KnowledgeBoat Logo
|

Mathematics

The mid-points of the sides BC, CA and AB of ΔABC are D(2, 1), E(-1, -3) and F(4, 5) respectively. Find the co-ordinates of A, B and C.

Section Formula

2 Likes

Answer

Let the vertices of ΔABC be A(x1, y1), B(x2, y2), and C(x3, y3).

The mid-points of the sides BC, CA and AB of ΔABC are D(2, 1), E(-1, -3) and F(4, 5) respectively. Find the co-ordinates of A, B and C. Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Given,

D(2, 1) is the midpoint of BC.

Substitute values we get,

⇒x2+x32=2⇒x2+x3=4 ….(1)⇒y2+y32=1⇒y2+y3=2 ….(2)\Rightarrow \dfrac{x2 + x3}{2} = 2 \\[1em] \Rightarrow x2 + x3 = 4 \text{ ….(1)} \\[1em] \Rightarrow \dfrac{y2 + y3}{2} = 1 \\[1em] \Rightarrow y2 + y3 = 2 \text{ ….(2)}

Given,

E(-1, -3) is the midpoint of CA.

⇒x3+x12=−1⇒x3+x1=−2 …..(3)⇒y3+y12=−3⇒y3+y1=−6 ….(4)\Rightarrow \dfrac{x3 + x1}{2} = -1 \\[1em] \Rightarrow x3 + x1 = -2 \text{ …..(3)} \\[1em] \Rightarrow \dfrac{y3 + y1}{2} = -3 \\[1em] \Rightarrow y3 + y1 = -6 \text{ ….(4)}

Given,

F(4, 5) is the midpoint of CA.

⇒x1+x22=4⇒x1+x2=8 ….(5)⇒y1+y22=5⇒y1+y2=10 ….(6)\Rightarrow \dfrac{x1 + x2}{2} = 4 \\[1em] \Rightarrow x1 + x2 = 8 \text{ ….(5)} \\[1em] \Rightarrow \dfrac{y1 + y2}{2} = 5 \\[1em] \Rightarrow y1 + y2 = 10 \text{ ….(6)}

Adding the three equations (1), (3) and (5), we get :

⇒ (x2 + x3) + (x3 + x1) + (x1 + x2) = 4 + (-2) + 8

⇒ 2x1 +2x2 + 2x3 = 10

⇒ 2(x1 +x2 + x3) = 10

⇒ (x1 +x2 + x3) = 102\dfrac{10}{2}

⇒ (x1 +x2 + x3) = 5 …..(7)

Subtract (Eq. 3) from (Eq. 7) :

⇒ (x1 +x2 + x3) - ( x2 + x3) = 5 - 4

⇒ (x1 +x2 + x3 -x2 - x3) = 5 - 4

⇒ x1 = 1.

Subtract (Eq. 2) from (Eq. 7) :

⇒ (x1 +x2 + x3) - ( x3 + x1) = 5 - (-2)

⇒ (x1 +x2 + x3 -x3 - x1) = 5 + 2

⇒ x2 = 7.

Subtract (Eq. 5) from (Eq. 7):

⇒ (x1 +x2 + x3) - ( x1 + x2) = 5 - 8

⇒ (x1 +x2 + x3 -x1 - x2) = -3

⇒ x3 = -3.

Adding equations (2), (4) and (5), we get :

⇒ (y2 + y3) + (y3 + y1) + (y1 + y2) = 2 + (-6) + 10

⇒ 2y1 +2y2 + 2y3 = 6

⇒ 2(y1 +y2 + y3) = 6

⇒ (y1 +y2 + y3) = 62\dfrac{6}{2}

⇒ (y1 +y2 + y3) = 3 …..(8)

Subtract (Eq. 2) from (Eq. 8):

⇒ (y1 +y2 + y3) - ( y2 + y3) = 3 - 2

⇒ (y1 +y2 + y3 -y2 - y3) = 1

⇒ y1 = 1.

Subtract (Eq. 4) from (Eq. 8):

⇒ (y1 + y2 + y3) - ( y3 + y1) = 3-(-6)

⇒ (y1 +y2 + y3 -y3 - y1) = 3 + 6

⇒ y2 = 9.

Subtract (Eq. 6) from (Eq. 8):

⇒ (y1 + y2 + y3) - ( y1 + y2) = 3 - 10

⇒ (y1 +y2 + y3 -y1 - y2) = -7

⇒ y3 = -7.

⇒ A = (x1, y1) = (1, 1), B = (x2, y2) = (7, 9), C = (x3, y3) = (-3, -7).

Hence, A = (1, 1), B = (7, 9) and C = (-3, -7).

Answered By

1 Like

Related Questions