Equation of the straight line perpendicular to $4 x + 3 y + 5 = 0$ and passing through $(1, 1)$ is

3 x + 4 y + 1 = 0

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3 x + 4 y - 5 = 0

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4 x - 3 y + 5 = 0

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3 x - 4 y + 1 = 0

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Solution

The correct option is D $3 x - 4 y + 1 = 0$
line $4 x + 3 y + 5 = 0$ & point (1, 1)
$3 y = - 4 x - 5$
$y = \frac{- 4 x}{3} - 5 / 3 (∵ y = m x + c)$
therefore slope $m_{1} = - 4 / 3$
slope of line perpendicular to the line $4 x + 3 y + 5 = 0$
is $m_{2} = 3 / 4$
equation of line pass (1, 1) & slope 3/4
$y - y_{1} = m (x - x_{1})$
$y - 1 = 3 / 4 (x - 1)$
$4 y - 4 = 3 x - 3$
$3 x - 4 y + 1 = 0$ Ans

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Equation of the straight line perpendicular to 4x+3y+5=0 and passing through (1,1) is

The correct option is D 3x−4y+1=0line 4x+3y+5=0 & point (1, 1)3y=−4x−5y=−4x3−5/3(∵y=mx+c)therefore slope m1=−4/3slope of line perpendicular to the line 4x+3y+5=0is m2=3/4equation of line pass (1, 1) & slope 3/4y−y1=m(x−x1)y−1=3/4(x−1)4y−4=3x−33x−4y+1=0 Ans

Equation of the straight line perpendicular to $4 x + 3 y + 5 = 0$ and passing through $(1, 1)$ is