MathematicsThe area of a triangle with vertices (-3,0),(3,0) and (0,k) is 9 sq. units. Then the value of k will be

The area of a triangle with vertices $(- 3, 0), (3, 0)$ and $(0, k)$ is $9$ sq. units. Then the value of $k$ will be

A
$9$
B
$3$
C
$- 9$
D
$6$

Solution:

It is given that the coordinates of vertices of a triangle are

A − 3, 0, B 3, 0

and

C 0, k .

We know that if

A x 1, y 1, B x 2, y 2, and C x 3, y 3

are the vertices of a triangle, then its area is given by,

Area = 12 x 1 y 2 − y 3 + x 2 y 3 − y 1 + x 3 y 1 − y 2 (1)

The area of the given triangle is 9 square units.
Therefore from equation (1), we have,

Area = \frac{1}{2} | ` 1 [- 3 (0 - k) + 3 (k - 0) + 0 (0 - 0))] |

\Rightarrow 9 = \frac{1}{2} | [3 k + 3 k + 0] |

\Rightarrow 9 = \frac{1}{2} | 6 k |

\Rightarrow 18 = 6 k

\Rightarrow k = \frac{18}{6}

\Rightarrow k = 3

Therefore, the value of

k = 3 .

Hence, option (2) is correct.