Question

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

Answer 1

1.
the slope of the line y=3x+2 is the coefficient of x, that is 3.

2.
the slope of any line perpendicular to the line y=3x+2, is m, such that:

$3*m=-1$

so $m= \frac{-1}{3}$

3. 
so any line with equation y=-1/3x+k is perpendicular to y=3x+2.

moreover, the line y=-1/3x (so k=0) is the line passing through the origin
(0, 0) and perpendicular to y=3x+2


4. This means that the closest point of y=3x+2 to the origin is the intersection point of lines y=3x+2 and y=-1/3x

5.
to find this point:

3x+2=(-1/3)x
x(3+1/3)=-2
x(9/3+1/3)=-2

x*10/3=-2

$x= \frac{-2*3}{10} = \frac{-6}{10}=-0.6$


thus y is :  (-1/3)(-6/10)=0.2



Answer: (-0.6, 0.2)

Answer 2

The point on the line $y=3x+2$ that is closest to the origin is $(-0.6, 0.2)$.

Given,  $y=3x+2$, whose slope is $3$.

Let $m$ is the slope of the line perpendicular to the above line.

So,

$3\times m=-1\\m=-1/3$

Then, the general equation of a line perpendicular line will be $y=-1/3x+k$.

So the equation line passing through the origin $(0,0)$ and perpendicular to the line $y=3x+2$ is $y=-1/3x$ , $(k=0)$.

Then the closest point will be the intersection point of these two lines, i.e.

$3x+2=-1/3x\\9x+6=-x\\10x+6=0\\x=-0.6$

and

$y=\dfrac{-0.6}{3} \\y=0.2$

Therefore, the point on the line $y=3x+2$ that is closest to the origin is $(-0.6, 0.2)$.

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