Question

In triangle ABC, P(-2,5) is the mid point of AB. Q(2,4) is the mid point of BC and R(-1,2)
is the mid point of AC. Calculate the co-ordinates of vertices A, B and C.
​

Answer 1

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) = {$\frac{x1+x2}{2}$ , $\frac{y1+y2}{2}$}

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).

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Answer 2

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) = {\frac{x1+x2}{2}

2

x1+x2

, \frac{y1+y2}{2}

2

y1+y2

}

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