Question

find the equation of a line whose y intercepts is 3/4 and which is perpendicular to 2x-3y+5=0​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Line is 2x-3y+5=0}$

$\underline{\textsf{To find:}}$

$\textsf{Equation of the line whose y interrcept is 3/4 and is}$

$\textsf{perpendicular to 2x-3y+5=0}$

$\underline{\textsf{Solution:}}$

$\textsf{We know that,}$

$\textsf{Any line perpendicular to ax+by+c=0 can be put}$

$\textsf{in the form bx-ay+k=0}$

$\mathsf{Now,}$

$\textsf{The line perpendicular to 2x-3y+5=0 can be written as}$

$\mathsf{-3x-2y+k=0}$

$\implies\mathsf{2y=-3x+k}$

$\implies\mathsf{y=\dfrac{-3x}{2}+\dfrac{k}{2}}$

$\textsf{Comparing with y=mx+c we get}$

$\mathsf{y-intercept=\dfrac{k}{2}}$

$\implies\mathsf{\dfrac{3}{4}=\dfrac{k}{2}}$

$\implies\mathsf{k=\dfrac{3}{2}}$

$\therefore\textsf{The required line is}$

$\mathsf{-3x-2y+\dfrac{3}{2}=0}$

$\mathsf{-6x-4y+3=0}$

$\boxed{\mathsf{6x+4y-3=0}}$

$\underline{\textsf{Find more:}}$

Find the equation of line perpendicular to the line 6x+5y+2=0 passing through the points (5,2)​

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Find the equation of the line perpendicular to 2x+3y-6=0 and passing through (-1,1)

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Answer 2

SOLUTION

TO DETERMINE

The equation of a line whose y intercepts is 3/4 and which is perpendicular to

2x - 3y + 5 = 0

EVALUATION

Here the given equation of the line is

2x - 3y + 5 = 0 ............. (1)

The equation of the line perpendicular to the line (1) is

$\sf{3x + 2y = k} \: \: \: .........(2)$

Now we express the equation (2) in intercept form as below :

$\displaystyle \sf{ \frac{3x}{k} + \frac{2y}{k} = 1 }$

$\implies\displaystyle \sf{ \frac{x}{ \frac{k}{3} } + \frac{y}{ \frac{k}{2} } = 1 }$

Now by the given condition

$\displaystyle \sf{ \frac{k}{2} = \frac{3}{4} }$

$\implies\displaystyle \sf{ {k} = \frac{3}{2} }$

Hence the required equation of the line is

$\displaystyle \sf{ 3x + 2y = \frac{3}{2} }$

$\implies\displaystyle \sf{ 6x + 4y = 3 }$

FINAL ANSWER

The required equation of the line is

$\boxed{\displaystyle \sf{ \: \: 6x + 4y = 3 } \: \: }$

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