Find the value of x for which the distance between the points P(-4,10) and Q(x,-2) is 13units
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By distance formula i.e.,`√[X2-X1]²+[Y2-Y1]²
13=√[X+4]²+[-12]²
13=√X²+8X+16+144
13²=√X²+8X+160
169=X²+8X+160
X²+8X-9=0
BY FACTORIZATION
X=1
13=√[X+4]²+[-12]²
13=√X²+8X+16+144
13²=√X²+8X+160
169=X²+8X+160
X²+8X-9=0
BY FACTORIZATION
X=1
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