Question

find the value of K so that the triangle with vertices (1-1), (4,2K) and -(k,-5) is 24sq unit​

Answer 1

Correct question -

Find the value of k so that the triangle with vertices (1,-1), (-4,2k) and (-k, - 5) is 24square unit

Solution -

Let A(1,-1), B(-4,2k) and C(-k, - 5)

And,

$(x_1,y_1) = (1, - 1) \: \: ,( x_{2}, y_{2}) = ( - 4,2k) \: \: and \: \: (x_{3}, y_{3} ) = ( - k, - 5)$

Now,

$Area \: \: of \: \: triangle \: \: ABC = \frac {1}{2} [x_1(y_2-y_3) +x_2(y_3-y_1) +x_3(y_1-y_2)] </p><p>$

Putting the values -

$24 = \frac{1}{2} [1(2k + 5) + ( - 4)( - 5 + 1) + ( - k)( - 1 - 2k)]$

$24 \times 2 = 2k + 5 + ( - 4)( - 4) - k( - 1 - 2k)$

$48 = 2k + 5 + 16 + k + 2 {k}^{2}$

$2 {k}^{2} + 3k + 21 - 48 = 0$

$2 {k}^{2} + 3k - 27 = 0$

By middle term splitting,

$2 {k}^{2} + (9 - 6)k - 27 = 0$

$2 {k}^{2} + 9k - 6k - 27 = 0$

$k(2k + 9) - 3(2k + 9) = 0$

$(2k + 9)(k - 3) = 0$

Either,

2k+9=0

2k= - 9

k =-9/2

Or,

k-3=0

k =3

Therefore the value of k is - 9/2 and 3