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 :

(a) Slope of BD

(b) equation of AC

(a) Given,

BD is equally inclined to both axes.

∴ Slope of BD = tan 45° = 1.

Hence, slope of BD = 1.

(b) Since, diagonals of a square are perpendicular and product of slope of perpendicular lines equals to -1.

∴ Slope of BD × Slope of AC = -1

⇒ 1 × Slope of AC = -1

⇒ Slope of AC = -1.

Since, diagonals meet at point (-2, -2).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = -1[x - (-2)]

⇒ y + 2 = -1[x + 2]

⇒ y + 2 = -x - 2

⇒ x + y + 2 + 2 = 0

⇒ x + y + 4 = 0.

Hence, equation of AC is x + y + 4 = 0.