Question

The distance of the point P on X-axis from point A(5, – 4) is 4 units. Then, the coordinates of P are ______________.

Answer 1

Let the point P which is on x - axis be ( x , 0 )

{ since on x axis y coordinate is zero }

Hence,

Distance,

AP = √ ( 0 + 4 )² + ( x – 5 )²

4 = √ 4² + x² + 25 – 10x

4² = 4² + x² – 10x + 25

x² – 10x + 25 = 0

x² – 5x – 5x + 25 = 0

x ( x – 5 ) – 5 ( x – 5 ) = 0

( x – 5 )² = 0

Hence x – 5 = 0

Implies x = 5

coordinates of P are ( 5 , 0 )

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Answer 2

AnsWer :

P( 5 , 0 )

SolutioN :

Condition :

• The distance of the point P on X-axis from point A is 4 units.
• When, y become 0.

Let,

• The point be Q( x , 0 )

So,

• A( 5 , - 4 )
• P( x , 0 )
• x1 = 5.
• x2 = - 4.
• y1 = x.
• y2 = 0.
• AP = 4 units.

$\tt \dagger \: \: \: \: \: AP = \sqrt{{\Big( x_2 -x_1 \Big)}^{2} +{\Big( y_2 -y_1 \Big)}^{2}}$

$\tt : \implies AP = \sqrt{{\Big( 5 -x \Big)}^{2} +{\Big( 0 + 4\Big)}^{2}}$

$\tt : \implies 4 = \sqrt{{\Big( 5 -x \Big)}^{2} +{\Big( 0 + 4\Big)}^{2}}$

$\tt : \implies 4 = \sqrt{\Big( 25 + {x}^{2} - 10x\Big) +{\Big( 16\Big)}}$

$\tt : \implies 4 = \sqrt{ 25 + {x}^{2} - 10x+{ 16}}$

$\tt : \implies 4 = \sqrt{{x}^{2} - 10x+41}$

$\tt : \implies {x}^{2} - 10x + 41 = 16.$

$\tt : \implies {x}^{2} - 10x + 25= 0.$

$\tt : \implies {x}^{2} - 5x - 5x + 25= 0.$

$\tt : \implies x(x - 5) - 5(x - 5)= 0.$

$\tt : \implies (x - 5) (x - 5)= 0.$

Either,

$\tt : \implies x - 5= 0.$

$\tt : \implies x = 5.$

Coordinates P( 5 , 0 )

Therefore, the value of x is 5 units.