Equation of line in Straight lines

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Let the perpendiculars from any point on the line $2 x + 11 y = 5$ upon the lines $24 x + 7 y - 20 = 0$ and $4 x - 3 y - 2 = 0$ have the lengths $p_{1}$ and $p_{2}$ respectively. then, $p_{1}$ - $p_{2}$ is equal to

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Solution

Let $(t, \frac{5 - 2 t}{11})$ be a point on the line $2 x + 11 y = 5$ .
Then,
$p_{1} = |\frac{24 t + 7 (\frac{5 - 2 t}{11}) - 20}{\sqrt{24^{2} + 7^{2}}}| = \frac{| 50 t - 37 |}{55}$
and $p_{2} = |\frac{4 t - 3 (\begin{matrix} 5 - 2 t \\ 11 \end{matrix}) - 2}{\sqrt{4^{2} + (- 3)^{2}}}| = \frac{| 50 t - 37 |}{55}$
Clearly, we have $p_{1}$ = $p_{2}$
$A LITER$ : Clearly, $2 x + 11 y = 5$ is the angle bisector of the two
lines. Therefore $p_{1}$ = $p_{2}$

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Equation of line in Straight lines

Remember concepts with our Masterclasses.

Let the perpendiculars from any point on the line 2x+11y=5 upon the lines 24x+7y−20=0 and 4x−3y−2=0 have the lengths p1 and p2 respectively. then, p1-p2 is equal to

Let t,5−2t11 be a point on the line 2x+11y=5.Then,p1=24t+75−2t11−20242+72=|50t−37|55and p2=4t−35−2t11−242+(−3)2=|50t−37|55Clearly, we have p1=p2ALITER: Clearly, 2x+11y=5 is the angle bisector of the two lines. Therefore p1=p2

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Let the perpendiculars from any point on the line $2 x + 11 y = 5$ upon the lines $24 x + 7 y - 20 = 0$ and $4 x - 3 y - 2 = 0$ have the lengths $p_{1}$ and $p_{2}$ respectively. then, $p_{1}$ - $p_{2}$ is equal to