Question

Determine the shortest distance from the point D(5, 4) to the line represented by 3x+5y-4 =0

Answer 1

Given Question :-

• Determine the shortest distance from the point D(5, 4) to the line represented by 3x+5y-4 =0.

Answer

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a \: line \: 3x + 5y - 4 = 0} \\ &\sf{a \: point \: (5, 4)} \end{cases}\end{gathered}\end{gathered}$

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

Concept Used :-

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

$\tt \:d \: = \: \dfrac{ |ap \: + \: bq \: + \: c| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

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

Given

A line 3x + 5y - 4 = 0 and point (5, 4).

Let d be the shortest distance between 3x + 5y - 4 = 0 and a point (5, 4).

Then,

$\bf \: d \: = \sf \: \dfrac{ |3 \times 5 + 5 \times 4 - 4| }{ \sqrt{ {3}^{2} + {5}^{2} } }$

$: \implies \bf \: d \: = \sf \: \dfrac{ |15 + 20 - 4| }{ \sqrt{9 + 25} }$

$: \implies \bf \: d \: = \sf \: \dfrac{ | \: 31 \: | }{ \sqrt{34} }$

$\large \boxed{ \red{ \bf \: Hence \: \tt \: d \: = \dfrac{31}{ \sqrt{34} \: \: }units} }$