Let two points be A(4,-2) and B(2,-4). A point P lies on the line 2x-y+5=0 such that |PA - PB| is maximum, then find the coordinates of P.
I figured out that this is actually a triangle with two fixed vertices and the third moving on a straight line and the point P will be such that the difference of the two sides connecting the vertices to the straight line has to be maximum. But I cant figure out how to maximise this quantity.
I tried using the fact that the third side of a triangle is always greater than the difference of the other two sides....but it doesnt seem to work.
Can anyone help me out?
Thanks in advance!!
We could solve this problem by expressing the difference $ |PA - PB| $ as a function of the coordinates of $P $, eliminating one coordinate in the function using the equation of the line where $P $ lies, then calculating the derivative of the squared function (to account for the absolute value), and thus finding the maximum.
However, to maximize the difference, it is sufficient to remind that the difference between two sides of a non-degenerate triangle is lower than the third side, and that it equals the third side only when the triangle reduces to a straight line. So the point $P $ that maximizes the difference is the crossing point between the line $y=2x+5 \, \,$ and that passing through $A $ and $B $, i.e., $y=x-6 \, \, $. This directly leads to the coordinates $(-11,-17) $.