Question

The area of a triangle with vertices(-3,0),(3,0) and (0,k) is 9 sq. units.The value of k will be(A)9(B)3(C)−9(D)6

Answer 1

Let the points be A(-3,0) , B(3,0) & C(0,k).

Here,

x1 = -3 & y1 = 0

x2 = 3 & y2 = 0

x3 = 0 & y3 = k

Using ,

ar.∆ = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

=> 9 = 1/2|(-3)(0-k) + 3(k-0) + 0(0-0)|

=> 9 = 1/2| 3k + 3k + 0 |

=> 9 = 1/2|6k|

=> 9 = 3k

=> 3k = 9

=> k = 9/3

=> k= 3 units.

Hence , the right option is (B).