Question

if theta is the acute angle between the lines given by equation 6x²+5xy-4y²+7x+13y-3=0 then equation of line passing through point of intersection of these lines and making angle theta with positive x- axis​

Answer 1

Answer:

Correct option is

B

11x−2y+13=0

Writing the given equation as a quadratic in x

we have 6x

2

+(5y+7)x−(4y

2

−13y+3)=0

⇒x=

12

−(5y+7)±

(5y+7)

2

+24(4y

2

−13y+3)

=

12

−(5y+7)±

121y

2

−242y+121

=

12

−(5y+7)±11(y−1)

=

12

6y−18

,

12

−16y+4

⇒2x−y+3=0 and 3x+4y−1=0

which are the two lines represented by the given equation and the point of intersection is (−1,1), obtained by solving these equations

[The point of intersection of the given lives is also obtained by solving 12x+5y+7=0 and 5x−8y+13=0(seeText,15.5(4)]

Also tanθ=

a+b

2

h

2

−ab

where a=6,b=−4,h=5/2

=

6−4

2

(5/2)

2

−6(−4)

=

4

121

=

2

11

So the equation of the required line is

y−1=

2

11

(x+1)⇒11x−2y+13=0