Number of its at unit distance from the lines 3x-4y+11=0 and 5x-12+7=0
There are only two points which are at unit distance from the lines 3x-4y+11=0 and 5x-12y+7=0. They can be found as follows:
Let the points to be found be (x1,y1) and (x2,y2).
By the formula for finding perpendicular distance of a point from a line,
3x1-4y1+11= 1
3x1-4y1+10 = 0 (I)
and
5x1-12y1+7 = 1
5x1-12y1+6 = 0 (II)
Solving I and II,
9x1-12y1+30 = 0
- 15x1-12y1+6 = 0 ==> -6x1+24 = 0
6x1 = 24
x1 = 4
=>y1 = 11/2.
Since Distance is taken with a modulus sign, x2 = -x1, and y2 = -y1.
Therefore, the required points are:
(x1 , y1) = (4 , 5.5)
(x2 , y2) = (-4 , -5.5)
Let the points to be found be (x1,y1) and (x2,y2).
By the formula for finding perpendicular distance of a point from a line,
3x1-4y1+11= 1
3x1-4y1+10 = 0 (I)
and
5x1-12y1+7 = 1
5x1-12y1+6 = 0 (II)
Solving I and II,
9x1-12y1+30 = 0
- 15x1-12y1+6 = 0 ==> -6x1+24 = 0
6x1 = 24
x1 = 4
=>y1 = 11/2.
Since Distance is taken with a modulus sign, x2 = -x1, and y2 = -y1.
Therefore, the required points are:
(x1 , y1) = (4 , 5.5)
(x2 , y2) = (-4 , -5.5)