Question

find the distance of the point (3,-3) from the line 3x-4y-26=0.

plz give step by step and so that i can clear my doubt..
and plz those who don't know the answer don't waste my question..​

Answer 1

$\huge{\underline{\underline{\sf{\red{Answer}}}}}$

Given:

Point = P(3,-3)

Line = 3x-4y-26=0

Formula:

Perpendicular Distance = $\bigg| \dfrac{ax_1+by_1+c}{\sqrt{a^2+ b^2}} \bigg|$

$\huge{\underline{\tt{SoluTion}}}$

By putting the Values in formula

$\implies \bigg| \dfrac{3\times 3-4\times (-3) -26}{\sqrt{3^2+ 4^2}} \bigg| \\ \\ \implies \bigg| \dfrac{9+12 -26}{\sqrt{25}} \bigg| \\ \\ \implies \bigg| \dfrac{21-26}{5} \bigg| \\ \\ \implies \bigg| \dfrac{-5}{5} \bigg| \\ \\ \leadsto \huge{\boxed{1}}$

Answer 2

$\huge{\fbox{\fbox{\mathfrak{\red{answer}}}}}$

Point = P(3,-3)

Line = 3x-4y-26=0

Formula:

Perpendicular Distance = \bigg| \dfrac{ax_1+by_1+c}{\sqrt{a^2+ b^2}} \bigg|∣∣∣∣a2+b2ax1+by1+c∣∣∣∣

\huge{\underline{\tt{SoluTion}}}SoluTion

By putting the Values in formula

\begin{lgathered}\implies \bigg| \dfrac{3\times 3-4\times (-3) -26}{\sqrt{3^2+ 4^2}} \bigg| \\ \\ \implies \bigg| \dfrac{9+12 -26}{\sqrt{25}} \bigg| \\ \\ \implies \bigg| \dfrac{21-26}{5} \bigg| \\ \\ \implies \bigg| \dfrac{-5}{5} \bigg| \\ \\ \leadsto \huge{\boxed{1}}\end{lgathered}⟹∣∣∣∣32+423×3−4×(−3)−26∣∣∣∣⟹∣∣∣∣259+12−26∣∣∣∣⟹∣∣∣∣521−26∣∣∣∣⟹∣∣∣∣5−5∣∣∣∣

⇝1