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co-ordinate geo

Expert replies
Source: — Data Sufficiency |

by rajataga » Tue Jan 06, 2009 7:07 am
IMO E

each statement provides us the information that it is perpendicular to the X axis, however no information is provided about line m, except that it passes through (1,-1)
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by vivek.kapoor83 » Tue Jan 06, 2009 7:18 am
pls explain
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by rajataga » Tue Jan 06, 2009 7:57 am
Now the question asks you to determine whether the 2 lines are perpendicular to each other.

that means, you should be able to find out if the angle between the 2 lines is 90 or not.

Now to find the slope of a line, you need atleast two points on the line. You have 3 for line k, but only one point of line m...Hence you can never find out the slope of line m, and hence, the angle it makes with line k.

Hence, E
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by vivek.kapoor83 » Tue Jan 06, 2009 8:52 am
OA is E.. i am looking for better explanation. I thought the way, as per upper post., but was doubting my reasoning was right.
Any alternate explanation????????????

1 more ques.
Can any1 tell
wht is the product of slopes when lines are perpendicular ( i think it is -1)
wht is the product of slopes when lines are parallel ( i think same slope)
can any1 confirm ?
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by vikz_316 » Tue Jan 06, 2009 9:04 am
Why dont you try plotting the points.

With the info given, you would be able to plot only 1 line for k, but many lines for n. Hence not sufficient. I hope this helps.
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by vivek.kapoor83 » Tue Jan 06, 2009 9:09 am
thanks all........
pls also explain my ques above regardin product of slopes
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by vikz_316 » Tue Jan 06, 2009 11:23 am
for perpendicular lines, m1 * m2 = -1
for parallel lines, m1=m2, hence the square(yet to come across a problem where this is asked though)

Another tip,
To find a slope of a line, just express the line as y=mx + c. Slope will be the value of m. Dont forget, there should be no coefficient for the variable y.
Hence, slope of ax+by+c = 0 is (-a/b)
and the line perpendicular to it will be (b/a)
The equation of the perpendicular line can be written as bx-ay+k = 0
Hope I did not confuse you.
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