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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

**Comments**

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

y - y_{1} = m(x - x_{1})

y_{1} = 3

x_{1} = -2

y - 3 = m_{1}(x - (-2))

y - 3 = m_{1}(x + 2) ------ eqn(2)

Since eqn(2) is perpendicular to eqn(1), the gradient or slope, m_{1} for eqn(2) = -1/gradient of eqn(1), m_{2}

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

m_{2} = gradient = -4/3

m_{1} = -1/m_{2}

m_{1} = -1/(-4/3)

m_{1} = 3/4

substitute m_{1} = 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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