Question

Find the values of x for which the distance between the points p(x,4) and Q(9,10) is 10 units

Answer 1

Answer:

$Value \: of \:x=17 \:or \: x=1$

Step-by-step explanation:

$Given \: two \: points \\P(x,4)\: and \: Q(9,10)\: and \: PQ = 10 \: units$

$Compare \: these \: points\\with \: (x_{1},y_{1}),\:(x_{2},y_{2})$

$x_{1}=x ,\: y_{1}=4;\\</p><p>x_{2}=9 ,\: y_{2}=10;\\</p><p>[tex]Using \: distance \: formula\\d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = PQ = 10$

$\implies d^{2}=10^{2}$

$\implies \left(9-x\right)^{2}+\left(10-4\right)^{2}=100$

$\implies 9^{2}+x^{2}-18x+6^{2}-100=0$

$\implies 81+x^{2}-18x+36-100=0$

$\implies x^{2}-18x+117-100=0$

$\implies x^{2}-18x+17=0$

Splitting the middle term, we get

$\implies x^{2}-17x-1x+17=0$

$\implies x(x-17)-1(x-17)=0$

$\implies (x-17)(x-1)=0$

$\implies x-17 = 0 \:or \: x-1=0$

$\implies x=17 \:or \: x=1$

Therefore,

$Value \: of \:x=17 \:or \: x=1$

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Answer 2

Step-by-step explanation:

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