The area of a triangle is 13 sq. units whose | vedclass

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

Question

(B) 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: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) 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: 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 $A(8, 2)$,$B(k, 4)$,and $C(6, 7)$ and Area $= 13$. Substituting the values: $13 = \frac{1}{2} |8(4 - 7) + k(7 - 2) + 6(2 - 4)|$ $26 = |8(-3) + 5k + 6(-2)|$ $26 = |-24 + 5k - 12|$ $26 = |5k - 36|$ This gives two cases: Case $1$: $5k - 36 = 26 \implies 5k = 62 \implies k = 12.4$ Case $2$: $5k - 36 = -26 \implies 5k = 10 \implies k = 2$ Since $k$ must be an integer,the value is $k = 2$.

The area of a triangle is $13$ sq. units whose vertices are $A(8, 2)$,$B(k, 4)$,and $C(6, 7)$. Then,the integer value of $k$ is . . . . . . .

Similar Questions

The sum of distinct values of $\lambda$ for which the system of equations
$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions,is

$\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{array} \right| = $

If $\left| \begin{array}{ccc} 3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8 \end{array} \right| = 0$,then the values of $x$ are

If $a \neq 1, b \neq -1, c \neq -1$ and the system of equations $x = a(y+z), y = b(z+x), z = c(x+y)$ has a non-trivial solution,then:

The roots of the equation $\left| \begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array} \right| = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module