Q.Find area of the triangle with vertices at the point given in each of the following :
(i) (1,0),(6,0),(4,3)
(ii) (2,7),(1,1),(10,8)
(iii) (−2,−3),(3,2),(−1,−8)
Sikkim CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question
Concept understanding — Area Of Triangle By Coordinates
Area of a Triangle from Coordinates
Given three vertices — say A(2,3), B(7,5), C(4,8) — you could try base × height, but a slanted triangle makes the height awkward to find. Coordinates give the area directly and exactly, because area is fundamentally a determinant.
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Here (x1,y1),(x2,y2),(x3,y3) are the vertices in any order, and the absolute value keeps the area positive.
Where it comes from
The expression inside the bars is the 3×3 determinant
x1x2x3y1y2y3111,
whose expansion is exactly x1(y2−y3)+x2(y3−y1)+x3(y1−y2). A 2×2 determinant gives the area of the parallelogram spanned by two sides, and a triangle is half of it — which is where the 21 comes from. The column of 1's lets the triangle sit anywhere, not just at the origin.
Keep the absolute value — area is never negative — and never drop the 21. Note the cyclic pattern: each xi multiplies the difference of the other twoy's, so writing the points in order avoids sign slips.
Tip
Collinearity test: if the three points lie on one line, the area comes out 0. Try (1,2),(3,4),(5,6) — you get 0. The same idea extends to any polygon (the shoelace formula).
This determinant-based technique for the area of a triangle is a recurring theme in the NCERT Class 11 Straight Lines and Class 12 Determinants chapters, and is frequently tested as a standalone 'area of triangle by coordinates' short-answer question in CBSE boards and JEE Main. Students searching 'area of triangle formula class 11 maths' or looking for a quick collinearity check will find this determinant form is exactly what most important-questions lists point to.
Use the coordinate area formula Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
The bracket came out −30 here only because the vertices were listed clockwise; the absolute value corrects the sign.
✓Final answer
7.5 sq units;
23.5 sq units;
15 sq units.
Method: Area of a Triangle From Three Coordinate Points
The standard technique for any "find the area given the vertices" question.
Steps
Step 1: Label the vertices consistently
Assign (x1,y1),(x2,y2),(x3,y3) to the three given points, in the order given.
Step 2: Apply the coordinate area formula
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
Step 3: Simplify the expression fully inside the modulus
Compute each bracketed difference, multiply, and add — only take the absolute value at the very end, never partway through.
Step 4: Report the area as a positive number with units
State the final area in square units; if the bracket came out negative, that only reflects the (clockwise) order the vertices were listed in, not a negative area.
Common Mistakes
Mistake 1: Dropping the absolute value and reporting a negative "area"
Why it's wrong: the bracketed expression x1(y2−y3)+x2(y3−y1)+x3(y1−y2) can come out negative depending on the order the vertices are listed in (clockwise vs anticlockwise), but area itself is never negative. Correct approach: always take the absolute value of the bracketed expression as the very last step, regardless of its sign.
Mistake 2: Mismatching which coordinate is x and which is y when substituting into the formula
Why it's wrong: transposing a point's coordinates (using y1 where x1 belongs, for instance) produces an entirely different — and wrong — numeric answer, even though the arithmetic afterward looks clean. Correct approach: write out (x1,y1),(x2,y2),(x3,y3) explicitly next to the given points before substituting, so each value goes into its correct slot.