line t has equation y = x - 3. find the distance between t and the point s(2, -6). round your answer to the nearest tenth.

Question

Accepted Answer

# Explanation:
## Step1: Convert the line equation to standard form
The line equation $y=x - 3$ can be rewritten as $x-y-3 = 0$. Here $A = 1$, $B=-1$, $C=-3$. The point $S(2,-6)$ has $x_0 = 2$ and $y_0=-6$.
## Step2: Use the distance formula
The distance $d$ from a point $(x_0,y_0)$ to a line $Ax+By + C=0$ is given by the formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}$.
Substitute $A = 1$, $B=-1$, $C=-3$, $x_0 = 2$, $y_0=-6$ into the formula:
$$
\begin{align*}
d&=\frac{\vert1\times2+(-1)\times(-6)-3\vert}{\sqrt{1^{2}+(-1)^{2}}}\
&=\frac{\vert2 + 6-3\vert}{\sqrt{1 + 1}}\
&=\frac{\vert5\vert}{\sqrt{2}}\
&=\frac{5}{\sqrt{2}}\
&=\frac{5\sqrt{2}}{2}\approx3.5
\end{align*}
$$

# Answer:
$3.5$

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QUESTION IMAGE

Question

Explanation:

Step1: Convert the line equation to standard form

Step2: Use the distance formula

Answer: