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AVprozaik [17]
4 years ago
9

HELP PLEASE???!!!!!!!

Mathematics
1 answer:
Fiesta28 [93]4 years ago
7 0

Answer:

<h2>A. (-4, -15)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}y=\dfrac{3}{4}x-12&(1)\\y=-4x-31&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\\dfrac{3}{4}x-12=-4x-31\qquad\text{multiply both sides by 4}\\\\3x-48=-16x-124\qquad\text{add 48 to both sides}\\\\3x=-16x-76\qquad\text{add}\ 16x\ \text{to both sides}\\\\19x=-76\qquad\text{divide both sides by 19}\\\\x=-4\\\\\text{Put it to (2):}\\\\y=-4(-4)-31\\y=16-31\\y=-15

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Norma-Jean [14]
First, coordinate Q is (-5,2) and Q' is (6,2). First, we know that this is a horizontal reflection becuase only the x-values change. This means that the line of reflection will be x=some value. I think the best way to go about this is to find the midpoint of Q and Q' using the formula: M=(x1+x2/2,y1+y2/2)
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