1.

1.If (k, 2), (2, 4), (3, 2) are the vertices of atriangle of area 4sq.units, then the value of k is write in solution​

Answer»

ANSWER :-

→ k = 7

To Find :-

Value of k

Used Formula :-

\triangle = \frac{1}{2} \bigg\{\sf{x_{1}(y_{2}-y_{3}) + x_{2}(y_{3} - y_{1} ) + x_{3}(y_{1} - y_{2}) \: } \bigg\} \\

Explanation :-

Let given POINTS are A(k,2) ,B(2,4) and C(3,2)

\star \sf{ \: A(k,2) \to x_{1} =k , y_{1} =2} \\ \\ \star \sf{ \: B(2,4) \to x_{2} =2 , y_{2} = 4}\\ \\ \star \sf{ \: C(3,2) \to \: x_{3} = 3, y_{3} = 2 \: }

Also given that area of triangle FORMED by the given vertices is 4 square units .

Hence ,apply the above values in the given formula -

\to \sf{ 4 = \frac{1}{2} \bigg\{ k(4 - 2) + 2 (2 - 2)+ 3(2 - 4) \bigg \}} \\ \\ \to \sf{ 8 = 2k \: - 6} \\ \: \\ \to \sf{2k = 8 + 6} \: \\ \: \\ \to \sf{ 2k = 14} \\ \\ \to \boxed{\sf{k = 7 \: }}

Hence value of k is 7 ,since POINT A is (7,2)



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