Question

Show that the points (8, 1) (3,-4) and (2,-5) are collinear using determinant.

Answer 1

The Principle

Three points, (x₁, y₁), (x₂, y₂), and (x₃, y₃), are collinear (lie on the same straight line) if and only if the area of the triangle they form is zero.

The area of a triangle with these vertices can be calculated using the following determinant formula:

Area = ½ | D |

Where D is the determinant:

$D = \begin{vmatrix} x₁ & y₁ & 1 \ x₂ & y₂ & 1 \ x₃ & y₃ & 1 \end{vmatrix}$

Therefore, the points are collinear if and only if D = 0.

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Step-by-Step Solution

Step 1: Assign the coordinates

Let the given points be:
(x₁, y₁) = (8, 1)
(x₂, y₂) = (3, -4)
* (x₃, y₃) = (2, -5)

Step 2: Set up the determinant

Substitute these coordinates into the determinant matrix:

$D = \begin{vmatrix} 8 & 1 & 1 \ 3 & -4 & 1 \ 2 & -5 & 1 \end{vmatrix}$

Step 3: Calculate the determinant

We will expand the determinant along the first row. The formula for this expansion is:

$D = a(ei - fh) - b(di - fg) + c(dh - eg)$

For our matrix:
$a=8, b=1, c=1$
$d=3, e=-4, f=1$
$g=2, h=-5, i=1$

$D = 8 \begin{vmatrix} -4 & 1 \ -5 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 1 \ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & -4 \ 2 & -5 \end{vmatrix}$

Now, calculate each of the 2x2 determinants:

• For the first term:
  $8 \times ((-4)(1) - (1)(-5)) = 8 \times (-4 - (-5)) = 8 \times (-4 + 5) = 8 \times (1) = \mathbf{8}$

• For the second term:
  $-1 \times ((3)(1) - (1)(2)) = -1 \times (3 - 2) = -1 \times (1) = \mathbf{-1}$

• For the third term:
  $1 \times ((3)(-5) - (-4)(2)) = 1 \times (-15 - (-8)) = 1 \times (-15 + 8) = 1 \times (-7) = \mathbf{-7}$

Step 4: Combine the results

Now, add the results from the three terms together:

$D = 8 + (-1) + (-7)$
$D = 8 - 1 - 7$
$D = 7 - 7$
$D = 0$

Conclusion

Since the value of the determinant is 0, the area of the triangle formed by the points (8, 1), (3, -4), and (2, -5) is zero. This proves that the points do not form a triangle and must lie on the same straight line.

Therefore, the points are collinear.