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In the xy-plane, line k passes through the point (1, 1) and line m [#permalink]

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27 Jan 2010, 09:48

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Difficulty:

45% (medium)

Question Stats:

67% (01:00) correct
33% (00:49) wrong based on 26 sessions

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Please, explain...In the xy-plane, line k passes through the point (1, 1) and line m passes through the point (1, -1). Are the lines k and m perpendicular to each other ?

(1) Lines k and m intersect at the point (1, -1) (2) Line k intersects the x-axis at the point (1, 0)

Re: In the xy-plane, line k passes through the point (1, 1) and line m [#permalink]

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27 Jan 2010, 14:03

imagine line K extending from (1,1) intersecting the x axis at 1 ,and reachig point (1,-1) and quadrant IV. now line M meets k at (1 , -1) but we dont know to which direction its sloping , could be horizontal at y = -1 or it could be -vely sloped . That doesnt tell us whether the 2 lines are perpindiclar

Re: In the xy-plane, line k passes through the point (1, 1) and line m [#permalink]

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27 Jan 2010, 16:50

To determine if the lines are perpendicular we need to know the slopes of the two lines. Two points are given for line k but only one point is given for line l so it is impossible to find the slope of line l. The two points help us find rise/run which is the slope.

The lines would be perpendicular if the slops of the two lines are negative reciprocals of each other.

Re: In the xy-plane, line k passes through the point (1, 1) and line m [#permalink]

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27 Jan 2010, 20:06

mirzohidjon wrote:

Please, explain...

I'll tell you what your mistake was.

You calculated line K's slope as -infinity and for line m, though it's the same point, you might not have recognized it in your hurry and solved for slope consider the same point as twice, giving you a +infinity as slope. So obviously, they seem perpendicular but actually, product of slopes of perpendicular intersecting lines is supposed to be -1, and we are not sure if product of two infinities is 1.
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