12.6k views
1 vote
The distance of the line 2x-3y=4 from the point(1,1) in the direction of the line x y=1 is

User Peter Long
by
8.3k points

1 Answer

0 votes
The required line is parallel to line x + y = 1 and passes through point (1, 1).
For parallel lines, slope is equal.
x + y = 1
y = -x + 1 => slope is -1
Line passing though point (1, 1) with slope = -1 is given by y - 1 = -(x - 1)
y - 1 = -x + 1
y + x = 1 + 1 = 2
i.e. Equation of the required lin is x + y = 2

Required distance is the distance between point (1, 1) and the point of intersection of lines x + y = 2 and 2x - 3y = 4.
x + y = 2 . . . (1)
2x - 3y = 4 . . . (2)

(1) x 2 => 2x + 2y = 4 . . . (3)

(2) - (3) => -5y = 0 => y = 0
From (1), x + 0 = 2 => x = 2.
i.e. the point of intersection of lines x + y = 2 and 2x - 3y = 4 is (2, 0).

Required distance is

d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ =√((0 - 1)^2+(2-1)^2)=√((-1)^2+1^2) \\ =√(1+1) \\ = √(2)
User Nikolay Rusev
by
9.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories