Question

3) If (3,-3) (3, -1) (1,1) are the mid points
of the sides BC, CA and AB of the triangle then
ABC are
a) A (1,3) B (+ 1-1) c(5,-5) b)A(1,-3) B (1,-1) C(5,5)
C) A (1,3) B(1,-1) C(5,5)
dJ AC1,3 18C1,1)C(5,5)​

Answer 1

Solution :-

Figure :-

   $\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A\ (x_1,y_1) }\put(0.5,-0.3){ \bf B }\put(5.2,-0.3){ \bf C }\put(1.93,1.45){ \large\bullet }\put(3.9,1.45){ \large\bullet }\put(2.9,-0.1){ \large\bullet }\put(0,-0.8){ \bf (x_2,y_2) }\put(4.8,-0.8){ \bf (x_3,y_3) }\put(0.4,1.5){ \bf(1,1) }\put(2.4,-0.6){ \bf(3,-3) }\put(4.3,1.5){ \bf(3,-1) }\end{picture}$

Midpoint of AC = ( 3 , -1 )

$\longrightarrow \sf \left( \dfrac{x_1+x_3}{2} \ , \ \dfrac{y_1+y_3}{2} \right) = ( 3 , -1 )$

$\bullet \sf \ \dfrac{x_1+x_3}{2} = 3$

$\longrightarrow \sf x_1+x_3 = 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(1)$

$\bullet \sf \ \dfrac{y_1+y_3}{2} = -1$

$\longrightarrow \sf y_1+y_3 = -2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(2)$

Midpoint of AB = ( 1 , 1 )

$\longrightarrow \sf \left( \dfrac{x_1+x_2}{2} \ , \ \dfrac{y_1+y_2}{2} \right)= (1,1)$

$\bullet \sf \ \dfrac{x_1+x_2}{2} = 1$

$\longrightarrow \sf x_1+x_2 = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(3)$

$\bullet \sf \ \dfrac{y_1+y_2}{2} =1$

$\longrightarrow \sf y_1+y_2 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(4)$

Midpoint of BC = ( 3 , -3 )

$\longrightarrow \sf \left( \dfrac{x_2+x_3}{2} \ , \ \dfrac{y_2+y_3}{2} \right)= (3 , -3 )$

$\bullet \sf \ \dfrac{x_2+x_3}{2}= 3$

$\longrightarrow \sf x_2+x_3 = 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(5)$

$\bullet \sf \ \dfrac{y_2+y_3}{2} =-3$

$\longrightarrow \sf y_2+y_3 =-6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(6)$

Adding eq(1), eq(3) and eq(5) :-

$\longrightarrow \sf 2(x_1+x_2+x_3) =2+6+6 \\\\\longrightarrow 2(x_1+x_2+x_3) = 14 \\\\\longrightarrow x_1+x_2+x_3 = 7$

Substitute $\sf x_2+x_3=6$,

$\longrightarrow \sf x_1+6 = 7$

$\longrightarrow\underline{\boxed{\bf x_1 = 1}}$

Substitute $\sf x_1 = 1$ in eq(1) :-

$\longrightarrow \sf 1+x_3 = 6$

$\longrightarrow \underline{\boxed{\bf x_3 = 5}}$

Substitute $\sf x_3 = 5$ in eq(5) :-

$\longrightarrow \sf x_2+5=6$

$\longrightarrow \sf \underline{\boxed{\bf x_2=1}}$

Adding eq(2), eq(4) and eq(6) :-

$\longrightarrow \sf 2(y_1+y_2+y_3) = -2+2-6 \\\\\longrightarrow 2(y_1+y_2+y_3) = -6 \\\\\longrightarrow y_1+y_2+y_3 = -3$

Substitute $\sf y_2+y_3 = -6$,

$\longrightarrow \sf y_1-6 = -3$

$\longrightarrow\underline{\boxed{\bf y_1 =3}}$

Substitute $\sf y_1 =3$ in eq(2) :-

$\longrightarrow \sf 3+y_3 = -2 \\\\\longrightarrow \underline{\boxed{\bf y_3=-5}}$

Substitute $\sf y_3 = -5$ in eq(6) :-

$\longrightarrow \sf y_2-5 = -6$

$\longrightarrow \underline{\boxed{\bf y_2 = -1}}$

• Coordinates of A = ( 1 , 3 )

• Coordinates of B = ( 1 , -1 )

• Coordinates of C = ( 5 , -5 )

Hence option A is correct.