Question

show that the points (3,-2) (-2,8) (0,4)
are collinear by using Heron's formula.​

Answer 1

Answer:

hope this helps u.......

Answer 2

A (3, -2)

B (-2, 8)

C (0, 4)

The points would be collinear only when the area of triangle formed by these points is equal to zero.

AB = $\sqrt{(-2-3)^{2} + (8+2)^{2} }$

     = $\sqrt{25+ 100}$

     = $\sqrt{125}$

     = $5\sqrt{5}$

BC = $\sqrt{(0+2)^{2} + (4-8)^{2} }$

     = $\sqrt{4+16}$

     = $\sqrt{20}$

     = $2\sqrt{5}$

CA = $\sqrt{(0-3)^{2} + (4+2)^{2} }$

     = $\sqrt{9+36}$

     = $\sqrt{45}$

     = $3\sqrt{5}$

So, now,  $S = \frac{5\sqrt{5} + 2\sqrt{5} + 3\sqrt{5} }{2}$

                   = $\frac{10\sqrt{5} }{2}$

                   = $5\sqrt{5}$

Now, using heron's formula,

$A = \sqrt{S (S-a) (S-b) (S-c)}$

   = $\sqrt{5\sqrt{5}(5\sqrt{5} - 5\sqrt{5}) (5\sqrt{5}-2\sqrt{5}) (5\sqrt{5}-3\sqrt{5}) }$

   = $\sqrt{5\sqrt{5} (0) (3\sqrt{5}) (2\sqrt{5})}$

   = $0$

Since, area of triangle is equal to zero, the points are collinear.