Question

Will someone please help on these three questions? For number 1 and 2 they have the same question but different problems.
Choose an equation for the line that is through the given point for questions 1 and 2. 1)y=5x+10;(2,14) a)y=1/5x+4 b)y=-1/5x+4 c)y=5x-68 d)y=5x+4
2)y=1/5x-10; (15,-16)
a)y=-5x+19 b)y=1/5x+91/5 c)y=1/5x-19 d)y=5x-19

5.choose the equation of a line that perpendicular to the given line and that passes through the given point. Ax-12y=2; (10,-1)
a)y=3x+29 b)y=-1/3x+29 c)y=-3x+29 d)y=-1/3x+7

Answer 1

Well the first one is 5x + 4 I just graphed it to get that and the second is -3x + 29 and I again graphed it :)

Answer 2

sorry the third one is -3x + 29 I didn't really see the second one

Answer 3

1. y = 5x + 10 -- In y = mx + b form, the slope is in the m position, therefore, the slope of this line is 5.

now use y = mx + b
slope(m) = 5
(2,14)..x = 2 and y = 14
now we sub, and since we already know the slope(m), we need to find the y intercept(b)
14 = 5(2) + b
14 = 10 + b
14 - 10 = b
4 = b
so your equation is : y = 5x + 4
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y = 1/5x - 10

y = mx + b
slope(m) = 1/5
(15,-16)..x = 15 and y = -16
time to sub and find the y intercept(b)
-16 = 1/5(15) + b
-16 = 3 + b
-16 - 3 = b
-19 = b
so your equation is : y = 1/5x - 19
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not sure about the 3rd one...sorry

Answer 4

well there ya go I answered the third one he gave you the first and second ;)

Answer 5

thanks y'all I really appreciate all y'alls help :)