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3 years ago

Can someone please explain how to find this answer? thanks

Mathematics

2 answers:

N76 [4]3 years ago

6 0

I would start with the standard form equation of the parallel line.
.. 5x -2y = 5(-2) -2(3) = -16
Then solve for y.
.. 2y = 5x +16 . . . add 2y+16; then divide by 2 for the next equation
.. y = (5/2)x +8 . . . . . . . corresponds to selection (B)

GrogVix [38]3 years ago

3 0

So, a line that is parallel to 5x-2y = -12, will have the same exact slope as that equation's, so what is it anyway?

$\bf 5x-2y=-12\implies 5x+12=2y\implies \cfrac{5x+12}{2}=y
\\\\\\
\cfrac{5x}{2}+\cfrac{12}{2}=y\implies \stackrel{slope}{\cfrac{5}{2}}x+6=y$

so, notice, from the slope-intercept form, it happens that the slope is 5/2, well, the line will have the same slope.

so we're looking for the equation of a line whose slope is 5/2 and runs through -2,3

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
&&(~ -2 &,& 3~)
\end{array}
\\\\\\
% slope = m
slope = m\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=\cfrac{5}{2}[x-(-2)]
\\\\\\
y-3=\cfrac{5}{2}(x+2)\implies y-3=\cfrac{5}{2}x+5\implies y=\cfrac{5}{2}x+8$

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equation for the perpendicular Bisector of the line segment whose endpoints are (-9,-8) and (7,-4)

Perpendicular bisector lies at the midpoint of a line

Lets find mid point of (-9,-8) and (7,-4)

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$\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}$

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Subtract 6 on both sides

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