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

Solution:

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)

Therefore, [(2 - m)/(m + 1), (5m + 2)/(m + 1)] is the point of intersection of line (1) and (2)

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

⇒ ((2 - m)/(m + 1) + 1)² + ((5m + 2)/(m + 1) - 2)² = 3²

⇒ ((2 - m + m + 1)/(m + 1) + 1)² + ((5m + 2 - 2m - 2)/(m + 1) - 2)² = 3²

⇒ 9/(m + 1)² + 9m²/(m + 1)² = 9

⇒ (1 + m²)/(1 + m)² = 1

⇒ 1 + m² = m² + 1 + 2m

⇒ 2m = 0

⇒ m = 0

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

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 16

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

Summary:

The direction in which a straight line must be drawn through the point (- 1, 2) so that its point of intersection with line x + y = 4 may be at a distance of 3 units from this point is parallel to x-axis