if the point P(2,2) is equidistant from the points A(-2,k) and B(-2k,-3) find k. also find the length of AP

Distance of P from A (2(2))2+(2k)2Distance of P from B (2(2k))2+(2(3))2P is equidistant(2(2))2+(2k)2=(2(2k))2+(2(3))2=>16+(2k)2=(2+2k)2+25Squaring=>16+(2k)2=(2+2k)2+25=>16+k24k+4=4k2+8k+4+25=>3k2+12k+9=0=>k2+4k+3=0=>k2+3k+k+3=0=>k(k+3)+1(k+3)=0=>(k+3)(k+1)=0Either, k+3=0=>k=3k=3 is not acceptable because it does not giveequidistant points.Or,k+1=0=>k=1Distance between P(2,2) between A(2,1)AP = (2(2))2+(2(1))2 = 25 =5 units

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