co-ordinate geometry .

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co-ordinate geometry .

by Md.Nazrul Islam » Fri Apr 06, 2012 8:00 pm
In the xy-plane line L and K intersect at the point (16/5,12/5 ).what is the slope of the line L .

1)THE product of the slopes of line L and K is -1.

2) line K pass through the origin .

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by Anurag@Gurome » Fri Apr 06, 2012 8:18 pm
Md.Nazrul Islam wrote:In the xy-plane line L and K intersect at the point (16/5,12/5 ).what is the slope of the line L .

1)THE product of the slopes of line L and K is -1.

2) line K pass through the origin .
Let us assume that the equation of line L is y = m1x + b1 and that of line K is y = m2x + b2
Since (16/5, 12/5) is the point of intersection of 2 lines, so this point lies on the lines also.
So, 12/5 = m1(16/5) + b1 ... Equation 1
and 12/5 = m2(16/5) + b2 ... Equation 2

We have to find the value of m1.

(1) The product of the slopes of line L and K is -1 implies m1 * m2 = -1, which means the two lines L and K are perpendicular lines. But we cannot find the value of m1; NOT sufficient.

(2) Line K pass through the origin.
So, 0 = m2(0) + b2 implies b2 = 0
This means 12/5 = m2(16/5) + 0
12 = 16 * m2
m2 = 3/4
Again we cannot find the value of m1 from here; NOT sufficient.

Next, combining (1) and (2), m1 * m2 = -1 and m2 = 3/4 implies m1 * (3/4) = -1
m1 = -4/3; SUFFICIENT.

The correct answer is C.
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