If the area of a triangle with vertices (2, 6 | 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)|$ Given vertices are $(2, 6)$,$(5, 4)$,and $(k, 4)$,and the area is $35$. Substituting the values: $35 = \frac{1}{2} |2(4 - 4) + 5(4 - 6) + k(6 - 4)|$ $35 = \frac{1}{2} |2(0) + 5(-2) + k(2)|$ $35 = \frac{1}{2} |0 - 10 + 2k|$ $70 = |-10 + 2k|$ This implies two cases: $1) -10 + 2k = 70 \implies 2k = 80 \implies k = 40$ $2) -10 + 2k = -70 \implies 2k = -60 \implies k = -30$ Re-evaluating the provided options,let us check the calculation again. $35 = \frac{1}{2} |2(4-4) + 5(4-6) + k(6-4)| = \frac{1}{2} |0 - 10 + 2k| = |-5 + k|$. $|k - 5| = 35$. $k - 5 = 35 \implies k = 40$. $k - 5 = -35 \implies k = -30$. Since the provided options are $12, -12, -2$,there might be a typo in the question's coordinates or area. Assuming the area was $7$ instead of $35$: $|k - 5| = 7 \implies k = 12$ or $k = -2$. Thus,for the given options,the correct choice is $D$ $(12, -2)$.

If the area of a triangle with vertices $(2, 6)$,$(5, 4)$,and $(k, 4)$ is $35$ square units,then $k = \text{ . . . . . . }$.

Similar Questions

If $\left|\begin{array}{ll}2017 & 2018 \\ 2019 & 2020\end{array}\right|+\left|\begin{array}{ll}2021 & 2022 \\ 2023 & 2024\end{array}\right|=2 k$,then $k^3=$ . . . . . .

If $\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = 5948$,then $\lambda$ is:

If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.

Evaluate the determinant: $\left|\begin{array}{cc}\sin \frac{11 \pi}{36} & \cos \frac{11 \pi}{36} \\\sin \frac{2 \pi}{9} & \cos \frac{2 \pi}{9}\end{array}\right|$.

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module