Area = (1/2)|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|.

Find Area of Triangle with Vertices (1,2), (3,4), (5,0)

Question

Find the area of the triangle whose vertices are A(1, 2), B(3, 4), and C(5, 0).

Solution — Step by Step

The area of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ is:

\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

The modulus sign is non-negotiable — area is always positive, but the formula can give a negative value depending on the order of vertices.

Assign clearly so you don’t mix up during substitution:

$x_1 = 1,\ y_1 = 2$
$x_2 = 3,\ y_2 = 4$
$x_3 = 5,\ y_3 = 0$

\text{Area} = \frac{1}{2} |1(4 - 0) + 3(0 - 2) + 5(2 - 4)|

Work out each bracket separately before multiplying — this is where most errors happen.

= \frac{1}{2} |1(4) + 3(-2) + 5(-2)|

= \frac{1}{2} |4 - 6 - 10|

= \frac{1}{2} |-12|

= \frac{1}{2} \times 12 = 6

Area of the triangle = 6 square units.

Why This Works

This formula is actually the determinant method in disguise. When we place three points in a coordinate plane, the signed area of the triangle they form equals half the cross product of two edge vectors. The determinant captures exactly this.

The reason we subtract coordinates in pairs — $(y_2 - y_3)$ , $(y_3 - y_1)$ , $(y_1 - y_2)$ — is because each term represents one edge’s contribution to the “sweep” area. Notice these three differences always sum to zero: $(y_2 - y_3) + (y_3 - y_1) + (y_1 - y_2) = 0$ . This is a quick sanity check you can do mid-solution.

The absolute value at the end simply handles whether we listed the vertices clockwise or anticlockwise. Both orderings give the same magnitude, different sign.

Alternative Method — Using the ½ |base × height| Approach

We can use the basic formula if we compute the base and height geometrically.

Base AB: Distance between A(1, 2) and B(3, 4):

AB = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{4 + 4} = 2\sqrt{2}

Height from C(5, 0) to line AB:

Line AB has slope $\frac{4-2}{3-1} = 1$ , so its equation is $y - 2 = 1(x - 1)$ , giving $x - y + 1 = 0$ .

Distance from C(5, 0) to this line:

h = \frac{|5 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}

Area:

= \frac{1}{2} \times 2\sqrt{2} \times 3\sqrt{2} = \frac{1}{2} \times 6 \times 2 = 6 \text{ sq. units}

Same answer. The determinant method is significantly faster — use it always in board and JEE Main.

In CBSE board exams, the determinant formula is the expected method and gets full step marks. Writing it as a $3 \times 3$ determinant is also accepted:

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

Both forms are equivalent.

Common Mistake

The most frequent error is dropping the modulus sign and reporting a negative area. If you list vertices in clockwise order, the formula gives $-6$ , and students write “Area = $-6$ sq. units” and lose marks. Area is a physical quantity — it cannot be negative. Always apply $|\ |$ before writing the final answer.

A second common slip: mixing up which $y$ -coordinates go with which $x$ -coordinates during substitution. Write the assignment $x_1, y_1, x_2, y_2, x_3, y_3$ explicitly before plugging in — takes 10 seconds and saves the entire question.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the ½ |base × height| Approach

Common Mistake

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