Question

What is the shortest distance from the line `8x-4y+7=0` to the point (1, 2)?

Answer 1

The shortest distance from a line `8x-4y+7=0` to a point `(1,2)` is `0.56` units

Explanation:

For a line `ax+by+c=0`, the shortest distance to a point `(x_1,y_1)` is given by

`|(ax_1+by_1+c)/sqrt(a^2+b^2)|`

Hence, the shortest distance from a line `8x-4y+7=0` to a point `(1,2)` is

`|(8*1-4*2+7)/sqrt(8^2+(-4)^2)|`

= `(8-10+7)/(4sqrt5)=5/(4sqrt5)=sqrt5/4=0.559~=0.56`

graph{(8x-4y+7)((x-1)^2+(y-2)^2-0.003)=0 [-2.5, 2.5, 0.89, 3.39]}