Question

The base at a triangle passes through a fixed point $(f,g)$ and its sides are respectively bisected at right angles by the lines $y^2-8xy-9x^2=0$. Find locus of its vertex.

Answer 1

Refer to the figure.

1) Given equation of pair of straight lines is,

$y^2-8xy-9x^2=0$

$\therefore y^2-9xy+xy-9x^2=0$

$\therefore y(y-9x)+x(y-9x)=0$

$\therefore (y-9x)(y+x)=0$

$\therefore y=9x$ and $y=-x$

2) Let $L_1:y=9x$ and $L_2:y=-x$

As shown in figure, line $L_1$ is perpendicular to line AB as well as bisecting line AB.

Thus, slope of line $AB$ is,

$m_1=\frac{-1}{9}$

Let coordinates of point A are $A(h,k)$ and point B are $B(x_1,y_1)$

Thus, coordinates of mid-point of line AB = $(\frac{h+x_1}{2},\frac{k+y_1}{2})$ (1)

Thus, equation of line AB is,

$y-y_1=m_1(x-x_1)$

$\therefore y-k=\frac{-1}{9}(x-h)$

$\therefore 9(y-k)=-1(x-h)$

$\therefore 9y-9k=-x+h$

$\therefore 9y+x=h+9k$

To find point of intersection of line $L_1$ with line AB, put $y=9x$ in above equation, we get,

$9\left(9x\right)+x=h+9k$

$\therefore 81x+x=h+9k$

$\therefore 82x=h+9k$

$\therefore x=\frac{h+9k}{82}$

$\therefore y=\frac{9(h+9k)}{82}$

Thus, coordinates of mid-point of line AB = $(x,y)=(\frac{h+9k}{82},\frac{9(h+9k)}{82})$ (2)

From (1) and (2), we get,

$\frac{h+x_1}{2}=\frac{h+9k}{82}$

$82(h+x_1)=2(h+9k)$

$\therefore 82h+82x_1=2h+18k$

$\therefore 82x_1=2h+18k-82h$

$\therefore 82x_1=-80h+18k$

$\therefore x_1=\frac{18k-80h}{82}$

$\therefore x_1=\frac{9k-40h}{41}$

Similarly, $\frac{k+y_1}{2}=\frac{9h+81k}{82}$

$\therefore k+y_1=\frac{9h+81k}{41}$

$\therefore y_1=\frac{9h+81k}{41}-k$

$\therefore y_1=\frac{9h+81k-41k}{41}$

$\therefore y_1=\frac{9h+40k}{41}$

Thus, coordinates of point B are $(\frac{9k-40h}{41},\frac{9h+40k}{41})$

3) line $L_2$ is perpendicular to line AC.

Thus, slope of line AC = $m_2=1$

Let coordinates of point C are $C(x_2,y_2)$

Thus, coordinates of mid-point of line AC = $(\frac{h+x_2}{2},\frac{k+y_2}{2})$ (3)

Now, equation of line AC is,

$y-k=1(x-h)$

$y-k=x-h$

$\therefore x-y=h-k$

To find point of intersection of line $L_2$ with line AC, put $y=-x$ in above equation, we get,

$x-(-x)=h-k$

$\therefore 2x=h-k$

$\therefore x=\frac{h-k}{2}$

$\therefore y=\frac{-(h-k)}{2}$

$\therefore y=\frac{k-h}{2}$

Thus, coordinates of mid-point of line AC are $(x,y)=(\frac{h-k}{2},\frac{k-h}{2})$ (4)

From (3) and (4), we get,

$\frac{h+x_2}{2}=\frac{h-k}{2}$

$\therefore h+x_2=h-k$

$\therefore x_2=-k$

Similarly, $\frac{k+y_2}{2}=\frac{k-h}{2}$

$\therefore k+y_2=k-h$

$\therefore y_2=-h$

Thus, coordinates of point C are $C(-k,-h)$

4) Equation of line BC=

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\therefore \frac{y-(-h)}{\left(\frac{9h+40k}{41}\right)-(-h)}=\frac{x-(-k)}{\left(\frac{9k-40h}{41}\right)-(-k)}$

$\therefore \frac{y+h}{\left(\frac{9h+40k}{41}\right)+h}=\frac{x+k}{\left(\frac{9k-40h}{41}\right)+k}$

$\therefore \frac{41(y+h)}{9h+40k+41h}=\frac{41(x+k)}{9k-40h+41k}$

$\therefore \frac{y+h}{50h+40k}=\frac{x+k}{50k-40h}$

$\therefore \frac{y+h}{10(5h+4k)}=\frac{x+k}{10(5k-4h)}$

$\therefore \frac{y+h}{5h+4k}=\frac{x+k}{5k-4h}$

$\therefore y+h=\frac{x+k}{5k-4h}(5h+4k)$

Line BC is passing through point $(f,g)$

$\therefore g+h=\frac{f+k}{5k-4h}(5h+4k)$

$\therefore (g+h)(5k-4h)=(f+k)(5h+4k)$

$\therefore 5gk-4gh+5kh-4h^2=5fh+4kf+5kh+4k^2$

$\therefore 4h^2-5hk+4gh-5gk+5fh+4kf+5kh+4k^2=0$

$\therefore 4h^2+4k^2+(4g+5f)h+(4f-5g)k=0$

Replace $h$ by $x$ and $k$ by $y$, we get,

$\therefore 4x^2+4y^2+(4g+5f)x+(4f-5g)y=0$

This is locus of vertex A of triangle