Q.Find area of the triangle with vertices at the point given in each of the following :
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.
Using it
For A(2,3), B(7,5), C(4,8):
Area=21∣2(5−8)+7(8−3)+4(3−5)∣=21∣−6+35−8∣=221=10.5 sq units.
Watch out
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 two y's, so writing the points in order avoids sign slips.
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)∣.
(i) (1,0),(6,0),(4,3): 21∣1(0−3)+6(3−0)+4(0−0)∣=21∣−3+18∣=21(15)=7.5.
(ii) (2,7),(1,1),(10,8): 21∣2(1−8)+1(8−7)+10(7−1)∣=21∣−14+1+60∣=21(47)=23.5.
(iii) (−2,−3),(3,2),(−1,−8): 21∣(−2)(2+8)+3(−8+3)+(−1)(−3−2)∣=21∣−20−15+5∣=21(30)=15.
- 7.5 sq units;
- 23.5 sq units;
- 15 sq units.
Applying the coordinate area formula gives (i) 7.5, (ii) 23.5, and (iii) 15 square units.
Three points fix a triangle, and its area comes straight from the coordinates — no need to hunt for a base and a perpendicular height.
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
The absolute value keeps the area positive whatever order you list the vertices in.
(i) (1,0),(6,0),(4,3)
21∣1(0−3)+6(3−0)+4(0−0)∣=21∣−3+18+0∣=21(15)=7.5.
(ii) (2,7),(1,1),(10,8)
21∣2(1−8)+1(8−7)+10(7−1)∣=21∣−14+1+60∣=21(47)=23.5.
(iii) (−2,−3),(3,2),(−1,−8)
21∣(−2)(2−(−8))+3(−8−(−3))+(−1)(−3−2)∣=21∣−20−15+5∣=21(30)=15.
The bracket came out −30 here only because the vertices were listed clockwise; the absolute value corrects the sign.
- 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.
- KCET 2022Set C-41 markMCQQ.If the vertices of a triangle are (−2,6),(3,−6) and (1,5), then the area of the triangle is (A) 15.5 sq. units (B) 30 sq. units (C) 35 sq. units (D) 40 sq. units
›Reveal solutionSolution
Plug the three vertices into the determinant/shoelace area formula and take half the absolute value.
Step 1 — The formula and why the modulus matters
For vertices (x1,y1),(x2,y2),(x3,y3) the area of the triangle is
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
which is just 21 times the absolute value of the determinant
x1x2x3y1y2y3111
The modulus is essential: the determinant is signed (its sign records whether the vertices were listed clockwise or anticlockwise), while an area must be positive.
Step 2 — Assign the coordinates
(x1,y1)=(−2,6),(x2,y2)=(3,−6),(x3,y3)=(1,5)
Step 3 — Compute each term carefully
x1(y2−y3)=−2(−6−5)=−2(−11)=22
x2(y3−y1)=3(5−6)=3(−1)=−3
x3(y1−y2)=1(6−(−6))=1(12)=12
Step 4 — Sum and halve
Area=21∣22+(−3)+12∣=21∣31∣=231=15.5
Step 5 — Sanity check
The sum came out positive and non-zero, so the three points are not collinear and a genuine triangle exists. A quick order-of-magnitude check: the points span roughly 5 units in x and 12 in y, so a bounding box of about 60 square units — a triangle of 15.5 sits comfortably inside that, which is consistent.
Area=15.5 sq. units
✓Final answerThe correct option is (A) — 15.5 sq. units.
ANSWER: A
- COMEDK 2024Set 2024-E1 markMCQQ.The area of the triangle whose vertices are (−2,a)(2,−6) and (5,4) is 35 sq units then the value of 'a' is (A) 4 (B) 353 (C) −323 (D) 3128
›Reveal solutionSolution
Using the determinant formula for the area of a triangle given three vertices, we set the absolute value equal to 35 and solve for a. The two possible values are a=4 and a=−3128; only a=4 appears among the options.
Concept & Intuition
The area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) can be found using the shoelace formula:
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
This works because the expression inside the absolute value is twice the signed area of the parallelogram formed by two sides of the triangle. The absolute value ensures we get a positive area regardless of vertex order. Here we know the area is 35, so we set the absolute value equal to 70 (twice the area) and solve for the unknown coordinate a.
Step-by-step solution
-
Label the points
Let A=(−2,a), B=(2,−6), C=(5,4).
-
Apply the area formula
Area=21∣(−2)(−6−4)+2(4−a)+5(a−(−6))∣.
-
Simplify inside the absolute value
- First term: (−2)(−10)=20
- Second term: 2(4−a)=8−2a
- Third term: 5(a+6)=5a+30 Sum: 20+8−2a+5a+30=58+3a.
-
Set up the equation
Since area = 35, we have
21∣58+3a∣=35⇒∣58+3a∣=70.
-
Solve the absolute value equation
This gives two cases:
- 58+3a=70 → 3a=12 → a=4.
- 58+3a=−70 → 3a=−128 → a=−3128.
-
Check against the options
The given choices are:
(A) 4
(B) 353
(C) −323
(D) 3128
Only a=4 matches an option. The other solution −3128 is not listed.
Watch outA common mistake is forgetting the absolute value and solving only 58+3a=70, missing the second solution. However, since this is multiple-choice, only one of the two solutions appears, so checking options is sufficient.
TipThe determinant formula is symmetric: swapping rows only changes the sign, not the absolute value. So you can always arrange points in any order — just remember the absolute value.
✓Final answerThe correct option is (A).
ANSWER: A
-
- KCET 2026Set UNKNOWN1 markMCQQ.The area of the triangle with vertices (3,8),(−4,2) and (5,1) is 4P, then the value of P is (A) 261 (B) 612 (C) 122 (D) 1221
›Reveal solutionSolution
Use the coordinate area formula for a triangle, then equate it to P/4 to solve for P.
Step 1 — Apply the area formula
For vertices (x1,y1)=(3,8), (x2,y2)=(−4,2), (x3,y3)=(5,1):
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Step 2 — Substitute values
=21∣3(2−1)+(−4)(1−8)+5(8−2)∣=21∣3(1)+(−4)(−7)+5(6)∣
=21∣3+28+30∣=21(61)=261
Step 3 — Solve for P
Given Area =4P:
4P=261⟹P=261×4=122
✓Final answerThe correct option is (C) — P=122.
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