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ser-zykov [4K]
3 years ago
8

What is an equation of the line that passes through the points (3,-3) and (-3,-3)?

Mathematics
1 answer:
victus00 [196]3 years ago
3 0
First find the gradient of the line
Change in y/change in x
-3–3/-3-3
0/-6
=0 ( so the gradient m is equal to zero)

Y=0x+c

Input the coordinates of one point to find c

-3=(0*3)+c
-3=c


So the equation is
Y= -3


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Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR
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Answer:

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Step-by-step explanation:

Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

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Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392  

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