Activity 2. Find the equation of the line using Two-Point form,
1. (1,3) and (-2,3)
2.(4,3) and (6,2)
3. (3,-3) and (0,-1)
4.(2, 2) and 4,-2)​

Respuesta :

Answer:

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

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

3) equation of line is: [tex]\mathbf{y+3=-\frac{4}{3}(x-3}[/tex]

4) equation of line is: [tex]\mathbf{y-2=-2(x-2)}[/tex]

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: [tex]y-y_1=m(x-x_1)[/tex] where m is slope.

The formula used to calculate slope is: [tex]Slope=\frac{y_2-y_1}{x_2-x_1}[/tex]

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

First finding slope

We have: [tex]x_1=1, y_1=3, x_2=-2, y_2=3[/tex]

[tex]Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{3-3}{-2-1}\\Slope=\frac{0}{-3}\\Slope=0\\[/tex]

So, equation of line will be:

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

[tex]y-y_1=m(x-x_1)\\y-3=0(x-1)\\y-3=0\\[/tex]

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

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

First finding slope

We have:[tex]x_1=4, y_1=3, x_2=6, y_2=2[/tex]

[tex]Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-3}{6-4}\\Slope=\frac{-1}{2}\\[/tex]

So, equation of line will be:

Using slope m=[tex]\frac{-1}{2}[/tex] and point (4,3)

[tex]y-y_1=m(x-x_1)\\y-3=\frac{-1}{2}(x-4)\\y-3=-\frac{1}{2}(x-4)[/tex]

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

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

First finding slope

We have: [tex]x_1=3, y_1=-3, x_2=0, y_2=1[/tex]

[tex]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}\\[/tex]

So, equation of line will be:

Using slope m=[tex]-\frac{4}{3}[/tex] and point (3,-3)

[tex]y-y_1=m(x-x_1)\\y-(-3)=-\frac{4}{3}(x-3)\\y+3=-\frac{4}{3}(x-3)\\[/tex]

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

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

First finding slope

We have: [tex]x_1=2, y_1=2, x_2=4, y_2=-2[/tex]

[tex]Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{-2-2}{4-2}\\Slope=\frac{-4}{2}\\Slope=-2\\[/tex]

So, equation of line will be:

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

[tex]y-y_1=-2(x-x_1)\\y-2=-2(x-2)\\[/tex]

So, equation of line is: [tex]\mathbf{y-2=-2(x-2)}[/tex]