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

User Prettyfly
by
8.2k points

2 Answers

2 votes
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
User Roi Tal
by
8.4k points
7 votes

Answer:


\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.

User Musab Bozkurt
by
8.1k points

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