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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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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
Ans C
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