-2019

The point-slope form:

[tex]y-y_1=m(x-x_1)[/tex]

m - slope

Let [tex]k:y=m_1x+b_1[/tex] and [tex]l:y=m_2x+b_2[/tex]

[tex]l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}[/tex]

We have

[tex]y-3=-4(x+2)\to m_1=-4[/tex]

Therefore

[tex]m_2=-\dfrac{1}{-4}=\dfrac{1}{4}[/tex]

We have the slope and the point (-5, 7). Substitute to the point-slope formula:

[tex]y-7=\dfrac{1}{4}(x-(-5))\\\\\boxed{y-7=\dfrac{1}{4}(x+5)}[/tex]

Answer Link
Dangerlol Dangerlol
  • 30-09-2021

Answer:

y-7=1/4(x+5)

Step-by-step explanation:

Answer Link

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