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trasher [3.6K]
2 years ago
14

Write an equation of the line that passes through (7,10) and is perpendicular to the line y=12x−9.

Mathematics
1 answer:
Ulleksa [173]2 years ago
8 0

Answer:

<em>y = - </em>\frac{1}{12}<em> x + </em>\frac{127}{12}<em> </em>

Step-by-step explanation:

y = 12x - 9

Slope of line perpendicular to given is m = - \frac{1}{12}

P(7, 10)

y - 10 = - \frac{1}{12} (x - 7)

<em>y = - </em>\frac{1}{12}<em> x + </em>\frac{127}{12}<em> </em>

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Which ordered pair describes a point that is located 1 unit to the left of the origin and 3 units above the x-axis?​
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4 0
3 years ago
Read 2 more answers
write the equation of the line, in standard form, that is perpendicular to y=3x-2 and passes through (-7,2)
34kurt

Answer:

x+3y=-1

Step-by-step explanation:

y=3x-2 so m=3 old slope

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y-y0=m*(x-x0)

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8 0
2 years ago
Write the equation of the circle with center (−1, −3) and (−7, −5) a point on the circle.
Pepsi [2]
So, we know the center is at -1, -3, hmmm what's the radius anyway?

well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ -1 &,& -3~) &#10;%  (c,d)&#10;&&(~ -7 &,& -5~)&#10;\end{array}&#10;\\\\\\&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2}&#10;\\\\\\&#10;r=\sqrt{36+4}\implies r=\sqrt{40}\\\\&#10;-------------------------------

\bf \textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad &#10;radius=\stackrel{\sqrt{40}}{ r}&#10;\\\\\\\&#10;[x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40
5 0
3 years ago
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