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

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

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13 Nov 2017, 12:57
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25% (medium)

Question Stats:

75% (01:25) correct 25% (01:27) wrong based on 116 sessions

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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. $$y = 3x – 4$$

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

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13 Nov 2017, 15:02
1
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

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

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13 Nov 2017, 17:05
3
1
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 [ 4.11 KiB | Viewed 2154 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.

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 [ 16.48 KiB | Viewed 2151 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

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

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02 Feb 2019, 20:48
1
generis wrote:
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

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.

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

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

Re: Which of the following equations defines a line for which no points on   [#permalink] 02 Feb 2019, 20:48
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