Question

write down the equation of the line AB through (3,2) and perpendicular to the line 2y=3x+5​

Answer 1

Given ,

The straight line AB passing through (3,2) is perpendicular to the line 3x - 2y + 5 = 0

We know that , if two lines are perpendicular to each other , then

$\boxed{ \tt{ m_{1} \times m_{2} = - 1}}$

and if the equation of straight line is ax + by + c = 0 then

$\boxed{ \tt{Slope \: (m) = - \frac{A}{B} }}$

And if equation of the line whose slope m is passing through (x,y) , then

$\boxed{ \tt{Slope \: (m) = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} }}$

Thus ,

Slope of the line 3x - 2y + 5 = 0 will be

$\tt \implies m_{1} = - \frac{3}{( - 2)} = \frac{3}{2}$

Now ,

$\tt \implies \frac{3}{2} \times m_{2} = - 1$

$\tt \implies m_{2} = - \frac{2}{3}$

Since , the equation of the line AB passing through (3,2)

Thus ,

$\tt \implies - \frac{ 2}{3} = \frac{2 - y}{3 - x}$

$\tt \implies - 6 + 2x = 6 - 3y$

$\tt \implies2x + 3y = 0$

Therefore , the equation of the line AB is 2x + 3y = 0

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Answer 2

Given ,

The straight line AB passing through (3,2) is perpendicular to the line 3x - 2y + 5 = 0

We know that , if two lines are perpendicular to each other , then

$\boxed{ \tt{ m_{1} \times m_{2} = - 1}} </p><p>$

and if the equation of straight line is ax + by + c = 0 then

$\boxed{ \tt{Slope \: (m) = - \frac{A}{B} }} </p><p></p><p>$

And if equation of the line whose slope m is passing through (x,y) , then

$\boxed{ \tt{Slope \: (m) = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} }}$

Thus ,

Slope of the line 3x - 2y + 5 = 0 will be

$\tt \implies m_{1} = - \frac{3}{( - 2)} = \frac{3}{2}$

Now ,

$\tt \implies \frac{3}{2} \times m_{2} = - 1 \\ </p><p></p><p>\tt \implies m_{2} = - \frac{2}{3}</p><p>$

Since , the equation of the line AB passing through (3,2)

Thus ,

$\tt \implies - \frac{ 2}{3} = \frac{2 - y}{3 - x}</p><p> \\ </p><p></p><p>\tt \implies - 6 + 2x = 6 - 3y \\ </p><p></p><p>\tt \implies2x + 3y = 0$

Therefore , the equation of the line AB is 2x + 3y = 0

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