Find the area of a triangle with vertices A(1,2), B(4,6), and C(7,2) using determinants. Then verify by an independent method.
Solution — Step by Step
The signed area is
Area=21x1x2x3y1y2y3111Area=21147262111=1⋅(6⋅1−1⋅2)−2⋅(4⋅1−1⋅7)+1⋅(4⋅2−6⋅7)=1(4)−2(−3)+1(−34)=4+6−34=−24Area=21∣−24∣=12 square units
Base AC runs from (1,2) to (7,2) along y=2, so length =6. Height is the perpendicular distance from B(4,6) to the line y=2, which is 6−2=4.
Area=21×6×4=12
Matches. ✓
Area =12 square units.
Why This Works
The determinant formula encodes the cross product of two side vectors. For vectors AB and AC, the magnitude of AB×AC equals twice the triangle’s area. The 3×3 determinant with the column of 1s is just a tidy way to compute this in 2D.
If the determinant is zero, the three points are collinear — a useful collinearity test.
Alternative Method
Shoelace (Gauss) formula: Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣. Plug in: 21∣1(4)+4(0)+7(−4)∣=21∣4−28∣=12. Same result, often faster for coordinate geometry MCQs.
Common Mistake
Forgetting the absolute value and the factor of 1/2. Some students report the determinant value −24 as the area. The signed determinant tells us orientation; the area is always positive and equals half the magnitude.
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