Question

What is the equation of the line that passes through (`5, -2)` and `(3, 4)`?

Answer 1

`y=-3x+13`

Explanation:

`"the equation of a line in "color(blue)"slope-intercept form"` is.

`•color(white)(x)y=mx+c`

`"where m is the slope and c the y-intercept"`

`"to calculate m use the "color(blue)"gradient formula"`

`•color(white)(x)m=(y_2-y_1)/(x_2-x_1)`

`"let "(x_1,y_1)=(5,-2)" and "(x_2,y_2)=(3,4)`

`m=(4-(-2))/(3-5)=6/(-2)=-3`

`y=-x+clarrcolor(blue)"is the partial equation"`

`"to find c substitute either of the 2 given points into"`
`"the partial equation"`

`"using "(3,4)" then"`

`4=-9+crArrc=4+9=13`

`y=-3x+13larrcolor(red)"is the equation of the line"`

Answer 2

The eqn. of `"line passes through "(5,-2) and (3,4)` is

`3x+y-13=0`

Explanation:

The eqn. of line passes through `A(x_1,y_1) and B(x_2,y_2)` is

`|(x,y,1),(x_1,y_1,1),(x_2,y_2,1)| =0`

We have , two points :`A(5,-2) and B(3,4)`

So, the eqn. of `"line passes through "A and B` is

`|(x,y,1),(5,-2,1),(3,4,1)| =0`

Expanding we get

`x(-2-4)-y(5-3)+1(20+6)=0`

`:.x(-6)-y(2)+1(26)=0`

`:.-6x-2y+26=0`

Divding each term by `(-2)`

`:.3x+y-13=0`

`:.`The eqn. of `"line passes through "(5,-2) and (3,4)` is

`3x+y-13=0`

Answer 3

`y = -3x+13`

Explanation:

There is a useful formula which can be used to find the equation of a line if two points are known. It is based on the slope formula.

`(y-y_1)/(x-x_1) = (y_2-y_1)/(x_2-x_1)`

The points are `(5,-2) and (3,4)`

`(y-(-2))/(x-5) = (4-(-2))/(3-5)`

`(y+2)/(x-5) = (4+2)/(3-5) =color(blue)( 6/-2 = -3/1)" "larr` (this is the slope)

`(y+2)/(x-5) = -3/1`

`y+2 = -3(x-5)`

`y = -3x+15-2`

`y = -3x+13`