Question

Y=-x-3 and Q(2,3)Find the distance

Answer 1

First, let's put the line equation in the form ax + by + c = 0:

`\begin{gathered} y=x-3 \\ x-y-3=0 \\ (a=1,b=-1,c=-3) \end{gathered}`

Now, to find the distance to a point (x1, y1), let's use the formula:

`\begin{gathered} d=\fracax1+by1+c{\sqrt[]{a^2+b^2^{}}} \\ (2,3)\colon \\ d=\frac{\sqrt[]{1^2+(-1)^2}} \\ d=\frac2-3-3{\sqrt[]{2}} \\ d=\frac{\sqrt[]{2}}=\frac{4}{\sqrt[]{2}}=\frac{4\sqrt[]{2}}{2}=2\sqrt[]{2}=2.83 \end{gathered}`

So the distance is 2.83.