#### If the coordinates of the vertex A of a square ABCD are (3, -2) and the equation of diagonal BD is 3x – 7y + 6 = 0, find the equation of the diagonal AC. Also, find the coordinates of the center of the square.

**Solution:**

⇒ y − y_{1}= m(x − x_{1})

⇒ y−(−2)= − (x−3)

⇒ 3(y + 2) =−7(x−3)

⇒ 3y + 6 =−7x + 21

⇒ 7x + 3y = 15

Now we will find the coordinates of O, the points of intersection of AC and BD

⇒ 3x – 7y = -6 ….(i)

⇒ 7x + 3y = 15 ….(ii)

Multiplying (i) by 3 and (ii) by 7, we get,

⇒ 9x – 21y = -18 ….(iii)

⇒ 49x + 21y = 105 …(iv)

Adding (iii) and (iv) we get,

⇒ 9x + 49x – 21y + 21y = -18 + 105

⇒ 58x = 87

⇒ x = .

⇒ 3 × − 7y=−6

⇒ −7y=−6

⇒ + 6 = 7y

⇒ = 7y

⇒ = 7y

⇒ y =

Hence, the equation of AC is 7x + 3y – 15 = 0 and coordinates of the center are

**More Solutions:**

- Straight line passing through the intersection of 2x + 5y – 4 = 0.
- Line perpendicular from the point (1, -2) on the line 4x – 3y – 5 = 0.
- The line through (0, 0) and (2, 3) is parallel.
- The points A (1, 3), B (3, -1) and C (-5, -5) is a right-angled triangle.
- A (-1, 3), B (4, 2), C (3, -2) are the vertices of a triangle.
- In the adjoining diagram, the coordinates of the points A, B and C.