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Write the equation of the line passing through (3,-2) andperpendicular to 3x + 4y = 20.

User Nulldroid
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1 Answer

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LINE 1 : 3x + 4y = 20

rearranging : 4y = -3x + 20

y= -3/4 + 5 (y =mx + b)

therefore, m1 (slope) = -3/4

slopes of perpendicular lines are negative reciprocals of each other

m1 = -1 / m2 and m2 = -1/ m1

slope of line 2 (m2) = -1 / m1 = -1 / (-3/4) = 4/3

LINE 2 :

POINT - SLOPE FORM : y- y1 = m2 (x-x1)

point (3, -2); x1 = 3 and y1 = -2

m2 = 4/3

substituting the values in the point-slope form

y-y1 = m2 (x-x1)

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

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

y+2 = 4/3 x - 4

y = 4/3 x -4 - 2

y= 4/3 x -6

multiplying both sides of the equation by 3

3y = 4x -18

transposing,

4x-3y-18 = 0

answer: 4x - 3y -18 = 0

User Steven Doggart
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