Question

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

Answer 1

y=-7/2x+1/2
1. find slope
2. plug in slope to y=mx+b
3. take point and plug it in (find a b)
4. plug in slope and b to get equation

Answer 2

`\text{The equation is }(14)/(4)x+y=1.05`

Explanation:

Given that line JK passes through points J(–3, 11) and K(1, –3).

we have to find the equation of JK in standard form.

`\text{The slope of line joining the points }(x_1,y_1)\text{ and }(x_2, y_2)\text{ is given by}`

`m=(y_2-y_1)/(x_2-x_1)`

Line JK passes through points J(–3, 11) and K(1, –3).

`m=(-3-11)/(1-(-3))`

`m=(-14)/(4)`

Put in slope intercept form

`y=mx+b`

`11=(-14)/(4)(-3)+b`

`11* (4)/(42)=b`

`b=1.05`

The equation is

`y=(-14)/(4)x+1.05`

`(14)/(4)x+y=1.05`

which is required form.