Answered

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

Sagot :

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

[tex]\begin{gathered} y=x-3 \\ x-y-3=0 \\ (a=1,b=-1,c=-3) \end{gathered}[/tex]

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

[tex]\begin{gathered} d=\frac{|ax1+by1+c|}{\sqrt[]{a^2+b^2^{}}} \\ (2,3)\colon \\ d=\frac{|1\cdot2+(-1)\cdot3+(-3)|}{\sqrt[]{1^2+(-1)^2}} \\ d=\frac{|2-3-3|}{\sqrt[]{2}} \\ d=\frac{|-4|}{\sqrt[]{2}}=\frac{4}{\sqrt[]{2}}=\frac{4\sqrt[]{2}}{2}=2\sqrt[]{2}=2.83 \end{gathered}[/tex]

So the distance is 2.83.

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