If the mid points of sides of ΔABC are (2,3),(3,7) and (−2,1), then the possible coordinates of its vertices are
A
(−1,5)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(−3,−3)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(7,9)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(5,−1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C(7,9) Let A(x1,y1),B(x2,y2),C(x3,y3) M1(2,3) is midpoint of side BC M2(3,7) is midpoint of side AC M3(−2,1) is midpoint of side AB
Adding equation (1),(2) and (3), 2(x1+x2+x3)=6⇒x1+x2+x3=3⋯(4)
Using equation (1) and (4), x3=7
Using equation (2) and (4), x1=−1
Using equation (3) and (4), x2=−3
Similarly, for y−coordinates y1+y2=2⋯(5)y2+y3=6⋯(6)y3+y1=14⋯(7)
Adding equation (5),(6) and (7) y1+y2+y3=11⋯(8)
Using equation (5) and (8), y3=9
Using equation (6) and (8), y1=5
Using equation (7) and (8), y2=−3
Hence, the coordinates of the triangle vertex are A=(−1,5),B=(−3,−3),C=(7,9)
Alternate Solution :
Let A(x1,y1),B(x2,y2),C(x3,y3) M1(2,3) is midpoint of side BC M2(3,7) is midpoint of side AC M3(−2,1) is midpoint of side AB
We know that, BM1M2M3 is a parallelogram.
And its diagonal bisect each other, (x2+32,y2+72)=(2−22,3+12)⇒x2=−3,y2=−3∴B=(−3,−3)
Similarly, CM1M3M2 is a parallelogram. (x3−22,y3+12)=(2+32,3+72)⇒x3=7,y3=9∴C=(7,9)
Similarly, AM3M1M2 is a parallelogram. (x1+22,y1+32)=(3−22,7+12)⇒x1=−1,y1=5∴A=(−1,5)