Question

Find the perpendicular distance of the point (1,2) from the line 3x+4y-13=0

Answer 1

GIVEN :–

• A line 3x + 4y - 13 = 0 and a point (1,2).

TO FIND :–

• Perpendicular distance of the point (1,2) from the line 3x + 4y - 13 = 0 is = ?

SOLUTION :–

• We know that Perpendicular distance of the point (m , n) from the line ax + by + c = 0 is –

$\\ \: \: \: { \huge{\star}} \: \large \: { \red{ \boxed{ \bold{Distance = \dfrac{ |a(m) + b(n) + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } }}}}$

• Here –

$\\ \: \: \: \: \: \: \: { \huge{.}} \: { \bold{ \: \: \: m = 1}}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: { \bold{ \: \: \: n = 2}}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: { \bold{ \: \: \: a = 3}}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: { \bold{ \: \: \: b = 4}}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: { \bold{ \: \: \: c = - 13}}$

• Now put the values –

$\\ \implies { \bold{Distance = \dfrac{ |3(1) + 4(2) - 13| }{ \sqrt{ {3}^{2} + {4}^{2} } } }}$

$\\ \implies { \bold{Distance = \dfrac{ |3 +8 - 13| }{ \sqrt{ 9 + 16 } } }}$

$\\ \implies { \bold{Distance = \dfrac{ |11 - 13| }{ \sqrt{ 9 + 16 } } }}$

$\\ \implies { \bold{Distance = \dfrac{ | - 2| }{ \sqrt{ 25 } } }}$

$\\ \implies \large { \green{ \boxed{ \bold{Distance = \dfrac{ 2 }{ 5 } } }}}$