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

Answer 1

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.

Answer 2

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