A line a drawn through A (4, – 1) parallel to the line 3x – 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

Given: (x1,y1) = A(4, – 1)

To find:


Coordinates of the two points on this line which are at a distance of 5 units from A.


Explanation:


Line 3x – 4y + 1 = 0


4y = 3x + 1


y


Slope


sin θ and cos θ


So, the equation of the line passing through A (4, −1) and having slope is


Formula Used:



3x – 4y = 16


Here, AP = r = ± 5
Thus, the coordinates of P are given by




x and y


x and y


x= ±4 + 4 and y = ±3–1


So x = 8, 0 and y = 2, – 4


Hence, the coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, −4).


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