Q. 86 - Precalculus Enhanced with Graphing Utilities

Question

Find the point on the line $y=x+1$ that is closest to the point $\left(4,1\right).$

Step-by-Step Solution

Answer

The point on the line $y=x+1$closest to $\left(1,4\right)$ is $\left(2,3\right).$

Step 1. Given information

We need to determine the point on the line $y=x+1$ that is closest to the point $\left(4,1\right).$ We need to express the distance $d$ from the point to the line as a function of $x$ and then find the minimum value of  $\left[d{\left(x\right)}^2\right].$

Step 2. Simplify

Let the line be perpendicular to the line $y=x+1$and $y=mx+c.$

Since the perpendicular lines have an inversely proportional slope.

So, $m=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1$}\right..$

$m=-1.$

The line passing through the point $\left(4,1\right)$ and having the slope $-1$ is,

$y-1=-1\left(x-4\right).$

$y=-x+5.$

Now put the point $\left(x,y\right)$ that is $\left(x,x+1\right)$ in $y=-x+5.$

$-x+1=-x+5.$

$2x=4.$

$x=2.$

The point on the line $y=x+1$is closest to $\left(1,4\right)$ is $\left(2,3\right).$