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maks197457 [2]
2 years ago
15

The slope-intercept form of the equation of a line that passes through point (-2,-13) is y=5x-3. What is the point-slope form of

the equation for this line?
Mathematics
1 answer:
marissa [1.9K]2 years ago
4 0

Answer:

y + 13 = 5(x + 2)

Step-by-step explanation:

<h3>Point - slope form of equation: </h3>

        \sf \boxed{\bf y - y_1=m(x-x_1)}

Slope intercept of the line : y =5x - 3

Comparing with y = mx +b , m =slope = 5

m = 5 and Point (-2 ,-13)

     y - [-13] = 5(x - [-2])

     y + 13 = 5(x + 2)

   

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1.What is the equation of the line perpendicular to  that passes through ? Write your answer in slope-intercept form. Show your
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Answer:

1. Use a compass to make arc marks which intersect above and below then connect.

2. y=\frac{1}{3}x + 2

Step-by-step explanation:

1. To construct a perpendicular line, use a compass to draw arc marks from one end of the segment through point P. Then repeat this again at the other end. This means at point P there will be two intersecting arc marks. Repeat the process down below with the same radius as used above. Then connect the two intersections.

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(y-1)=\frac{1}{3}(x+3)  

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(y-1)=\frac{1}{3}(x+3)\\(y-1)=\frac{1}{3}x+1\\y=\frac{1}{3}x + 2


3 0
3 years ago
Write a translation rule that maps point D(7,-3) onto point D'(2,5).
Sergio039 [100]
We are given D (7, -3) and D'(2, 5).

SAppy the transformation
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Answer: D (x-5, y+8) → D'
6 0
3 years ago
Read 2 more answers
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