In triangle ABC the co ordinates of A are (3,2) and the coordinates of the midpoint of AC and AB are (2,-1)and (1,2) respectively find the co ordinates of mid point BC
Answers
Answer:
The coordinates of vertices A is (-5, 3), B is (1, 7) and C is (3, 1).
Step-by-step explanation:
It is given that,
ABC is a triangle where P(-2,5) is the midpoint of AB, Q(2,4) is the midpoint of BC and R(-1,2) is the midpoint of AC.
Let the coordinates of the vertices be
A = (x1, y1)
B = (x2, y2)
C = (x3, y3)
The formula for the midpoint (x, y) between two points (x1, y1) & (x2, y2) is given by
(x, y) = { , }
Therefore, based on the above formula, we can write
-2 = (x1+x2)/2 ⇒ x1 + x2 = -4 …… (i)
5 = (y1+y2)/2 ⇒ y1 + y2 = 10 …… (ii)
2 = (x2+x3)/2 ⇒ x2 + x3 = 4 ……. (iii)
4 = (y2+y3)/2 ⇒ y2 + y3 = 8 …… (iv)
-1 = (x1 + x3)/2 ⇒ x1 + x3 = -2 …… (v)
2 = (y1 + y3)/2 ⇒ y1 + y3 = 4 …… (vi)
Now,
Adding (i), (iii) & (v), we get
2 (x1 + x2 + x3) = -4 + 4 – 2 = -2
⇒ (x1 + x2 + x3) = -1 …… (vii)
And,
Adding (ii), (iv) & (vi), we get
2 (y1 + y2 + y3) = 10 + 8 + 4 = 22
⇒ (y1 + y2 + y3) = 11 …… (viii)
So,
From (vii) & (i), we get x3 = -1 – (-4) = -1+4 = 3
From (vii) & (iii), we get x1 = -1 – 4 = -5
From (vii) & (v), we get x2 = -1 – (-2) = -1 + 2 = 1
From (viii) & (ii), we get y3 = 11 – 10 = 1
From (viii) & (iv), we get y1 = 11 – 8 = 3
From (viii) & (vi), we get y2 = 11 – 4 = 7
Thus, the coordinates of vertices of the triangle ABC are A(-5, 3), B(1, 7) & C(3, 1).