The aim of this problem is to find a point that is nearest to the origin. We are given a linear equation which is only a straight line in the xy-plane. The nearest point from the origin will be the vertical distance from the origin to that line. For this, we need to be aware of the distance formula between two points and the derivation.
The nearest distance of a point to a line will be the smallest vertical distance from that point to any random point on the straight line. As concerned above, it is the perpendicular distance of the point to that line.
To solve this problem, we will have to figure out an equation of the perpendicular from (0,0) on y = 4x + 3. This equation is actually the slope intercept form i.e. y = mx + c.
Expert Answer
Let’s assume $P$ to be the point that is on the line $y = 4x+3$ and closest to the origin.
Suppose the $x$-coordinate of $P$ is $x$ and $y$-coordinate is $4x+3$. So the point is $(x, 4x+3)$.
We have to find the distance of point $P (x, 4x+3)$ to the origin $(0,0)$.
Distance formula between two points $(a, b)$ and $(c, d)$ is given as:
\[D=\sqrt{(a + b)^2+(c + d)^2 }\]
Solving it for $(0,0)$ and $(x, 4x+3)$:
\[D=\sqrt{(x-0)^2+(4x+3 -0)^2 }\]
\[=\sqrt{x^2+(4x+3)^2 }\]
We have to minimize the $x$ to find the minimal distance from point $P$ to the origin.
Now let:
\[f(x)=\sqrt{x^2 + (4x+3)^2 }\]
We have to find the $x$ that makes $f(x)$ minimum by implementing a derivation.
If we minimize $x^2 + (4x+3)^2$, it will automatically minimize the $\sqrt{x^2 + (4x+3)^2 }$ so assuming $x^2 + (4x+3)^2$ to be $g(x)$ and minimizing it.
\[g(x)=x^2 + (4x+3)^2\]
\[g(x)=x^2+16x^2+9+24x\]
\[g(x)=17x^2+24x+9\]
To find the minimum, let’s take the derivative of $g(x)$ and put it equals to $0$.
\[g'(x)=34x + 24\]
\[0 = 34x + 24\]
$x$ comes out to be:
\[x=\dfrac{-12}{17}\]
Now put $x$ into the point $P$.
\[P=(x, 4x+ 3)\]
\[=(\dfrac{-12}{17} , 2(\dfrac{-12}{17})+ 3)\]
Point $P$ comes out to be:
\[P=(\dfrac{-12}{17},\dfrac{27}{17})\]
Numerical Result
$(\dfrac{-12}{17},\dfrac{27}{17})$ is the point on the line $y = 4x+3$ that is closest to the origin.
Example
Find a point on a straight line $y = 4x + 1$ that is nearest to the origin.
Let’s assume $P$ to be the point $(x, 4x+1)$.
We have to find the smallest distance of point $P (x, 4x+1)$ from the origin $(0,0)$.
\[D=\sqrt{x^2 + (4x+1)^2 }\]
Now let,
\[f(x)=\sqrt{x^2 +(4x+1)^2 }\]
We have to find the $x$ that makes $f(x)$ minimum by the derivative process.
Let’s assume,
\[g(x)=x^2 + (4x+1)^2 \]
\[g(x)= x^2 + 16x^2+ 1 + 8x \]
\[g(x) = 17x^2 +8x + 1\]
Taking derivative of $g(x)$ and put it equals to $0$.
\[g'(x) = 34x + 8\]
\[0 = 34x + 8 \]
$x$ comes out to be:
\[x = \dfrac{-4}{17} \]
Now put $x$ into the point $P$.
\[P=(x, 4x+ 1) \]
Point $P$ comes out to be:
\[P=( \dfrac{-4}{17} , \dfrac{1}{17})\]