Let the points of intersection of the lines x | vedclass

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

Question

(D) The intersection points of the given lines are the midpoints of the sides of triangle $ABC$. Let these points be $D(1, 2)$,$E(7, 5)$,and $F(2, 3)$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The intersection points of the given lines are the midpoints of the sides of triangle $ABC$. Let these points be $D(1, 2)$,$E(7, 5)$,and $F(2, 3)$.The area of the triangle formed by the midpoints $(\Delta DEF)$ is given by:$\Delta DEF = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$\Delta DEF = \frac{1}{2} |1(5 - 3) + 7(3 - 2) + 2(2 - 5)|$$\Delta DEF = \frac{1}{2} |1(2) + 7(1) + 2(-3)|$$\Delta DEF = \frac{1}{2} |2 + 7 - 6| = \frac{1}{2} |3| = 1.5$The area of the original triangle $ABC$ is $4$ times the area of the triangle formed by joining the midpoints:$	ext{Area}(ABC) = 4 	imes \Delta DEF = 4 	imes 1.5 = 6$.

Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .... .

Similar Questions

The four points whose coordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form:

If a straight line $L$ is perpendicular to the line $4x - 2y = 1$ and forms a triangle of area $4 \text{ sq unit}$ with the coordinate axes,then the equation of the line $L$ is

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively, is:

The bisectors of the angles of a parallelogram enclose a

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module