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

If $p + q + r = 0$ and $a + b + c = 0$,then the value of the determinant $\left| \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} \right|$ is

The cubic equation $\left| \begin{array}{ccc} 0 & a-x & b-x \\ -a-x & 0 & c-x \\ -b-x & -c-x & 0 \end{array} \right| = 0$ has a repeated root in $x$. Then:

The roots of the equation $\left| \begin{matrix} x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x \end{matrix} \right| = 0$ are equal to

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,where $0 \le \theta < 2\pi$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module