Question

If $(\alpha ,\beta )$ is the centroid of the triangle formed by the lines $a{x}^2+2hxy+b{y}^2=0$ and $lx+my=1$, then prove that $\frac{\alpha }{bl-hm}=\frac{\beta }{am-hl}=\frac{2}{3(b{l}^2-2hlm+a{m}^2)}$.

Answer 1

The given pair of st lines passes through origin intersect equation of line side $AB:lx+my=1$ or $y=\frac{1}{n}(1-lx)$

Putting this $y$ in equation of Pair of st lines will give coordinates of $A$ and $B$.

$\Rightarrow a{x}^2+2nx\left[\frac{1-lx}{m}\right]+b{\left[\frac{1-lx}{m}\right]}^2$ so,

$a{m}^2{x}^2+2mnx(1-ln)+b(1-ln{)}^2$ so,

$x^2[a{m}^2-2mnl+b{l}^2]+x[2mn-2l]+b$ So

Solution are $x_1$ and $x_2\Rightarrow x_1+x_2=\frac{2(lb-nm)}{a{m}^2-2mnl+b{l}^2}$ .....(1)

$A$ lies on $AB$ $A$ lies on $AB$

$l{x}_1+m{y}_1-1$ $l{x}_2+m{y}_2-1$

Adding $l(x_1+x_2)+m(y_1+y_2)=2$

Using equation (1)

$\frac{2l(lb-hm)}{a{m}^2-2mnl+b{l}^2}+m(y_1+y_2)=2$

$m(y_1+y_2)=\frac{(2a{m}^2-4mnl+2b{l}^2-2b{l}^2+2lhm)}{(a{m}^2-2mhl+b{l}^2)}$

$d\left(\right(y_1+y_2\left)\right)=\frac{(2am-2mhl)}{a{m}^2-2mnl+b{l}^2}$

$(y_1+y_2^2)=\frac{2(am-hl)}{a{m}^2-2mhl+b{l}^2}$

$G=[\frac{x_1+x_2+0}{3},\frac{y_1+y_2+o}{3}]\le \alpha ,\beta$

$\alpha =\frac{x_1+x_2}{3}$ $\beta =\frac{y_1+y_2}{3}$

$\Rightarrow \alpha =\frac{2(lb-hm)}{a{m}^2-2hml+b{l}^2}$ $\beta =\frac{2(am-hl)}{3[a{m}^2-2mhl+b{l}^2]}$

$\frac{\alpha }{(lb-hm)}=\frac{\beta }{am-hl}=\frac{2}{3[a{m}^2-2mhl+b^2]}$

Hence proved.