Write the equation of the line perpendicular to 2x-3y=12 which passes through the point (16,-7).

1 answer:

3 0

Y= mx+b

(16, -7) = (x,y)

2(16) - 3y = 12
32 - 3y = 12
32 - 12 = 3y
20/3 = y
6.6 = y

2x-3(-7) = 12
2x+21 = 12
2x = 12 - 21
x = -9/2

Insert the values into y = mx + b
solve for m and then solve for b

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Read 2 more answers

[Geometry Basics] I'm just checking if this is correct:

Leno4ka [110]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}$

$\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}$

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Ne4ueva [31]

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