Question

Activity 2. Find the equation of the line using Two-Point form,

Answer 1

Answer:

1) equation of line is: y-3=0

2) equation of line is: $\mathbf{y-3=-\frac{1}{2}(x-4)}$

3) equation of line is: $\mathbf{y+3=-\frac{4}{3}(x-3}$

4) equation of line is: $\mathbf{y-2=-2(x-2)}$

Step-by-step explanation:

We need to find the equation of the line using Two-Point form.

The general equation of two-point form is: $y-y_1=m(x-x_1)$ where m is slope.

The formula used to calculate slope is: $Slope=\frac{y_2-y_1}{x_2-x_1}$

1. (1,3) and (-2,3)

First finding slope

We have: $x_1=1, y_1=3, x_2=-2, y_2=3$

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{3-3}{-2-1}\\Slope=\frac{0}{-3}\\Slope=0$

So, equation of line will be:

Using slope m=0 and point (1,3)

$y-y_1=m(x-x_1)\\y-3=0(x-1)\\y-3=0$

So, equation of line is: y-3=0

2. (4,3) and (6,2)

First finding slope

We have:$x_1=4, y_1=3, x_2=6, y_2=2$

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-3}{6-4}\\Slope=\frac{-1}{2}$

So, equation of line will be:

Using slope m=$\frac{-1}{2}$ and point (4,3)

$y-y_1=m(x-x_1)\\y-3=\frac{-1}{2}(x-4)\\y-3=-\frac{1}{2}(x-4)$

So, equation of line is: $\mathbf{y-3=-\frac{1}{2}(x-4)}$

3) (3,-3) and (0,1)

First finding slope

We have: $x_1=3, y_1=-3, x_2=0, y_2=1$

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{1-(-3)}{0-3}\\Slope=\frac{1+3}{-3}\\Slope=-\frac{4}{3}$

So, equation of line will be:

Using slope m=$-\frac{4}{3}$ and point (3,-3)

$y-y_1=m(x-x_1)\\y-(-3)=-\frac{4}{3}(x-3)\\y+3=-\frac{4}{3}(x-3)$

So, equation of line is: $\mathbf{y+3=-\frac{4}{3}(x-3)}$

4) (2,2) and (4,-2)

First finding slope

We have: $x_1=2, y_1=2, x_2=4, y_2=-2$

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{-2-2}{4-2}\\Slope=\frac{-4}{2}\\Slope=-2$

So, equation of line will be:

Using slope m=-2 and point (2,2)

$y-y_1=-2(x-x_1)\\y-2=-2(x-2)$

So, equation of line is: $\mathbf{y-2=-2(x-2)}$