Points A (-3, -2), B (-6, a), C (-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.

Given:

Points A(-3, -2), B(-6, a), C(-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD and AB = CD.

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Distance between points A(-3, -2) and B(-6, a):

$= \sqrt{((-6) - (-3))^2 + (a - (-2))^2}\\[1em] = \sqrt{(-3)^2 + (a - (-2))^2}\\[1em] = \sqrt{9 + a^2 + 4 + 4a}\\[1em] = \sqrt{13 + a^2 + 4a}$

Distance between points C (-3, -4) and D(0, -1):

$= \sqrt{(0 - (-3))^2 + ((-1) - (-4))^2}\\[1em] = \sqrt{3^2 + 3^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}$

Since AB = CD,

$⇒ \sqrt{13 + a^2 + 4a} = \sqrt{18}\\[1em] ⇒ 13 + a^2 + 4a = 18\\[1em] ⇒ 13 + a^2 + 4a - 18 = 0\\[1em] ⇒ a^2 + 4a - 5 = 0\\[1em] ⇒ a^2 + 5a - 1a - 5 = 0\\[1em] ⇒ (a^2 + 5a) - (1a + 5) = 0\\[1em] ⇒ a(a + 5) - 1(a + 5) = 0\\[1em] ⇒ (a + 5)(a - 1) = 0\\[1em] ⇒ a = -5 \text{ or } 1$

Since it is given that a is negative, we select a = -5.

Hence, the value of a = -5.