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iogann1982 [59]
3 years ago
13

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

Mathematics
1 answer:
Allushta [10]3 years ago
3 0
<span>J(–3, 11) and K(1, –3)
Slope = (11 + 3)/(-3 - 1) = 14/-4 = -7/2

Equation
y + 3 = -7/2(x - 1)
2y + 6 = -7x + 7
2y = -7x + 1
7x + 2y = 1 ....This is standard form (Ax  + By = C)

Hope it helps.</span>
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Step-1 : Multiply the coefficient of the first term by the constant   3 • 24 = 72

Step-2 : Find two factors of  72  whose sum equals the coefficient of the middle term, which is   -22 .

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     -36    +    -2    =    -38

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Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -18  and  -4

                    3x2 - 18x - 4x - 24

Step-4 : Add up the first 2 terms, pulling out like factors :

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             Add up the last 2 terms, pulling out common factors :

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Step-5 : Add up the four terms of step 4 :

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 (x - 6) • (3x - 4)  = 0

STEP 3:

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When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

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