Application of Determinants to Coordinate Geometry

Determinants can be used in coordinate geometry mainly to:

: find the area of a triangle
: check whether three points are collinear
: find the equation of a line using collinearity

Consistent Solution: A system is consistent if it has at least one solution.

Inconsistent Solution: A system is inconsistent if it has no solution.

If the vertices of a triangle are \[ (x_1, y_1), \quad (x_2, y_2), \quad (x_3, y_3), \]

then \[ \boxed{\text{Area} = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \]

Important Remarks

• Area is always non-negative, so the absolute value of the determinant is taken.
• If the area is given, the determinant may have positive or negative value during calculation.
• If three points are collinear, the area of the triangle formed by them is zero.

Three points \[ (x_1, y_1), \quad (x_2, y_2), \quad (x_3, y_3) \]

are collinear if \[ \boxed{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0} \]

because the area of the triangle formed by collinear points is zero.

Consider the system:

\[ a_1x + b_1y + c_1z = d_1 \]

\[ a_2x + b_2y + c_2z = d_2 \]

\[ a_3x + b_3y + c_3z = d_3 \]

Let

\[ A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}. \]

Then the system can be written as \[ \boxed{AX = B} \]

Case I: A is Non-Singular

\[ |A| \neq 0 \]

Then \[A^{-1}\] exists and: \[ \boxed{X = A^{-1}B} \]

System has a unique solution.

Case II: A is Singular

\[ |A| = 0 \]

Then \[A^{-1}\] does not exist. Further consistency is checked using \[(\text{adj } A)B\].

Solve the system of equations

Solution: The system of equations can be written in the form

Now,

Note that

Therefore

i.e.

Hence

The sum of three numbers is 6. If we multiply the third number by 3 and add the second number to it, we get 11. By adding the first and third numbers, we get twice the second number. Represent it algebraically and find the numbers using the matrix method.

Solution: Let the first, second and third numbers be denoted by\[ x, y\] and\[ z \], respectively.

Then, according to the given conditions, we have

This system can be written as

Here

Hence \[ adj \text{ A} = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \]