# Find the point on the line y = 3x +

Find the point on the line y = 3x + 4 that is closest to the origin.
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Susan Yang

Let's consider an arbitrary point p(x,y)
The distance from P to the origin is:
$d=\sqrt{{x}^{2}+{y}^{2}}$
we notice that this the objective function. since p is on the line $y=3x+4$ the coordinates of P satisties this equation. This is the constrain for the objective function in (1).
Therefore, the objective function can be written as a function of one variable:
$d=d\left(x\right)=\sqrt{{x}^{2}+{\left(3x+4\right)}^{2}}$
From the first devirative tes, we have that
${d}^{\prime }\left(x\right)=0$
$⇒\frac{d}{2d\left(x\right)}\cdot \frac{d}{dx}\left({x}^{2}+{\left(3x+4\right)}^{2}\right)=0$
$⇒2x+2\left(3x+4\right)\left(3\right)=0$
$⇒2x+18x+24=0$
$⇒20x=-24$
$⇒x=\frac{-24}{20}=\frac{-6}{5}$
Since ${d}^{\prime }\left(x\right)<0$ for $x<\frac{-6}{5}$ and ${d}^{\prime }\left(x\right)>0$ for $x\succ \frac{6}{5}$, the minimum distance from p to the origin is achieved when $x=-\frac{6}{5}$
Then, $y=3x+4=3\left(-\frac{6}{5}\right)+y$
$⇒y=-\frac{18+20}{5}=\frac{2}{5}$
we found that, $P\left(-\frac{6}{5},\frac{2}{5}\right)$ is the point on the line $y=3x+4$ closest to the origin.