Question

Find the perpendicular distance of the straight line 5x - 12y + 7 = 0 from the point
(3, 4).

Answer 1

GIVEN :–

• A Point (3,4) perpendicular to the line 5x - 12y + 7 = 0.

TO FIND :–

• Perpendicular Distance = ?

SOLUTION :–

• We know that if a point is $\bf (x_1 , y_1)$ and straight line ax + by = c , then Perpendicular Distance –

$\\ \implies \large { \boxed{\bf Perpendicular \: \: Distance = \bigg| \dfrac{ax_1 + by_1 - c}{ \sqrt{ {a}^{2} + {b}^{2} }} \bigg|}}$

• Now put the values –

$\\ \implies \bf Distance = \bigg| \dfrac{5(3)- 12(4) + 7}{ \sqrt{ {(5)}^{2} + {(12)}^{2}}}\bigg|$

$\\ \implies \bf Distance = \bigg|\dfrac{15-48+ 7}{ \sqrt{ {(5)}^{2} + {(12)}^{2}}}\bigg|$

$\\ \implies \bf Distance = \bigg|\dfrac{15-48+ 7}{ \sqrt{25+144}}\bigg|$

$\\ \implies \bf Distance = \bigg|\dfrac{22-48}{ \sqrt{169}}\bigg|$

$\\ \implies \bf Distance = \bigg|\dfrac{- 26}{ \sqrt{169}}\bigg|$

$\\ \implies \bf Distance =\bigg| \dfrac{26}{ \sqrt{169}}\bigg|$

$\\ \implies \bf Distance = \cancel \dfrac{26}{13}$

$\\ \implies \large{ \boxed{ \bf Distance =2 \: \: unit}}$

Answer 2

Given

• A point ( 3 , 4 ) is perpendicular to line 5x - 12y - 7 = 0.

Required to Find

Perpendicular Distance = ?

Solution

Distance from ( x , y ) to the line ax + by + c = 0

$\large \sf \: d = \dfrac{l \: ax_{1} + \: by_{1} + cl }{ \sqrt{ {a}^{2} + {b}^{2} } }$

Therefore,

$\\ \large \sf \: d = \dfrac{15.3 - 124 + 71}{ \sqrt{ {5}^{2} + {12}^{2} } }$

$\\ \sf \large \mapsto \dfrac{l15 - 48 + 71}{ \sqrt{169} }$

$\\ \large\mapsto \sf \dfrac{1 - 261}{13}$

$\\ \large \mapsto \sf \dfrac{26}{13} = 2$

$\\ \color{red} \sf \large \mapsto \: d = 2 unit$