How do you find the distance from the line with equation #y=9x-3# to the point at (-3,2)?

Question

Delilah Aguilar · Accepted Answer

The answer is #32/sqrt82# The distance from the point #(x_0,y_0)# to the line #ax+by+c=0# is #d=(∣ax_0+by_0+c∣)/(sqrt(a^2+b^2))# Here the line is #9x-y-3=0# and the point is #(-3,2)# #:. d=(∣-27-2-3∣)/sqrt(81+1)=32/sqrt82#

Oliver Atkinson · Answer

undefined To find the distance from the line with equation $ y = 9x - 3 $ to the point (-3, 2), you can use the formula for the distance between a point and a line.

The formula is:

$$ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $$

Where (x₁, y₁) is the point, and A, B, and C are the coefficients of the line equation in the form Ax + By + C = 0.

In this case, the line equation is given as $ y = 9x - 3 $, so the coefficients are A = 9, B = -1, and C = 3.

Substituting the values into the formula and plugging in the coordinates of the point (-3, 2), we get:

$$ \text{Distance} = \frac{|9(-3) + (-1)(2) + 3|}{\sqrt{9^2 + (-1)^2}} $$

$$ \text{Distance} = \frac{|-27 - 2 + 3|}{\sqrt{81 + 1}} $$

$$ \text{Distance} = \frac{|-26|}{\sqrt{82}} $$

$$ \text{Distance} = \frac{26}{\sqrt{82}} $$

$$ \text{Distance} \approx 2.868 $$

So, the distance from the line $ y = 9x - 3 $ to the point (-3, 2) is approximately 2.868 units.