The line x + y = 1 cuts the coordinate axes at P and Q and a line perpendicular to it meet the axes in R and S. The equation to the locus of the point of intersection of the lines PS and QR is

Given that $x+y=1$ intersect the axes at two points $P,Q$, so that $P\left(1,0\right)$ and $Q\left(0,1\right)$

A line perpendicular to the line $x+y=1$ can be taken as $x-y+k=0$

The points of intersection of the above line with axes are $R\left(-k,0\right)S\left(0,k\right)$

Equation of line passing through the points $P and S$ is $y=\frac{k}{-1}\left(x-1\right)\Rightarrow y=-k\left(x-1\right)$

Equation of line passing throhg the points $Q$ and $R$ is $y-1=\frac{1}{k}x\Rightarrow y=\frac{x}{k}+1$

To find the locus of the point of intersection of the above two lines, find the equaiton from both equations by eliminating  $k$

The equation is $-\frac{y}{x-1}=\frac{x}{y-1}$

cross multiply 

$x^2-x+y^2-y=0\Rightarrow x^2+y^2-x-y=0$

Therefore, the equation of the required locus is $\boxed{x^2+y^2-x-y=0}$

In the above figure, the point $S\left(0,k\right)$ is ortho center of the triangle $PQR$

The base of the triangle $QR$ is perpendicular to $PS$

The slope of $AQ$$\times$ slope of $AP$ is $-1$

hence, $\frac{y}{x-1}·\frac{y-1}{x}=-1$

it implies that $y^2-y+x^2-x=0\Rightarrow x^2+y^2-x-y=0$

$\perp \text{ line }\mathrm{x}-\mathrm{y}=\mathrm{\lambda }\Rightarrow \text{ eq of }\mathrm{PS}\text{ is }\frac{\mathrm{y}-0}{\mathrm{x}-1}=\frac{\mathrm{\lambda }}{1}$

$\text{ eq of }\mathrm{QR}\text{ is }\frac{\mathrm{y}-1}{\mathrm{x}-0}=\frac{-1}{\mathrm{\lambda }}$

$\therefore \text{ eq }\frac{\mathrm{y}(\mathrm{y}-1)}{\mathrm{x}(\mathrm{x}-1)}=1$