Question

Find the equation of the line which is at right angles to 3x +4y = 12, such that its perpendicular distance from the origin is equal to the length of the perpendicular from (3,2) on the given line. How to solve this?​

Answer 1

SOLUTION

TO DETERMINE

The equation of the line which is at right angles to 3x + 4y = 12, such that its perpendicular distance from the origin is equal to the length of the perpendicular from (3,2) on the given line.

EVALUATION

Here the given equation of the line is

$\sf{3x + 4y = 12} \: \: \: \: - - - - - (1)$

Now the equation of the line perpendicular to the given line is

$\sf{4x - 3y = k \: \: \: \: \: - - - - (2)}$

Now perpendicular distance of the line 2 from the origin is

$\displaystyle \sf{ = \bigg| \frac{0 + 0 - k}{ \sqrt{ {4}^{2} + {3}^{2} } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k}{ \sqrt{16 + 9} } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k}{5} \bigg| }$

Again length of perpendicular from (3,2) to the line 1

$\displaystyle \sf{ = \bigg| \frac{(3 \times 3 )+ (4 \times 2) -12 }{ \sqrt{ {3}^{2} + {4}^{2} } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{9 + 8 - 12 }{ \sqrt{ 16 + 9 } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{5}{5} \bigg| }$

$= 1$

So by the given condition

$\displaystyle \sf{ \bigg| \frac{k}{5} \bigg| = 1}$

$\displaystyle \sf{ \implies \: \frac{k}{5} = \pm \: 1}$

$\displaystyle \sf{ \implies \: k = \pm \: 5}$

Hence the required equation of the line is

$\displaystyle \sf{ 4x - 3y = \pm \: 5}$

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