Question

Line 1 thru (3,2) and (5,-1)

Answer 1

see explanation

Explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slopw and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (3, 2) and (x₂, y₂ ) = (5, - 1)

m =
`(-1-2)/(5-3)` = -
`(3)/(2)`, thus

y = -
`(3)/(2)` x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (3, 2), then

2 = -
`(9)/(2)` + c ⇒ c = 2 +
`(9)/(2)` =
`(13)/(2)`

y = -
`(3)/(2)` x +
`(13)/(2)` ← equation of line

Answer 2

The equation of the line passing through the given points is

`y=-(3)/(2)x+(13)/(2)`

Explanation:

GIven two points are (3,2) and (5,-1)

To find the equation of the line passing through these two points

Let
`(x_(1),y_(1))` and
`(x_(2),y_(2))` be the two points (3,2) and (5,-1) respectively

Using the two points formula for finding slope:

`m=(y_(2)-y_(1))/(x_(2)-x_(1))`

`m=(-1-2)/(5-3)`

`m=(-3)/(2)`

Therefore
`m=-(3)/(2)`

By using formula:

`y=mx+c`

Here Let (x,y) be (3,2)

`y=mx+c`

`2=-{(3)/(2)}* 3+c`

`2=(-9)/(2)+c`

`c=(9)/(2)+2`

`c=(9+4)/(2)`

`c=(13)/(2)`

Therefore substitute values of m and c in

`y=mx+c`

`y=-(3)/(2)x+(13)/(2)`

Therefore the equation of the line passing through the given points is

`y=-(3)/(2)x+(13)/(2)`