Posted by **karan** on Saturday, March 2, 2013 at 5:08am.

find k if one of the lines given by kx^2+10xy+8y^2=0 is perpendicular to 2x-y=5

- geometry -
**Reiny**, Saturday, March 2, 2013 at 8:09am
The slope of 2x-y=5 is 2

So the slope of a perpendicular line is -1/2

kx^2 + 10xy + 8y^2 = 0 is supposed to be the intersection of 2 straight lines, so let's solve it for y

y = (-10x ± √(100x^2 - 32kx^2) )/16

= (-10x ± x√100-68k)/16

= x(-10 ± √(100-32k) )/16

So the slope of the two lines is (-10 ± √(100-32k) )/16

thus:

(-10 ± √(100-32k) )/16 = -1/2

-20 ± 2√100-32k = -16

±2√100-32k = 4

±√100-32k = 2

square both sides

100-32k = 4

32k=96

k=3

check my arithmetic.

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