Question

Find the point on the line y = 5x + 4 that is closest to the origin. (x, y) = $− 10 13​, 2 13​‍

Answer 1

we know that

 the point on the line y = 5x + 4 that is closest to the origin will be on the perpendicular line through the origin,

the line 
y=5x+4---------> the slope is m1=5

if two lines are perpendicular 
m1*m2=-1-----------> m2=-1/m1

so
m2=-1/5
y = (-1/5)x

so will satisfy 
(-1/5)x = 5x +4 
(-26/5)x = 4 
x = 4*(-5/26) = -10/13 
y = (-1/5)x = 2/13


The answer is
the point  (-10/13, 2/13)

see the attached figure

Answer 2

The point on the line $y=5x+4$ that is closest to the origin is $\boxed{\left({ - \frac{{10}}{{13}},\frac{2}{{13}}} \right)}.$

Further explanation:

The formula for distance between the two points can be expressed as follows,

$\boxed{{\text{Distance}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }$

Given:

The line is $y = 5x + 4.$

Explanation:

The coordinate of the origin is $\left({0,0}\right).$

The first point is $\left( {x,y} \right)$ and the second point is $\left({0,0}\right).$

The distance between the two points can be calculated as follows,

$\begin{aligned}{\text{Distance}} &= \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}}  \\ &=\sqrt{{x^2}+{{\left( {5x + 4} \right)}^2}}\\&= \sqrt {{x^2} + 25{x^2}+ 40x + 16}\\ &= \sqrt {26{x^2} + 40x + 16}\\\end{aligned}$

Differentiate the above equation with respect to $x$.

$\begin{aligned}\frac{d}{{dx}}\left({{\text{Distance}}}\right)&=\frac{d}{{dx}}\left( {\sqrt {26{x^2} + 40x + 16} }\right)\\&=\frac{1}{{2\times\sqrt {26{x^2} + 40x + 16} }} \times \left( {52x + 40}\right)\\&=\frac{{\left( {52x + 40} \right)}}{{2 \times \sqrt {26{x^2} + 40x + 16} }}\\\end{aligned}$

Substitute the first derivative equal to zero.

$\begin{aligned}\frac{d}{{dx}}\left( {{\text{Distance}}} \right)&= 0\\\frac{{\left( {52x + 40} \right)}}{{2 \times \sqrt {26{x^2} + 40x + 16} }}&= 0 \\\left( {52x + 40} \right)&= 0\\52x&=- 40\\x &= \frac{{ - 40}}{{52}}\\x&= - \frac{{10}}{{13}}\\\end{aligned}$

Substitute $x=- \dfrac{{10}}{{13}}$ in equation $y=5x+4$ to obtain the value of $y$.

$\begin{aligned}y&= 5 \times \left({ - \frac{{10}}{{13}}} \right) + 4\\ &=- \frac{{50}}{{13}}+4\\&= \frac{{ - 50 + 52}}{{13}}\\&= \frac{2}{{13}}\\\end{aligned}$

The point on the line $y=5x+4$ that is closest to the origin is $\boxed{\left({ - \frac{{10}}{{13}},\frac{2}{{13}}} \right)}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Application of derivatives

Keywords: Derivative, attains, maximum, value of x, function, differentiate, minimum value, closest point, line, y=5x+4.