Question

( x+y-3=0 quad ) Griven point ( =(1,1) )
Dist ( =frac{1}{sqrt{2}} ) Let required point be ( P equiv(h, k) )
( x+y-3=0 quad Rightarrow quad y=-x+3 )
( S_{0}, m^{2}-1=tan theta )
So, ( theta=135^{circ} )
( sin theta=frac{1}{sqrt{2}} quad cos theta=frac{-1}{sqrt{2}} )
Using parametrie form :
( h=x pm 0 cos theta quad k=frac{y}{2} pm 0 sin theta )
( h=1 pm frac{1}{sqrt{2}}left(frac{-1}{sqrt{2}}right) quad k=18 pm frac{1}{sqrt{2}}left(frac{1}{sqrt{2}}right) )
( h=1 mpleft(frac{1}{2}right) quad k=x / pm frac{1}{2} )
( operatorname{sen}(h, k) equivleft(frac{1}{2}, frac{3}{2}right) quad ) or ( quadleft(frac{3}{2}, frac{1}{2}right) )

# The coordinates of a point which is at a distance of units from (1, 1) in the direction of the line x + y - 3 = 0 is The family of lines (3 + 2x + (1 + 52)y - 7(1 + 2) = 0

Solution