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

Question

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

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Nevaeh 1 week 2021-10-04T15:41:40+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-04T15:42:56+00:00

    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.

    0
    2021-10-04T15:42:57+00:00

    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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