Question

what is the shortest distance between the line y = 3x + 5 and the point at (0, 0)? round to the nearest tenth.
a) 3.9
b) 2.5
c) 2.0
d) 1.6

Explanation:

Step1: Recall the distance formula from a point \((x_0,y_0)\) to the line \(Ax + By+C = 0\)

The formula for the distance \(d\) from a point \((x_0,y_0)\) to the line \(Ax+By + C=0\) is \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\).
First, rewrite the line \(y = 3x+5\) in the general form \(3x - y+5 = 0\). Here \(A = 3\), \(B=-1\), \(C = 5\), and the point \((x_0,y_0)=(0,0)\).

Step2: Substitute the values into the formula

Substitute \(x_0 = 0\), \(y_0 = 0\), \(A = 3\), \(B=-1\), and \(C = 5\) into the formula:

$$
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Wait, there is a mistake. Let's start over.
The formula for the distance \(d\) from a point \((x_0,y_0)\) to the line \(Ax+By + C = 0\) is \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\).
Rewrite the line \(y=3x + 5\) as \(3x-y-5=0\). Here \(A = 3\), \(B=-1\), \(C=-5\), and the point \((x_0,y_0)=(0,0)\)

$$
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Another way:
The formula for the distance from a point \((x_0,y_0)\) to the line \(y=mx + c\) is \(d=\frac{\vert mx_0 - y_0 + c\vert}{\sqrt{m^{2}+1}}\)
Here \(m = 3\), \(c = 5\), \(x_0=0\), \(y_0 = 0\)

$$
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Wait, no. The formula for the distance from \((x_0,y_0)\) to \(y=mx + b\) is \(d=\frac{\vert mx_0 - y_0 + b\vert}{\sqrt{m^{2}+1}}\)
If the line is \(y = 3x+5\) (i.e., \(m = 3\), \(b = 5\)) and the point \((0,0)\)

$$
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Answer:

D. 1.6