The area of a triangle with vertices (-3,0), (3,0) and (0,k) is 9 sq. units. The value(s) of k is/are:

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B -3
We know that area of triangle whose vertices are
$(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is given by

$= ⎛ ⎜ ⎝ \frac{1}{2} ⎛ ⎜ ⎝ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \Rightarrow ⎛ ⎜ ⎝ \frac{1}{2} ⎛ ⎜ ⎝ \begin{matrix} - 3 & 0 & 1 3 & 0 & 1 0 & k & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 9 \Rightarrow - 3 (- k) - 0 + 1 (3 k) = \pm 18 \Rightarrow 6 k = \pm 18 \Rightarrow k = \pm 3$

Suggest Corrections

Similar questions

The are of $△ A B C$ with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14 sq units (b) 28 sq units (c) 8 sq units (d) 6 sq units

P A

P B

x^{2} + y^{2} = 4

P (3, 0) .

△ P A B

(0, 4), (0, 0)

(3, 0)

x^{2} - 3 y^{2} = 3

(\sqrt{3}, 0)

Join BYJU'S Learning Program