Question

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​

Answer 1

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.