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 ?

Answer 1

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