Distance of a Point From a Line
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The distance between the parallel lines y = 2x + 4
and 6x = 3y + 5 isMarks:1
Let the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2), (1, 1 ) to a variable line be zero, then the line passes through a fixed point whose coordinates areMarks:1
Two lines are given (x – 2y)2 + k (x – 2y) = 0. The value of k, so that the distance between them is 3, isMarks:1
k = 35.
The lines are given by
(x – 2y)2 + k (x – 2y) = 0
(x - 2y)(x - 2y + k) = 0
That is x – 2y = 0 and x – 2y + k = 0
These are parallel. The distance between the two lines
The equation of a straight line, which passes through the point (a, 0) and whose perpendicular distance from the point (2a, 2a) is a, isMarks:1
3x – 4y – 3a = 0 and x – a = 0.
Equation of line passing through (a, 0) is y = m(x – a)
mx - y - ma = 0 ...(i)
Its distance from the point (2a, 2a) is
|(2am - 2a - ma)/(m2 + 1)| = a (given)
(m - 2)2 = (m2 + 1)
m2 - 4m + 4 = m2 + 1
- 4m + 3 = 0
m = 3/4, .
[If a = 0 then one root of ax2 + bx + c = 0 is infinite]
This result must be taken into care otherwise one root will be lost.
The required equation of lines is, from (i)
3x – 4y – 3a = 0
and y = (x – a)
x – a = 0.
The distance of the line 2x + y = 3 from the point (–1, 3) isMarks:1