Question

If x is a positive integer such that the distance between points P(x, 2) and Q(3, -6) is 10 units, then value of x is *
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Answer 1

Answer:

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Answer 2

★GIVEN★

• x is a positive integer.
• Distance between points P(x, 2) and Q(3, -6) is 10 units.

★To Find★

The value of x.

★SOLUTION★

We know the distance formula that,

$\large{\green{\underline{\boxed{\bf{Distance\:formula=\sqrt{{(x_2 - x_1)}^{2}+{(y_2-y_1)}^{2}}}}}}}$

where,

• $\large{\bf{(x_1,x_2)=Coordinates\:of\:the\:first\:point}}$
• $\large{\bf{(y_1,y_2)=Coordinates\:of\:the\:second\:point}}$

So,

• $\large{\sf{Let\:(x_1,x_2)\:be\:(x, 3)\:respectively.}}$
• $\large{\sf{Let\:(y_1,y_2)\:be\:(2, -6)\:respectively.}}$

According to the question,

$\large\implies{\sf{Distance\:formula=\sqrt{{(x_2 - x_1)}^{2} + {(y_2-y_1)}^{2}}}}$

$\large\implies{\sf{10=\sqrt{{(3-x)}^{2}+{(-6-2)}^{2}}}}$

$\large\implies{\sf{10=\sqrt{{(3-x)}^{2}+{(-8)}^{2}}}}$

Squaring both the sides,

$\large\implies{\sf{{10}^{2}={(3-x)}^{2}+{(-8)}^{2}}}$

$\large\implies{\sf{100=9-6x+{x}^{2}+64}}$

$\large\implies{\sf{100=-6x+{x}^{2}+73}}$

$\large\implies{\sf{100-73=-6x+{x}^{2}}}$

$\large\implies{\sf{27=-6x+{x}^{2}}}$

$\large\implies{\sf{0={x}^{2}-6x-27}}$

$\large\implies{\sf{{x}^{2}-6x-27=0}}$

Splitting the middle term,

$\large\implies{\sf{{x}^{2}-9x+3x-27=0}}$

$\large\implies{\sf{x(x-9)+3(x-9)=0}}$

$\large\implies{\sf{(x-9)(x+3)=0}}$

=> x - 9 = 0

=> x = 9.

and

=> x + 3 = 0

=> x = -3.

It is said that x is positive integer so,

Value of x is 9.