The distance between the points P(x,-1) and Q (3, 2) is 5 units. Find the value of x​

Question

The distance between the points P(x,-1)
and Q (3, 2) is 5 units. Find the value of x​

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Margaret 2 weeks 2021-11-15T10:06:22+00:00 1 Answer 0 views 0

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    2021-11-15T10:07:36+00:00

    Given: Two points P(x,-1)  and Q (3, 2)

    To find: The value of x?

    Solution:

    • Now we have given two points P(x,-1)  and Q (3, 2). The distance between them is 5 units.
    • Now we know the distance formula as:

                       PQ = √{ ( x2 – x1 )^2 + ( y2 – y1 )^2 }

    • So putting the values in formula, we get:

                       5 = √{ ( 3 – x )^2 + ( 2 – (-1) )^2 }

                       5 = √{ ( 3 – x )^2 + ( 2 + 1) )^2 }

    • Squaring on both sides, we get:

                       25 =  ( 3 – x )^2 + 9

                       16 =  ( 3 – x )^2

                       ( 3 – x ) = 4,-4

                       x = -1 or x = 7

    Answer:

                So the value of x is -1 or 7.

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