Here we are going to see how to find a point or points on a straight line which is at a particular distance from the line.
Example 1 :
A straight line is passing through the point A(1, 2) with slope 5/12. Find points on the line which are 13 units away from A.
Solution :
Let us draw a rough diagram to represent the given details.
Distance between AP = 13
A (1, 2) P (x, y)
AP = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}
√(1 + x)^{2} + (2 + y)^{2 } = 13
Taking squares on both sides, we get
(x - 1)^{2} + (y - 2)^{2 } = 169 ----(1)
Using the given point A(1, 2) and slope (m) = 5/12, we may find equation of the line.
(y - y_{1}) = m(x - x_{1})
(y - 2) = (5/12)(x - 1)
12(y - 2) = 5(x - 1)
y - 2 = (5/12) (x - 1)
(x - 1)^{2} + (y - 2)^{2 } = 169
(x - 1)^{2} + (25/144)(x - 1)^{2 } = 169
(x - 1)^{2} [169/144] = 169
(x - 1)^{2} = 144
x - 1 = ±12
x - 1 = 12 x = 13 y - 2 = 5 y = 7 (13, 7) |
x - 1 = -12 x = -11 y - 2 = -5 y = -3 (-11, -3) |
So, the required points are (13, 7) and (-11, -3)
Example 2 :
A 150m long train is moving with constant velocity of 12.5 m/s. Find (i) the equation of the motion of the train, (ii) time taken to cross a pole. (iii) The time taken to cross the bridge of length 850 m is?
Solution :
(i) Time = Distance /speed
Let "x" be the time taken by the trian.
Let "y + 150" be the distance covered by the train.Here y stands for length of pole.
Speed = 12.5
x = (y + 150)/12.5
12.5x = y + 150
y = 12.5x - 150 ----(1)
(ii) Time taken to cross the pole
x = y/12.5
x = 150/12.5
x = 12 seconds
(iii) The time taken to cross the bridge of length 850 m is?
Here the train is crossing the bridge, so we have to take (1)
y = 12.5x - 150
y = 850
850 = 12.5x - 150
x = (850 + 150)/12.5
x = 80 seconds
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