Find a point P on the line `3x+2y+10=0` such that `|PA - PB| `is minimum where A is (4,2) and B is (2,4)

The co-ordinates of a point P on the line 2x-y+5=0 such that |PA-PB| is maximum where A is (4,-2) and B is (2,-4) will be

The coordinates of the point P on the line 2x + 3y +1=0 such that |PA-PB| is maximum where A is (2,0) and B is (0,2) is

If A (3,1) and B (-5,7) are any two given points, If P is a point one the line y = x such that PA + PB is minimum then P is

Consider two points A=(-3,0) and B=(0,4). If a point P on the line 2x-3y-12=0 is such that |PA-PB| is maximum,then P is

STATEMENT-1: The point P(-22,28) lies on the line 3x+2y+10=0 such that for the points A(4,2)and B(2,4),|PA-PB| is maximum. STATEMENT-2: A point P on ax+by+c=0 such that |PA-PB| maximum is determined by the point of intersection of the line AB and Ax+by+c=0.

The locus of the moving point P such that 2PA=3PB, where A is (0.0) and B is (4,-3) , is

Statement I Let A-= (0,1) and B -= (2,0) and P be a point on the line 4x+3y+9=0 then the co - ordinates of P such that |PA -PB| is maximum is (-12/5,17/5) Statement II |PA - PB | le |AB|

P is a point lying on line y = x then maximum value of |PA - PB| , (where A-=(1, 3), B-=(5, 2) ) is

A(-3,2),B(2,5) be two points if the point P(0,k) lies on y- axis such that |PA-PB| is maximum. Then (k-1)/(5)=