Question

find the distance from point x to yz. if necessary, round your answer to the nearest tenth place
y(-4,3)
x(2,1)
z(1,-2)

Explanation:

Step1: Find the equation of line YZ

The slope $m$ of the line passing through $Y(-4,3)$ and $Z(1, - 2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 3}{1-(-4)}=\frac{-5}{5}=-1$.
Using the point - slope form $y - y_1=m(x - x_1)$ with the point $Z(1,-2)$, we get $y+2=-(x - 1)$, which simplifies to $x+y+1 = 0$.

Step2: Use the distance formula from a point to a line

The distance $d$ from the point $X(2,1)$ to the line $Ax+By + C=0$ (here $A = 1$, $B = 1$, $C = 1$) is given by the formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$.
Substitute $x_0 = 2$, $y_0=1$ into the formula: $d=\frac{\vert1\times2 + 1\times1+1\vert}{\sqrt{1^2+1^2}}=\frac{\vert2 + 1+1\vert}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\approx2.8$.

Answer:

$2.8$ units