Question

find the distance from point a to line c. round your answers to the nearest tenth. 19. 20. 21.

Explanation:

Step1: Recall the distance formula

The distance $d$ from a point $(x_0,y_0)$ to a line $Ax + By+C = 0$ is given by $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$. First, find the equation of the line $c$ in the general - form $Ax + By + C=0$.

Step2: For the first graph (problem 19)

The line $c$ has two - point form using two points on the line, say $( - 2,0)$ and $(0,-2)$. The slope $m=\frac{-2 - 0}{0+2}=-1$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(0,-2)$, we have $y+2=-1(x - 0)$ or $x + y+2 = 0$. The point $A=(0,2)$. Substitute $x_0 = 0,y_0 = 2$ into the distance formula: $d=\frac{\vert0 + 2+2\vert}{\sqrt{1^2+1^2}}=\frac{\vert4\vert}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\approx2.8$.

Step3: For the second graph (problem 20)

The line $c$ has two points $(0,0)$ and $(2,2)$. The slope $m = 1$. Using the point - slope form with $(0,0)$, we get $y=x$ or $x - y=0$. The point $A=( - 1,3)$. Substitute $x_0=-1,y_0 = 3$ into the distance formula: $d=\frac{\vert-1-3\vert}{\sqrt{1^2+( - 1)^2}}=\frac{\vert-4\vert}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\approx2.8$.

Step4: For the third graph (problem 21)

The line $c$ has two points $(0,1)$ and $(2,3)$. The slope $m=\frac{3 - 1}{2 - 0}=1$. Using the point - slope form with $(0,1)$, we have $y-1=x$ or $x - y+1 = 0$. The point $A=( - 1,2)$. Substitute $x_0=-1,y_0 = 2$ into the distance formula: $d=\frac{\vert-1-2 + 1\vert}{\sqrt{1^2+( - 1)^2}}=\frac{\vert-2\vert}{\sqrt{2}}=\sqrt{2}\approx1.4$.

Answer:

1. $2.8$
2. $2.8$
3. $1.4$