Question

find the value of x for which the distance between points P(2, -3) and Q(x, 5) is 10 units​

Answer 1

Solution

Given :-

• Distance between Two point are P(2,-3) and Q(x,5) is 10

Find :-

• Value of x

Explanation

Let,

If A(x,y) and B(x',y') be any two point , then distance between then, be

$\dag\boxed{\tt{\red{\:Distance\:=\:\sqrt{(x'-x)^2+(y'-y)^2}}}}$

So, Now

==> Distance = √[(x - 2)² + (5 + 3)²]

==> 10 = √[(x-2)² + 8²]

Squaring both side

==> 10² = (x-2)² + 64

==> x² - 4x + 4 + 64 = 100

==> x² - 4x + 68 - 100 = 0

==> x² - 4x - 32 = 0

==> x² - 8x + 4x - 32 = 0

==> x(x - 8)+4(x - 8) = 0

==> (x - 8)(x + 4) = 0

==> x - 8 = 0 , x + 4 = 0

==> x = 8 , x = -4

Hence

• Value of x will be = 8 & -4

__________________

Answer 2

We know that,

$\mathfrak{\boxed{\purple{Distance = \sqrt{(x - x)^2 + (y - y)^2}}}}$

⠀⠀⠀⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━━━━

⠀

$\sf{\underline{According\: to\: the\: question}}$

⠀

$\sf{\dashrightarrow{Distance = \sqrt{(x - 2)^2 + (5 + 3)^2}}}$

⠀

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

⠀

$\sf{\dashrightarrow{10^2 = (x - 2)^2 + 64}}$

⠀

$\sf{\dashrightarrow{x^2 - 4x + 4 + 64 = 100}}$

⠀

$\sf{\dashrightarrow{x^2 - 4x + 68 - 100 = 0}}$

⠀

$\sf{\dashrightarrow{x^2 - 4x - 32 = 0}}$

⠀

$\sf{\dashrightarrow{x^2 - 8x + 4x - 32 = 0}}$

⠀

$\sf{\dashrightarrow{x(x - 8) + 4(x - 8) = 0}}$

⠀

$\sf{\dashrightarrow{(x - 8)(x + 4) = 0}}$

⠀

$\sf{\dashrightarrow{x - 8 = 0 , x + 4 = 0}}$

⠀

$\mathfrak{\dashrightarrow{\boxed{\pink{x = 8 , x = -4}}}}$

⠀

Hence,

• The value of x is 8 and -4.