Math, asked by kushwahaamit422, 11 months ago

Show that the angle between the lines represented by
3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 is tan-
and these lines intersect at the point of (1, -2).

Answers

Answered by Swarup1998

Let us consider the general equation of a pair of straight lines as follows,

$\quad ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$

If $\phi$ be the angle between the straight lines represented by the above pair, then

$\quad tan\phi =\frac{2\sqrt{h^{2}-ab}}{a+b}$ .

The given equation is

$\quad 3x^{2}+10xy+8y^{2}+14x+22y+15=0$

Comparing with the general equation of second degree, we get

$\quad h=5,\:a=3,\:b=8$ .

If $\phi$ be the required angle between the lines represented by the given equation, then

$\quad tan\phi=\frac{\sqrt{5^{2}-3\times 8}}{3+8}$

$\Rightarrow tan\phi=\frac{\sqrt{25-24}}{11}$

$\Rightarrow tan\phi=\frac{\sqrt{1}}{11}$

$\Rightarrow tan\phi=\frac{1}{11}$

$\Rightarrow \phi=tan^{1}(\frac{1}{11})$

Hence the required angle is $tan^{-1}(\frac{1}{11})$ .

This proves the first part.

The given point is $(1,-2)$ .

We can say this point is the mentioned intersecting point if it satisfies the given equation.

Let, $f(x,y)=3x^{2}+10xy+8y^{2}+14x+22y+15$ .

Now $f(1,-2)$

$=3(1)^{2}+10(1)(-2)+8(-2)^{2}+14(1)+22(-2)+15$

$=3-20+32+14-44+15$

= 0

Since $f(1,-2)$ is $0\:(zero)$ , we can say that the straight lines represented by the given pair intersect at the point $(1,-2)$ .

This proves the second part.

HENCE PROVED.