If the area of triangle with vertices (3,2) , (-1,4) and (6,k) is 7 sq. units then find the smallest possible value of k ?

Question

If the area of triangle with vertices (3,2) , (-1,4) and (6,k) is 7 sq. units then find the smallest possible value of k ?

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Kaylee 3 weeks 2021-10-04T02:33:50+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-04T02:35:48+00:00

    given that :-

    • vertices of triangle :-
    1. A(-1,4)
    2. B(3,2)
    3. C(6,k)
    • area of triangle = 7 unit².

    To find :-

    • smallest possible value of k .

    Formula used :-

    area of triangle = \sf \dfrac{1}{2}   \bigg[ x_1( y_2 -  y_3) + x_2 (y_3 - y_1) + x_3 (y_1 -  y_2 )\bigg]

    Solution :-

    area of triangle = \sf \dfrac{1}{2}   \bigg[ x_1( y_2 -  y_3) + x_2 (y_3 - y_1) + x_3 (y_1 -  y_2 )\bigg]

     \sf  \blue  \implies\dfrac{1}{2}   \bigg[  - 1( 2 -  k) + 3 (k -4) + 6 (4-  2 )\bigg] \\  \\  \sf  \blue  \implies\dfrac{1}{2}  \bigg[  - 2 + k + 3k - 12 + 12\bigg]  \\  \\ \sf  \blue  \implies\dfrac{1}{2}  \bigg[  - 2 + k + 3k \cancel {- 12} \:  \: \cancel{ + 12}\bigg]  \\  \\ \sf  \blue  \implies\dfrac{1}{2}  \times 4k - 2 \\  \\ \sf  \blue  \implies4k = 16 \\  \\ \sf  \blue  \implies \: k =  \dfrac{16}{4}  \\  \\ \bf  \blue  \implies  \fbox{ \bf \: k = 4}

    therefore the possible smallest value of k is 4

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