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e-lub
3 months ago
10

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:
Leona [12.6K]3 months ago
6 0
One equation is Y - 11 = -7/2(x + 3), while the other is y + 3 = -7/2(x - 1).
tester [12.3K]3 months ago
5 0

Answer:

7x + 2y = 1

Step-by-step explanation:

Greetings

Utilize the given points to calculate the slope (A)

A=\frac{-3-11}{1-(-3)} \\A= \frac{-14}{4} \\\\\\A=-\frac{7}{2} \\\\\\

With the slope and a chosen point, we can use J(-3, 11).

y-y_{0} = A (x-x_{0} )\\\\y-11=-\frac{7}{2} (x+3)\\y-11=-\frac{7x}{2} -\frac{21}{2} \\y+\frac{7x}{2} =-\frac{21}{2} +11\\\\y+\frac{7x}{2} =0.5\\\\\\

It should be multiplied by two to clear the fraction and then reorganized.

2*(y+\frac{7x}{2}) =2*0.5\\7x+2y =1

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