Find the direction in which a straight line must be drawn through the point `(1,2)` so that its point of intersection with the line `x+ y =4` may be at a distance of 3 units from this point.

Explanation

Let the slope of line be m. Then line through point P(-1,2) is
y-2 = m(x+1)
`" or " y=mx+2+m " " (1)`
The given line is
`x+y=4 " " (2)`
Solving (1) and (2), we get the point of intersection as
`Q ((2-m)/(m+1), (5m+2)/(m+1))`
Given PQ=3. Therefore,
`sqrt(((2-m)/(m+1)+1)^(2) + ((5m+2)/(m+1)-2)^(2)) = 3`
`"or " (9)/(m+1)^(2) + (9m^(2))/(m+1)^(2) = 9`
`"or " 1+m^(2) = m^(2) + 1+2m`
or 2m=0
or m=0
Thus, the slope of the required line must be zero, i.e., the line must be parallel to the x-axis.