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Which of the following equations defines a line for which no points on

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Question Stats:

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Which of the following equations defines a line for which no points on the line lie in quadrant III?

A.\(y = –3x + 4\)

B.\(y = 3x + 4\)

C.\(y= \frac{X}{3}+4\)

D.\(y = –3x – 4\)

E. \(y = 3x – 4\)
[Reveal] Spoiler: OA

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Gnpth wrote:
Which of the following equations defines a line for which no points on the line lie in quadrant III?

A.\(y = –3x + 4\)

B.\(y = 3x + 4\)

C.\(y= \frac{X}{3}+4\)

D.\(y = –3x – 4\)

E. \(y = 3x – 4\)



The question simply asks line that does not have either x or y with NEGATIVE sign


A.\(y = –3x + 4\).......If x is negative then y is always positive............Keep....There is no points on the line lie in quadrant III.

B.\(y = 3x + 4\)........x = -2, then y = -2.........................................Eliminate

C.\(y= \frac{X}{3}+4\)........x = -36, then y = - 9.......... .Eliminate

D.\(y = –3x – 4\)................................x = -1 , then y = -1.............Eliminate

E. \(y = 3x – 4\).................................x = -1 , then y = -7.............Eliminate

Answer: A
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Which of the following equations defines a line for which no points on [#permalink]

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Gnpth wrote:
Which of the following equations defines a line for which no points on the line lie in quadrant III?

A.\(y = –3x + 4\)

B.\(y = 3x + 4\)

C.\(y= \frac{X}{3}+4\)

D.\(y = –3x – 4\)

E. \(y = 3x – 4\)


Conventional: find the intercepts and graph the line
Attachment:
lines.png
lines.png [ 4.11 KiB | Viewed 695 times ]

Set x equal to zero, then y equal to zero to find x- and y-intercepts.
These equations are in point slope form, and the numbers aren't bad.

A. \(y = –3x + 4\)
\(x = 0: y = 4\)
\(y = 0: 3x = 4, x = \frac{4}{3}\)
POINTS \((0,4) , (\frac{4}{3}, 0)\)

B. \(y = 3x + 4\)
\(x = 0: y = 4\)
\(y = 0: -3x = 4, x = -\frac{3}{4}\)
POINTS \((0, 4) and (-\frac{3}{4}, 0)\)

C. \(y= \frac{X}{3}+4\)
\(x = 0: y = 4\)
\(y = 0: \frac{x}{3} = -4, x = -12\)
POINTS \((0, 4) and (-12, 0)\)

D. \(y = –3x – 4\)
\(x = 0: y = -4\)
\(y = 0: 3x = -4, x = -\frac{4}{3}\)
POINTS (\(0, -4) and ( -\frac{4}{3}, 0)\)

E. \(y = 3x – 4\)
\(x = 0: y = -4\)
\(y = 0: 3x = 4, x = \frac{4}{3}\)
POINTS \((0, -4) and (\frac{4}{3}, 0)\)

See diagram. Only A avoids Q III.

ANSWER A

Sketch generally: watch slope and y-intercept
This way is a lot faster: Figure out whether the line's slope and y-intercept should be positive or negative.

Sketch the two absolute cases: a line with a positive slope through the origin, and one with a negative slope through the origin. Go from there.
Attachment:
lines2.png
lines2.png [ 16.48 KiB | Viewed 692 times ]

1. Sketch a line with a positive slope that passes through the origin: lines with positive slopes always pass through Quadrants I and III
Reject this one. We need to avoid Quadrant III. We need a negative slope

2) Sketch a line with a negative slope that passes through the origin
Lines with negative slopes always pass through Quadrants II and IV. How to avoid Q III?

3) Shift the line with the negative slope from the origin so that it does not pass through Q III
Note the y-intercept: it is positive

4) We need: a line with NEGATIVE slope and POSITIVE y-intercept

Check the options: Answers, B, C, and E are eliminated. Their slopes are positive

A. \(y = –3x + 4\)
Slope is -3
y-intercept is 4 (if x = 0, y = 4) KEEP

D. \(y = –3x – 4\)
Slope is -3
y-intercept is -4 (if x = 0, y = -4) REJECT

ANSWER A
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