Question

The co-ordinates of the vertex A of a square ABCD are (1, 2) and the equation of the diagonal BD is x + 2y = 10. Find the equation of the other diagonal and the co-ordinates of the centre of the square. google doesn't have proper answer so write on own pls

Answer 1

Given :-

• The co-ordinates of the vertex A of a square ABCD are (1, 2).
• The equation of the diagonal BD is x + 2y = 10.

To Find :-

• The equation of the other diagonal ?
• The co-ordinates of the centre of the square.

Solution :-

As we know that, slope of a line Ax + By + C = 0 is (-A/B) .

So,

→ Slope of the Diagonal BD with equation x + 2y - 6 = 0 is (-1/2).

Now, we also know that :-

• Diagonals of a square are perpendicular to each other.
• Product of the slopes of the perpendicular lines is (-1).

So, Diagonal BD is Perpendicular to Diagonal AC.

Than,

→ Slope of Diagonal BD * Slope of Diagonal AC = (-1)

→ (-1/2) * Slope of Diagonal AC = (-1)

→ Slope of Diagonal AC = 2.

Now, Diagonal AC passes through the vertex A(1, 2).

Therefore, Equation of AC :-

→ (y - y1) = m(x - x1)

→ (y - 2) = 2(x - 1)

→ y - 2 = 2x - 2

→ 2x - y - 2 + 2 = 0

→ 2x - y = 0

Hence, The equation of the other diagonal AC is 2x - y = 0 .

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Now, we have to find The co-ordinates of the centre of the square, or, the point of intersection of AC and BD.

Solve Both Equations of Diagonals :-

→ x + 2y = 10 ------- (1)

→ 2x - y = 0 --------- (2)

from (2) , we get,

→ 2x = y

Putting this value of y in Equation (1) , we get,

→ x + 2(2x) = 10

→ x + 4x = 10

→ 5x = 10

→ x = 2.

Therefore,

→ y = 2x = 2*2 = 4.

Hence, The co-ordinates of the centre of the square are (2,4).

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