Find the point on the line y=2x+3 that is closest to the origin.

This problem aims to find a point that is closest to the origin. A linear equation is given, which is just a simple line in the xy-plane. The closest point from the origin will be the vertical distance from the origin to that line. For this, we need to be familiar with the distance formula between two points and the derivatives.

The distance from a line to a point is the smallest distance from a point to any arbitrary point on a straight line. As discussed above, it is the perpendicular distance of the point to that line.

We need to figure out an equation of the perpendicular from (0,0) on y = 2x + 3. This equation is of the slope intercept form i.e. y = mx + c.

Expert Answer

Let’s assume $P$ to be the point that is on the line $y=2x+3$ and closest to the origin.

Suppose the $x$-coordinate of $P$ is $x$ and $y$-coordinate is $2x+3$. So the point is $(x,2x+3)$.

We have to find the distance of point $P(x,2x+3)$ to the origin $(0,0)$.

Distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given as:

$$D=\sqrt{(x_1+x_2{)}^2+(y_1+y_2{)}^2}$$

Solving it for $(0,0)$ and $(x,2x+3)$:

$$D=\sqrt{(x-0{)}^2+(2x+3-0{)}^2}$$

$$=\sqrt{x^2+(2x+3{)}^2}$$

We have to minimize the $x$ to find the minimal distance from point $P$ to the origin.

Now let:

$$f(x)=\sqrt{x^2+(2x+3{)}^2}$$

We have to find the $x$ that makes $f(x)$ smallest by usual derivative process.

If we minimize $x^2+(2x+3{)}^2$, it will automatically minimize the $\sqrt{x^2+(2x+3{)}^2}$ so assuming $x^2+(2x+3{)}^2$ to be $g(x)$ and minimizing it.

$$g(x)=x^2+(2x+3{)}^2$$

$$g(x)=x^2+4x^2+9+12x$$

$$g(x)=5x^2+12x+9$$

To find the minimum lets take the derivative of $g(x)$ and put it equals to $0$.

$$g^{\prime }(x)=10x+12$$

$$0=10x+12$$

$x$ comes out to be:

$$x={\displaystyle \frac{-6}{5}}$$

Now put $x$ into the point $P$.

$$P=(x,2x+3)$$

$$=({\displaystyle \frac{-6}{5}},2({\displaystyle \frac{-6}{5}})+3)$$

Point $P$ comes out to be:

$$P=({\displaystyle \frac{-6}{5}},{\displaystyle \frac{3}{5}})$$

Numerical Result

$({\displaystyle \frac{-6}{5}},{\displaystyle \frac{3}{5}})$ is the point on the line $y=2x+3$ that is closest to the origin.

Example

Let’s assume $P$ to be the point is $(x,4x+5)$.

We have to find the distance of point $P(x,4x+5)$ to the origin $(0,0)$.

$$D=\sqrt{x^2+(4x+5{)}^2}$$

Now let:

$$f(x)=\sqrt{x^2+(4x+5{)}^2}$$

We have to find the $x$ that makes $f(x)$ smallest by usual derivative process.

Let’s assume,

$$g(x)=x^2+(4x+5{)}^2$$

$$g(x)=x^2+16x^2+25+40x$$

$$g(x)=17x^2+40x+25$$

To find the minimum lets take the derivative of $g(x)$ and put it equals to $0$.

$$g^{\prime }(x)=34x+40$$

$$0=34x+40$$

$x$ comes out to be:

$$x={\displaystyle \frac{-20}{17}}$$

Now put $x$ into the point $P$.

$$P=(x,4x+5)$$

Point $P$ comes out to be:

$$P=({\displaystyle \frac{-20}{17}},{\displaystyle \frac{5}{17}})$$

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