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# If line L passes through point (m, n) and (– m, – n), where m and n

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If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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31 Mar 2011, 05:45
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If line L passes through point (m, n) and (– m, – n), where m and n are not 0, which of the following must be true?

I. The slope of L is positive
II. The slope of L is negative
III. L exactly passes through 2 quadrants

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

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Re: If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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31 Mar 2011, 06:20
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1
Pkit wrote:
157. If line L passes through point (m, n) and
(– m, – n), where m and n are not 0, which of the following must be true?
I. The slope of L is positive
II. The slope of L is negative
III. L exactly passes through 2 quadrants
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

Equation of the line with slope m passing through point $$(x_1,y_1)$$ $$&$$ $$(x_2,y_2)$$
$$(y-y_1)=m(x-x_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

We have the line that is passing through $$(m,n)$$ $$&$$ $$(-m,-n)$$

$$Slope=\frac{-n-n}{-m-m}=\frac{n}{m}$$

Equation of the line will be:
$$y-n=\frac{n}{m}(x-m)$$
$$y-n=\frac{n}{m}x-n$$
$$y=\frac{n}{m}x$$

Thus we know that this line passes through the origin as the y-intercept is 0. This makes III true.

If we put, n=1 and m=1
$$y=x$$; this line has a positive slope. We can count II out.

Put n=-1 and m=1
$$y=-x$$; this line has negative slope. We can count I out.

Ans: "C"
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Re: If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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31 Mar 2011, 06:28
Let (m,n) = (2,3)

(-m,-n) = (-2,-3)

Slope = (3+3)/(2+2) = 6/4 = 3/2

So II may not be true

Let (m,n) = (2,-3) then (-m,-n) = (-2,3)

So slope = (3 +3)/(-2-2) = -3/2

so I may not be true

So such a line would be -> (y - 3) = 3/2(x - 2)

=> 2y - 6 = 3x - 6

=> 2y - 3x = 0, hence no x or y intercept, so it passes through origin.

III is true.

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Re: If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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31 Mar 2011, 06:35
@fluke and @PKit, should this be moved to PS forum, doesn't look like a DS question.
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Re: If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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31 Mar 2011, 08:56
1
this means that L exactly passes through 2 quadrants.
<see picture;>
and L:y=x (for quadrants 1&3)==>slop=1 (positive)
and L:y=-x (for quadrants 2&4)==>slop=-1 (negative)
therfore, the correct answer is C;
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If line L passes through point (m, n) and (– m, – n), where m and n  [#permalink]

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03 Sep 2017, 00:38
Since we know nothing about the signs of n and m, we can say nothing for sure about positions of two points on the coordinate plane.
Hence options I and II can't be proved.

Lets analyze the III option:

For the line to pass only two quadrants it must either pass trough the origin or be parallel to X or Y-axis.
Since the coordinates of two points have different signs, we can eliminate the possibility of parallelism.

Now, try to prove that line passes trough the origin.
The equation of a line is y=s*x+b. In order to the line to pass thru the origin, y-intercept (e.g. b in the equation) must be zero.

The question becomes whether y-s*x=o?

Lets prove it.
Find the slope --> s=(n-(-n))/(m-(-m)) --> s=n/m

Put two given points into the equation
First point (m, n) --> n-n/m*m =0 True
Second point (-m,-n) --> -n - ( n/m*(-m))=0 True

Agree?
If line L passes through point (m, n) and (– m, – n), where m and n   [#permalink] 03 Sep 2017, 00:38
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