Question

Find the angle between the lines joining the origin to the points of intersection of the
curve x2+2xy+y2+2x+2y-5=0 and the line 3x-y+1=0.
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Answer 1

Given:

A curve x2+2xy+y2+2x+2y-5=0 and a line 3x-y+1=0.

To Find:

Find the angle between the lines joining the origin to the points of intersection of the  curve x2+2xy+y2+2x+2y-5=0 and the line 3x-y+1=0.

Solution:

here, given line is -   $3x-y+1=0$

implies, $y=3x+1$

on putting, value of y in the equation of curve- $x^{2} +2xy+y^2+2x+2y-5=0$

we get,

$x^{2} +2x(3x+1)+(3x+1)^2+2x+2(3x+1)-5=0\\\\x^2+6x^{2} +2x+9x^{2} +1+6x+2x+6x+2-5=0\\\\16x^2+14x-2=0\\\\8x^{2} +7x-1=0\\\\8x^{2} -x+8x-1=0\\\\x(8x-1)+1(8x-1)=0\\\\(8x-1)(x+1)=0\\\\\Rightarrow x=-1 \ , or \ x=1/8$

And, $y=-2 \ , or \ y=11/8$

therefore,

$\dfrac{y}{x}=2 \ , \ or \ \dfrac{y}{x}=11$

thus, the angle between the line joining the origin to the points of intersection of the curve $x^{2} +2xy+y^2+2x+2y-5=0$ and the line $3x-y+1=0$ is given by;

$tan\theta=|\dfrac{11-2}{1-22}|=|9/21|=|3/7|\\\\\theta= tan^{-1}(3/7) \\\\\theta = 23.20 \ degree$

Answer 2

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