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vodomira [7]
3 years ago
12

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

Mathematics
2 answers:
iren [92.7K]3 years ago
8 0
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
Arisa [49]3 years ago
3 0

Answer:

\text{The equation is }\frac{14}{4}x+y=1.05

Step-by-step 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=\frac{y_2-y_1}{x_2-x_1}

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

m=\frac{-3-11}{1-(-3)}

m=\frac{-14}{4}

Put in slope intercept form

y=mx+b

11=\frac{-14}{4}(-3)+b

11\times \frac{4}{42}=b

b=1.05

The equation is

y=\frac{-14}{4}x+1.05

\frac{14}{4}x+y=1.05

which is required form.

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