Telangana Board Solutions for Chapter: Pair of Straight Lines, Exercise 1: Exercise 4(a)

Find the equation of the bisector of the obtuse angle between the lines $x+y-5=0$ and $x-7y+7=0$.

Show that the lines represented by ${\left(lx+my\right)}^2-3{\left(mx-ly\right)}^2=0$ and $lx+my+n=0$ form an equilateral triangle with area $\frac{n^2}{\sqrt{3}\left(t^2+m^2\right)}$.

Show that the straight lines represented by $3x^2+48xy+23y^2=0$ and $3x-2y+13=0$ form equilateral triangle of area $\frac{13}{\sqrt{3}}$ sq.units.

Show that the equation of the pair of lines bisecting the angles between the pair of bisectors of the angles between the pair of lines $a{x}^2+2hxy+b{y}^2=0$ is $\left(a-b\right)\left(x^2-y^2\right)+4hxy=0$.

If one line of the pair of lines $a{x}^2+2hxy+b{y}^2=0$ bisects the angle between the coordinate axis prove that ${\left(a+b\right)}^2=4h^2$.

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

Prove that the distance from the origin to the orthocentre of the triangle formed by the line $\frac{x}{\alpha }+\frac{y}{\beta }=1$ and $a{x}^2+2hxy+b{y}^2=0$ is ${\left({\alpha }^2+{\beta }^2\right)}^{\frac{1}{2}}\left|\frac{\left(a+b\right)\alpha \beta }{a{\alpha }^2-2h\alpha \beta +b{\beta }^2}\right|$.

The straight line $lx+my+n=0$ bisects an angle between the pair of lines of which one is $px+qy+r=0$. Show that the other line is $(px+qy+r)\left(l^2+m^2\right)-2(lp+mq)(lx+my+n)=0$.