find the angle between the lines joining the origin to the of intersection 7xsquare-4xy+8ysquare+2x-4y-8=0and the line 3x-y=2
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Step-by-step explanation:
7X^2 - 4XY + 8Y^2+2X-4Y-8
= 7x^2 -4x(3x-2) + 8(3x -2)^2 + 2x -4(3x-2)-8
= 7x^2 - 12x^2 + 8x + 8(9x^2-12x+4) + 2x -12x +8–8
= 67x^2 -98x + 32 = 0.
Solving for x, x = (49/67) ± √(49^2 -(32)(67))/67 = (49 ±√(257))/(67).
y = 3x - 2 = (13 ± 3√(257))/(67).
The points of intersection are:
P {(49 +√(257))/(67), (13 + 3√(257))/(67)},
Q {(49 -√(257))/(67), (13 - 3√(257))/(67)}.
The slopes of OP and OQ are m1 = (13 + 3√(257))/(49 +√(257)
and m2 = (13 - 3√(257))/(49 -√(257). Let the angle between OP and OQ be ø.
tanø = (m1 - m2)/(1 - (m1)(m2)) = 7.31215×10^8 (almost ∞).
The angle ø between OP and OQ is 90º.
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