If the area of a triangle is 35 sq. units wit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3

Vedclass · Accepted Answer

$12$,-$2$

Vedclass · Answer

(D) The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 35$.
Substituting the given vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$ into the formula:
$\frac{1}{2} |2(4-4) + 5(4 - (-6)) + k(-6-4)| = 35$
$\frac{1}{2} |2(0) + 5(10) + k(-10)| = 35$
$|50 - 10k| = 70$
This gives two possible cases:
Case $1$: $50 - 10k = 70 \implies -10k = 20 \implies k = -2$.
Case $2$: $50 - 10k = -70 \implies -10k = -120 \implies k = 12$.
Thus, the possible values for $k$ are $12$ and $-2$.

If the area of a triangle is $35$ sq. units with vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$, then $k$ is . . . . . . .

Similar Questions

Evaluate $\Delta = \begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix}$

$\left|\begin{array}{lll}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right|=$ . . . . . . .

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

If $\left|\begin{array}{lll}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{array}\right|=k(a-b)(b-c)(c-a)$ then $k=$

If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module