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. 

Let y = mx + c be the line through point (-1, 2).

Accordingly, 2 = m(-1) + c

⇒ 2 = -m + c

⇒ c = m + 2

∴ y = mx + m + 2                ... (1)

The given line is

x + y = 4                          ... (2)

On solving equations (1) and (2), we obtain

x = ^{(2 - m)}/_{(m + 1)} and y = ^{(5m + 2)}/_{(m + 1)}

∴ (^{(2 - m)}/_{(m + 1)}, ^{(5m + 2)}/_{(m + 1)}) is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (-1, 2), according to the distance formula,

√((^{(2 - m)}/_{(m + 1)} + 1)² + (^{(5m + 2)}/_{(m + 1)} - 2)²) = 3

⇒ (^{(2 - m + m + 1)}/_{(m + 1)})² + (^{(5m + 2 - 2m - 2)}/_{(m + 1)})² = 3²

⇒ ⁹/_{(m + 1)²} + ^9m2/_{(m + 1)²} = 9

⇒ ^{(1 + m2)}/_{(m + 1)²} = 1

⇒ 1 + m² = m² + 1 + 2m

⇒ 2m = 0

⇒ m = 0

Therefore, the slope of the required line must be zero, i.e., the line must be parallel to the x-axis.