Question

In triangle abc (3,1), (5,6), (-3,2) are midpoints of ab bc and cd respectively. Find the co-ordinates of a,b and c.


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Answer 1

Answer

Co-ordinates of A = (-5, 3)

Co-ordinates of B = (11, 5)

Co-ordinates of C = (-1, 7)

$\rule{200}2$

Explanation

(Refer the attachment for figure)

Let the coordinates of A be (x, y), of B be (p, q) and that of C be (r, s)

Then, applying section formula, we get

For AB,

3 = $\sf \frac{1(x) + 1(p)}{1 + 1}$

(since, the ratio in which (3, 1) divides AB is 1)

→ 3 = (x + p)/2

→ x + p = 6 ..... (i)

Similarly, for AC,

-3 = (x + r)/2

→ x + r = -6.....(ii)

And for BC,

5 = (p + r)/2

→ p + r = 10.....(iii)

Now, subtracting (ii) from (i),

x + p - (x + r) = 6 - (-6)

→ x + p - x + r = 12

→ p - r = 12......(iv)

Adding (iii) and (iv)

→ p - r + p + r = 10 + 12

→ 2p = 22

→ p = 11

Putting value of p in (iii)

p + r = 10

→ r = 10 - p

→ r = 10 - 11

→ r = - 1

Putting value of r in (ii),

x + r = -6

→ x - 1 = -6

→ x = -6 + 1

→ x = -5

Now, we got the x-coordinate of A, B and C. Now finding the y-coordinate.

Applying section formula for y coordinate.

For AB,

1 = (y + q)/2

→ y + q = 2.......(a)

For AC,

2 = (y + s)/2

→ y + s = 4 .........(b)

For BC,

6 = (q + s)/2

→ q + s = 12 ........(c)

Subtracting (b) from (a)

y + q - (y + s) = 2 - 4

→ q - s = -2.....(d)

Adding (c) and (d)

→ q + s + q - s = 12 - 2

→ 2q = 10

→ q = 5

Putting value of q in (c),

q + s = 12

→ s = 12 - q

→ s = 12 - 5

→ s = 7

Putting value of s in (b),

y + s = 4

→ y = 4 - s

→ y = 4 - 7

→ y = -3.

$\rule{50}1$

Hence, the coordinates of A = (x, y) = (-5, -3)

Hence, the coordinates of A = (x, y) = (-5, -3) The coordinates of B = (p, q) = (11, 5)

Hence, the coordinates of A = (x, y) = (-5, -3) The coordinates of B = (p, q) = (11, 5) The coordinates of C = (r, s) = (-1, 7)