Question

Square ABCD lies in the third quadrant of a XY plane such that its vertex A is at (–3, –1) and the diagonal DB produced is equally inclined to both the axes. The diagonals AC and BD meets at P(–2, –2). Find the:

1. slope of BD
2. equation of AC

Answer 1

Given:

Square ABCD with A = (–3, –1).

Diagonals AC and BD meet at P = (–2, –2) the centre / midpoint of both diagonals.

Step-wise calculation:

1. P is midpoint of AC.

So, C = (2·P_x – A_x, 2·P_y – A_y)

= (2(–2) – (–3), 2(–2) – (–1)) 

= (–1, –3)

2. Slope of AC = `(y_C - y_A)/(x_C - x_A)`

= `(-3 - (-1))/(-1 - (-3))` 

= `(-2)/2`

= –1

3. In a square the diagonals are perpendicular, so slope(BD) = negative reciprocal of slope(AC) = –1’s negative reciprocal = 1.

4. Equation of AC: slope = –1 and passing through A(–3, –1):

y – (–1) = –1(x – (–3))

⇒ y + 1 = –1(x + 3)

⇒ y = –x – 4 or x + y + 4 = 0

Slope of BD = 1.

Equation of AC: y = x + y + 4 = 0.