Question

Find the distance from the point (9, –2) to the line y = 3∕2x + 4. Choices are in the attachment...

Answer 1

Answer: Choice B)  $\sqrt{117} \text{ units}$

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Work Shown:

y = (3/2)x + 4

2y = 3x + 8 .... multiply both sides by 2

0 = 3x + 8 - 2y

3x-2y+8 = 0

The original equation transforms to 3x-2y+8 = 0. It is in the form Ax+By+C = 0. We see that A = 3, B = -2, C = 8. This form is very useful to help find the distance from a point to this line.

The formula we will use is

$d = \frac{|A*p+B*q+C|}{\sqrt{A^2+B^2}}$

where A,B,C were the values mentioned earlier. The p,q are the x and y coordinates of the point given. So p = 9 and q = -2

Plugging all that in gives...

$d = \frac{|A*p+B*q+C|}{\sqrt{A^2+B^2}}\\\\d = \frac{|3*9+(-2)*(-2)+8|}{\sqrt{3^2+(-2)^2}}\\\\d = \frac{|27+4+8|}{\sqrt{9+4}}\\\\d = \frac{|39|}{\sqrt{13}}\\\\d = \frac{39}{\sqrt{13}}\\\\d = \frac{39\sqrt{13}}{\sqrt{13}\sqrt{13}} \ \text{ rationalizing denominator}\\\\d = \frac{39\sqrt{13}}{13}\\\\d = 3\sqrt{13}\\\\d = \sqrt{9}*\sqrt{13}\\\\d = \sqrt{9*13}\\\\d = \sqrt{117}$