# Find the point on the line 4x + 7y - 3 = 0 which is closest to the point (-3,1) .

## Question:

Find the point on the line 4x + 7y - 3 = 0 which is closest to the point (-3,1) .

## Closest Point :

Closest point from a point to any line is the point where the perpendicular from the given point meets the given line.

So the point of intersection of the perpendicular from point gives shortest distance.

Given line {eq}\displaystyle 4x+7y-3=0 {/eq}

Slope of line for {eq}\displaystyle ax+by+c=0 {/eq} is {eq}\displaystyle \frac{-a}{b} {/eq}

So slope of {eq}\displaystyle 4x+7y-3=0 {/eq} is {eq}\displaystyle \frac{-4}{7} {/eq}

Consider a parametric point of line that is

Let {eq}\displaystyle x=t {/eq} then {eq}\displaystyle 4t+7y-3=0 \Rightarrow y=\frac{3-4t}{7} {/eq}

So slope of two points {eq}\displaystyle (-3,1) {/eq} and {eq}\displaystyle (t,\frac{3-4t}{7}) {/eq} is {eq}\displaystyle \frac{\frac{3-4t}{7}-1}{t+3} {/eq}

Simplifying we get

Slope = {eq}\displaystyle \frac{-4t-4}{7(t+3)} {/eq}

Since line and perpendicular from {eq}\displaystyle (-3,1) {/eq} should to perpendicular to have shortest distance.

That gives {eq}\displaystyle \frac{-4(t+1)}{7(t+3)} \left( \frac{-4}{7} \right) =-1 {/eq}

Simplifying we get

{eq}\displaystyle 16t+16 =-49t-147 \Rightarrow 65t=-163 {/eq}

That gives {eq}\displaystyle t=2.5 {/eq}

So point is {eq}\displaystyle \left( 2.5, \frac{3-4(2.5)}{7} \right) {/eq}

Closest point is {eq}\displaystyle \left( frac{5}{2} ,-1 \right) {/eq} 