The image of the line  $\mathrm{x}-\mathrm{y}+3=0$ on  x-axis is  $\mathrm{px}+\mathrm{qy}+\mathrm{r}=0$ then  the value of $\mathrm{p}+\mathrm{q}+\mathrm{r}=$

detailed solution

Correct option is D

To get the image of any line L1=0  with respect to the other line  L2=0  then find the equation of the line passing through images of any two points on the line  L1=0 with respect to the line  L2=0Two points on the line  x−y+3=0 are 0,3  and  1,4The image of a point  a,b on   x-axis is  a,−bHence the images of   0,3 and  1,4 on   x-axis are  0,−3 and  1,−4Equation of the line passing through the points  0,−3  and  1,−4 is y+3=−4+31−0x−0y+3=−xx+y+3=0 Compare this with  px+qy+r=0, Hence p=1,q=1,r=3.Therefore p+q+r=1+1+3=5