Question

The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k.

Answer 1

Let A(1, 2), B(-5, 6), C(7, -4) and D(k, -2) be the given points

Firstly, area of ∆ABC is given by

= \(\frac{1}{2}\)|(1)(6 + 4) – 5(-4 + 2) + 7(2 – 6)|

= \(\frac{1}{2}\)|10 + 30 – 28|

= \(\frac{1}{2}\) x 12

= 6

Now, the area of ∆ACD is given by

= \(\frac{1}{2}\)|(1)(-4 + 2) + 7( – 2 – 2) + k(2 + 4)|

= \(\frac{1}{2}\)|- 2 + 7x(-4) + k(6)|

= (- 30 + 6k)/2

= -15 + 3k

= 3k – 15

Thus, the area of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)

= 6 + 3k – 15

= 3k – 9

But, given area of quadrilateral is O.

So, 3k – 9 = 0

k = \(\frac{9}{3}\) = 3