NCERT Solutions
Class 11 Maths
Straight Lines

Ex.Misc.Q.16
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 the straight line is y = mx + c
This line passes through the point (-1, 2)
So, 2 = -m + c
=> c = m + 2
Now y = mx + m + 2 ......... (1)
Given equation of line is:
x + y = 4 .......... (2)
Solving equation (1) and (2), we get
x = (2 - m) ÷ (m + 1) and y = (5m + 1) ÷ (m + 1)
Now this point is at a distance of 3 unit from the point (-1, 2)
From distance formula
√ [{(2 - m) ÷ (m + 1) + 2}2 + {(5m + 2) ÷ (m + 1) - 2}2] = 3
Squaring both sides, we get
{(2 - m) ÷ (m + 1) + 1}2 + {(5m + 2) ÷ (m + 1) - 2}2 = 9
=> {(2 – m + m + 1) ÷ (m + 1)}2 + {(5m + 2 - 2m - 2) ÷ (m + 1)}2 = 9
=> (9 ÷ (m + 1)2) + (9m2 ÷ (m + 1)2) = 0
=> (1 + m2) ÷ (m + 1)2 = 1
(divide by 9 on both side)
=> 1 + m2 = (m + 1)2
=> 1 + m2 = (m2 + 2m +1)
=> 1 + m2 = m2 + 2m + 1
=> 2m = 0
=> m = 0
Since the slope of the line is zero.
Hence, the line is parallel to x-axis.