Question

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 the line x + y = 4 may be at a distance of 3 units from this point.

Class X1 - Maths -Straight Lines Page 234

Answer 1

The line passing through (-1,2) and having slope m is
y - 2 = m(x +1 )
y = mx + m + 2 __________(1)
[ because equation of line (y-y1)=m(x-x1) where , x1 = -1 and y1 = 2 ]


now, we have to find point of intersection of line AB and line PQ .
put equation (1) in x + y = 4, we get
x + mx + m + 2 = 4
(1 + m)x + m = 2
x = (2 - m)/(m + 1) put it in equation x + y = 4
y = 4 - x
= 4 - (2 - m)/(m + 1)
= { 4(m + 1) - (2 - m)}/(m+1)
= {4m +4 - 2 + m}/(m +1)
= ( 5m + 2)/(m +1)

so, co-ordinate of Q = { (2-m)/(m+1),(5m+2)/(m+1)}

A/C to question,
distance between P and Q = 3
PQ = 3
take square both sides,
PQ² = 9
(x2 - x1)² + (y2 - y1)² = 9 [ by using distance formula]
{(2-m)/(m+1)+1}² + {(2+5m)/(m+1)-2}²=9
{3/(m+1)}² + {3m/(m+1)}² = 9
{9 + 9m²} = 9(m + 1)²
9 + 9m² = 9m² + 18m + 9
m = 0

hence, slope of PQ is zero .it means line PQ is parallel to x -axis .

Answer 2

Answer:

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