Find the equation of the line through point (4,−7) and parallel to y=−23x+32.

3 answers
Question:

Find the equation of the line through point (4,−7) and parallel to y=−23x+32.

Answers

Step-by-step explanation:

m1 = m2 = -2/3

[tex]y - y1 = m2(x - x2)[/tex]

y - ( -7) = -2/3 ( x - 4)

y + 7 = -2/3x + 8/3

y = -2/3x + 8/3 -7

y = -2/3x -13/3

Step-by-step explanation:

Slope of parallel lines are equal.

1) m = -2/3

(4,-7)

[tex]y-y_{1}=m(x-x_{1})\\\\y-[-7]=\frac{-2}{3}(x-4)\\\\y+7=\frac{-2}{3}*x-\frac{-2}{3}*4\\\\y+7=\frac{-2}{3}x+\frac{8}{3}\\\\y=\frac{-2}{3}x+\frac{8}{3}-7\\\\y=\frac{-2}{3}x+\frac{8}{3}-\frac{7*3}{1*3}\\\\y=\frac{-2}{3}x+\frac{8}{3}-\frac{21}{3}\\\\y=\frac{-2}{3}x-\frac{13}{3}[/tex]

2) 5x + y = 4

  y = -5x + 4

slope m = -5

(-1,4)

y - 4 = -5(x -[-1])

y- 4 = -5(x+1)

y - 4 = -5x - 5

y = -5x - 5 + 4

y = -5x -1

If two lines are perpendicular, m2 = -1/m1

m1 = -4/3

[tex]m_{2}=\frac{-1}{\frac{-4}{3}}=-1*\frac{-3}{4}=\frac{3}{4}\\[/tex]

(5,4)

[tex]y-4=\frac{3}{4}(x - 5)\\\\y-4=\frac{3}{4}x-\frac{3}{4}*5\\\\y-4=\frac{3}{4}x-\frac{15}{4}\\\\y=\frac{3}{4}x-\frac{15}{4}+4\\\\y=\frac{3}{4}x-\frac{15}{4}+\frac{4*4}{1*4}\\\\y=\frac{3}{4}x-\frac{15}{4}+\frac{16}{4}\\\\y=\frac{3}{4}x+\frac{1}{4}[/tex]

y = -23x + 85

Step-by-step explanation:

for an equation to be parallel to another, they must have the same slope. So plugging in (4, -7) we get -7 = -23(4) + P. P being the new y intercept we must find. Now simple algebra add -23(4) = -92 to both sides and we get 85 = P. Plug in to y = mx + b and we get

y = -23x + 85

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