A) 6
B) 3
C) 4
D) 5
Correct Answer: C
Solution :
Since, the point Q (0, 1) is equidistant from P \[(5,-3)\]and R(x, 6). \[\therefore \]\[QP=QR\]\[\Rightarrow \]\[Q{{P}^{2}}=Q{{R}^{2}}\] \[\Rightarrow \]\[{{(5-0)}^{2}}+{{(-3-1)}^{2}}={{(x-0)}^{2}}+{{(6-1)}^{2}}\] [\[\because \]Distance\[=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]] \[25+16={{x}^{2}}+25\]\[\Rightarrow \]\[{{x}^{2}}=16\]\[x=\sqrt{16}=4\]
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