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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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Answer:

a) 4x -y = 2; (1/2, 0), (0, -2)

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The form you have is ...

y = -Ax +C . . . . . where B = 1

You can simply rearrange the coefficients to get the form you want.

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"Standard form" requires the leading coefficient be positive, so you now need to multiply the entire equation by -1:

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I suspect either of the above answers will satisfy your grader, but I like the latter form better. So, the answers below have the equation multiplied by -1 when A is negative to start with.

Since the given equation is already in slope-intercept form, the y-intercept is read directly from the equation as C.

The x-intercept is the value C/A, where you do need to pay attention to the signs.

Again, for (a), the y-intercept is already known to be -2; the x-intercept is ...

C/A = -2/-4 = 2/4 = 1/2.

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a) A=-4, C=-2, x-intercept = C/A = -2/-4 = 1/2, y-intercept = C = -2. Equation in standard form is 4x -y = 2

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b) A=3, C=5, x-intercept = C/A = 5/3, y-intercept = C = 5. Equation in standard form is 3x+y=5

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c) A=-5, C=3, x-intercept = C/A = 3/-5 = -5/3, y-intercept = C = 3. Equation in standard form is 5x -y = -3

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d) A=-1, C=-7, x-intercept = C/A = -7/-1 = 7, y-intercept = C = -7. Equation in standard form is x -y = 7

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e) A=8, C=-12, x-intercept = C/A = -12/8 = -3/2, y-intercept = C = -12. Equation in standard form is 8x +y = -12

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