Question

find the distance from point a to $overleftrightarrow{xz}$. round your answer to the nearest tenth. a(3, 3) z(4, -1) y(2, -1.5) x(-4, -3) the distance is about units.

Explanation:

Step1: Find the equation of line XZ

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. The slope $m$ between two points $X(-4,-3)$ and $Z(4,-1)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-1-(-3)}{4 - (-4)}=\frac{-1 + 3}{4 + 4}=\frac{2}{8}=\frac{1}{4}$.
Using the point - slope form $y - y_1=m(x - x_1)$ with the point $Z(4,-1)$:
$y+1=\frac{1}{4}(x - 4)$
$y+1=\frac{1}{4}x-1$
$y=\frac{1}{4}x-2$.

Step2: Use the distance formula from a point $(x_0,y_0)$ to a line $Ax+By + C = 0$

Rewrite the line $y=\frac{1}{4}x-2$ in the general form $x - 4y-8 = 0$.
The distance $d$ from the point $A(3,3)$ to the line $Ax+By + C = 0$ is given by the formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$.
Here, $A = 1$, $B=-4$, $C=-8$, $x_0 = 3$, and $y_0 = 3$.
$d=\frac{\vert1\times3+(-4)\times3+(-8)\vert}{\sqrt{1^2+(-4)^2}}=\frac{\vert3-12 - 8\vert}{\sqrt{1 + 16}}=\frac{\vert-17\vert}{\sqrt{17}}=\frac{17}{\sqrt{17}}=\sqrt{17}\approx4.1$.

Answer:

$4.1$