FIND A POINT GIVEN DISTANCE FROM ANOTHER POINT ALONG A LINE

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  =  √(x₂ - x₁)² + (y₂ - y₁)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 - y₁)  =  m(x - x₁)

(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

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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