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4. Find the equations of the following lines:

a. having slope of -2 and passing through the point (-4, 2)
b. parallel to y = 6x + 3 and passing through the point (-1, 13)
c. perpendicular to 3x – 5y = 5 and having an x-intercept of 3
d. passing through (-2, -6) and (4, 2)
e. having an x-intercept of -3 and a y-intercept of 6

This is frustrating me to no end!!! Please Help!

User Matt West
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1 Answer

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Answer:

Explanation:

The general equation for a line is given by: y = mx + n (m: slope, n: y-intercept)

a)

given: m = -2, x = -4, y = 2

equation becomes: 2 = -2 * (-4) + n

solve for n: n = -6

solution: y = -2x - 6

b)

given: m = 6, x = -1, y = 13

equation becomes: 13 = 6 * (-1) + n

solve for n: n = 19

solution: y = 6x + 19

c)

given: perpendicular to: y = 3/5x - 1 => m = -5/3, x = 3, y = 0

equation becomes: 0 = (-5/3) * 3 + n

solve for n: n = 5

solution: y = -5/3x + 5

d) get the slope m from the two points.

given: m = Δy/Δx = 8/6 = 4/3, x = 4, y = 2

equation becomes: 2 = (4/3) * 4 + n

solve for n: n = -10/3

solution: y = (4/3)x - (10/3)

e)

given: point1 (-3,0), point2(0,6)

m = Δy/Δx = 6/3 = 2, x = 0, y = 6

equation becomes: 6 = 0 + n

solve for n: n = 6

solution: y = 2x + 6

User Matthew Dresser
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