If area of a triangle with vertices (3,2), (-1,4) and (6,k) is 7 sq.units then possible values of k are

Question

If area of a triangle with vertices (3,2), (-1,4) and (6,k) is 7 sq.units then possible values of k are

in progress 0
2 months 2021-12-03T06:10:29+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-12-03T06:11:39+00:00

    Answer:

    \sf{The \ value \ of \ k \ is \ -3.}

    Given:

    • If area of a triangle with vertices (3,2), (-1,4) and (6,k) is 7 sq units

    To find:

    • The value of k.

    Solution:

    \boxed{\sf{Area \ of \ triangle=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] }}

    \sf{Here \ x_{1}=3, \ x_{2}=-1, \ x_{3}=6, \ y_{1}=2,}

    \sf{y_{2}=4, \ and \ y_{3}=k.}

    \sf{According \ to \ the \ given \ condition.}

    \sf{7=\frac{1}{2}[3(4-k)-1(k-2)+6(2-4)]}

    \sf{\therefore{14=12-3k-k+2+12-24}}

    \sf{\therefore{14-2=-4k}}

    \sf{\therefore{-4k=12}}

    \sf{\therefore{k=\frac{-12}{4}}}

    \sf{\therefore{k=-3}}

    \sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ -3.}}}

Leave an answer

Browse

14:4+1-6*5-7*14:3+5 = ? ( )