Quadrilateral ABCD has vertices at A(0,0), B(0,2), C(6,2), and D(6,0). Find the vertices of the quadrilateral after a dilation w
1 answer:
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Here are a couple I found:
<u>Similarities</u>:
- They have the same y-intercept of (0,5).
- They are both in slope-intercept form.
<u>Differences</u>:
- The line of y = -13x + 5 "falls" from left to right. The line of y = 2x + 5 "rises" from left to right.
- They have different x-intercepts. (y = 2x + 5 intersects (-
, 0) while y = -13x + 5 intersects at (
, 0)
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<u>Explanation</u>:
Slope-intercept form is y = mx + b, and by looking at the equations, they both already fit that format, with m as their slope and b as their y-intercept. Also, since they both have a 5 as that "b," their y-intercepts are the same: (0,5).
As for differences, we can see that the coefficient in place of that "m" is positive in y = <u>2x</u> + 5 and negative in y = <u>-13x</u> + 5. Therefore, one line would rise due to their slope being positive and one would fall due to their slope being negative. They also have two different x-intercepts, which we can calculate by substituting 0 in place of the y, then isolating x.