A line through point A(1,3) and parallel to the line x-y+1 = 0 meets the line 2x-3y + 9 = 0 at point P. Find distance AP without finding point P.

Explanation

Slope of line x-y + 1 = 0 is 1, i.e., tan `45^(@).`
So, the equation of line through A(1,3) and having slope I is
`(x-1)/("cos" 45^(@)) = (y-3)/("sin" 45^(@))`
`"or " (x-1)/(1//sqrt(2)) = (y-3)/(1//sqrt(2)) r (say)`
This line meets the given line 2x-3y + 9 = 0 at point P.
Let AP= r. Than P`=(1+(r)/(sqrt(2)), 3 + (r)/(sqrt(2))).`
But point P lies on the line 2x-3y + 9 = 0.
`therefore 2(1+(r)/(sqrt(2))) - 3(3+(r)/(sqrt(2)))+9 = 0`
`rArr (r)/(sqrt(2)) = 2`
`rArr r= 2sqrt(2)`