Let the points of intersections of the lines x - y + 1 = 0 ,   x - 2 y + 3 = 0 and 2 x - 5 y + 11 = 0 are the mid points of the sides of a triangle ABC . Then the area of the triangle ABC is

Question

Let the points of intersections of the lines x - y + 1 = 0 , &#160; x - 2 y + 3 = 0 and 2 x - 5 y + 11 = 0 are the mid points of the sides of a triangle ABC . Then the area of the triangle ABC is

Aicharya · Accepted Answer

If $P, Q, R$ are mid-points of sides of $	riangle ABC$ then $4 	imes Area(	riangle PQR) = Area(	riangle ABC)$ Area of $	riangle ABC = 4 	imes \frac{1}{2} \left| \frac{\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix}^2}{C_1 C_2 C_3} ight|$ Where $C_1, C_2, C_3$ are cofactors of $c_1, c_2, c_3$ $	riangle ABC = 4 	imes \frac{1}{2} \left| \frac{\begin{vmatrix} 1 & -1 & 1 \ 1 & -2 & 3 \ 2 & -5 & 11 \end{vmatrix}^2}{(-1)(3)(-1)} ight|$ $= 2 	imes \frac{9}{3}$ $= 6$

Let the points of intersections of the lines x - y + 1 = 0 ,   x - 2 y + 3 = 0 and 2 x - 5 y + 11 = 0…

Solution

Let the points of intersections of the lines x - y + 1 = 0 , &#160; x - 2 y + 3 = 0 and 2 x - 5 y + 11 = 0…

Solution

Let the points of intersections of the lines x - y + 1 = 0 , x - 2 y + 3 = 0 and 2 x - 5 y + 11 = 0…