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Find the equation of a straight line through (-2, 3) and per... 0 Find the equation of a straight line through (-2, 3) and perpendicular to 4x +3y - 5 = 0

A. 5x - 2y - 11 = 0

B. 3x + 2y - 18 = 0

C. 3x - 4y + 18 = 0

D. 4x + 5y + 3 = 0

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University of Lagos Nigeria
04 May 2020

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Advertisement (continue below) James_mark answeredrep:728
School not set Nigeria
12 December 2020 0 4x +3y - 5 = 0 ------ eqn(1)

y - y1 = m(x - x1)

y1 = 3

x1 = -2

y - 3 = m1(x - (-2))

y - 3 = m1(x + 2) ------ eqn(2)

Since eqn(2) is perpendicular to eqn(1), the gradient or slope, m1 for eqn(2) = -1/gradient of eqn(1), m2

From eqn(1), 4x +3y - 5 = 0

3y = -4x + 5

y = -4x/3 + 5/3 -------eqn(3)

compare eqn(3) to the general stright line equation, y = mx + c

m2 = gradient = -4/3

m1 = -1/m2

m1 = -1/(-4/3)

m1 = 3/4

substitute m1 = 3/4 into eqn(2)

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

y - 3 = 3x/4 + 6/4

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

y = 3x/4 + 3/2 + 3

y = 3x/4 + 9/2

multiply through by 4 to remove fraction

4y = 3x + 18

3x - 4y + 18 = 0

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