Question

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)​

Answer 1

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

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

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

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

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=(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=(y_2-y_1)/(x_2-x_1)\\Slope=(3-3)/(-2-1)\\Slope=(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=(y_2-y_1)/(x_2-x_1)\\Slope=(2-3)/(6-4)\\Slope=(-1)/(2)\\`

So, equation of line will be:

Using slope m=
`(-1)/(2)` and point (4,3)

`y-y_1=m(x-x_1)\\y-3=(-1)/(2)(x-4)\\y-3=-(1)/(2)(x-4)`

So, equation of line is:
`\mathbf{y-3=-(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=(y_2-y_1)/(x_2-x_1)\\Slope=(1-(-3))/(0-3)\\Slope=(1+3)/(-3)\\Slope=-(4)/(3)\\`

So, equation of line will be:

Using slope m=
`-(4)/(3)` and point (3,-3)

`y-y_1=m(x-x_1)\\y-(-3)=-(4)/(3)(x-3)\\y+3=-(4)/(3)(x-3)\\`

So, equation of line is:
`\mathbf{y+3=-(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=(y_2-y_1)/(x_2-x_1)\\Slope=(-2-2)/(4-2)\\Slope=(-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)}`