Nobody1278
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What is an equation of the line perpendicular to y=-x-2 and through (-2, 4)?

I would really appreciate the help, I've been stuck on this problem for awhile.​

Sagot :

Snor1ax

Answer: y = x + 6

Step-by-step explanation:

[tex]\mathrm{Find\:the\:line\:}\mathbf{y=mx+b}\mathrm{\:perpendicular\:to\:}y=-x-2\mathrm{\:that\:passes\:through\:}\left(-2,\:4\right)[/tex]

[tex]\mathrm{For\:a\:line\:equation\:for\:the\:form\:of\:}\mathbf{y=mx+b}\mathrm{,\:the\:slope\:is\:}\mathbf{m}[/tex]

[tex]m=-1[/tex]

[tex]\mathrm{The\:perpendicular\:slope\:is\:the\:negative\:reciprocal\:of\:the\:given\:slope}[/tex]

[tex]\left(-1\right)m_p=-1[/tex]

[tex]\frac{\left(-1\right)m_p}{-1}=\frac{-1}{-1}[/tex]

[tex]m_p=1[/tex]

[tex]\mathrm{Compute\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:for\:slope\:m=}1\mathrm{\:and\:passing\:through\:}\left(-2,\:4\right)[/tex][tex]\mathrm{Plug\:the\:slope\:}1\mathrm{\:into\:}y=mx+b[/tex]

[tex]y=x+b[/tex]

[tex]\mathrm{Plug\:in\:}\left(-2,\:4\right)\mathrm{:\:}\quad \:x=-2,\:y=4[/tex]

[tex]4=\left(-2\right)+b[/tex]

[tex]-2+b=4[/tex]

[tex]\mathrm{Add\:}2\mathrm{\:to\:both\:sides}[/tex]

[tex]-2+b+2=4+2[/tex]

[tex]\text{Simplify}[/tex]

[tex]b=6[/tex]

[tex]\mathrm{Construct\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:where\:}\mathbf{m}=1\mathrm{\:and\:}\mathbf{b}=6[/tex]

[tex]y=x+6[/tex]

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