Question

Which is the distance between the point with the coordinates (-2, 3) and the line with the equation 6x-y=-3 HURRY

Answer 1

Given Question :-

• What is the distance between the point with the coordinates (-2, 3) and the line with the equation 6x - y = - 3.

Answer

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a \: line \: 6x - y = - 3} \\ &\sf{a \: point \: (-2, 3)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find  \begin{cases} &\sf{perpendicular \: distance}  \end{cases}\end{gathered}\end{gathered}$

Concept Used :

Let us consider a line ax + by + c = 0 and a point (p, q), then shortest or perpendicular distance is given by

$\boxed{ \red{ \bf \: d \: = \sf \: \dfrac{ |ap + bq + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } }}$

$\large\underline\purple{\bold{Solution :- }}$

Given

A line 6x - y + 3 = 0 and a point (-2, 3).

Let 'd' be the shortest or perpendicular distance between 6x - y + 3 = 0 and (-2, 3)

Then,

$\bf \: d \: = \sf \: \dfrac{ |6( - 2) - 3 + 3| }{ \sqrt{ {6}^{2} + {( - 1)}^{2} } }$

$: \implies \bf \: d \: = \sf \: \dfrac{ | - 12 - \cancel3 \: + \cancel3| }{ \sqrt{36 + 1} }$

$: \implies \bf \: d \: = \sf \: \dfrac{ | - 12| }{ \sqrt{37} }$

$: \implies \bf \: d \: = \sf \: \dfrac{ 12 }{ \sqrt{37} }$

$: \implies \large\boxed{ \pink{ \bf \: d \: = \rm \: \dfrac{ 12}{ \sqrt{37} \: }units \: }}$