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.
- GUJCET 2025Set 031 markMCQQ.Find the area of a triangle given that midpoints of its sides are (2,7), (1,1) and (10,8). (A) 447 (B) 47 (C) 94 (D) 247
›Reveal solutionSolution
Original area =4× area of the triangle formed by the midpoints.
Midpoint triangle area with (2,7),(1,1),(10,8):
21∣2(1−8)+1(8−7)+10(7−1)∣=21∣−14+1+60∣=247.
Original =4×247=94.
✓Final answer(C) 94
ANSWER: (C)
- GUJCET 2023Set 091 markMCQQ.If A(K,1), B(2,4) and C(1,1) are the vertices of the △ABC, such that area of the △ABC is 6 unit, then K= ______. (A) −5 and 3 (B) 5 and −3 (C) 3 and −1 (D) 5 and 3
›Reveal solutionSolution
[!TLDR]
λ≈1.53 A˚.
Concept
An electron accelerated from rest through potential difference V gains kinetic energy eV, giving momentum p=2meV and de Broglie wavelength
λ=2meVh=V12.27 A˚(V in volts).
Solution
λ=6412.27=812.27=1.53 A˚.
[!ANSWER]
(C)
- GUJCET 2021Set 151 markMCQQ.A(1,3), B(0,0) and C(k,0) are vertices of △ABC. If area of △ABC is 3 units, then k is . (A) 2 (B) 0 (C) −2 (D) ±2
›Reveal solutionSolution
With base BC on the x-axis, area =21∣k∣× height.
Concept: B(0,0) and C(k,0) lie on the x-axis, so BC=∣k∣; the height is the y-coordinate of A, i.e. 3.
Area=21∣k∣(3)=3⇒∣k∣=2⇒k=±2.
✓Final answer(D) ±2
ANSWER: (D)
- GUJCET 2024Set 131 markMCQQ.If area of △PQR is 3 sq. units with vertices P(k,1),Q(2,4) and R(1,1). Then value of k is __________. (A) 1,3 (B) −3,1 (C) −1,3 (D) 0,2
›Reveal solutionSolution
Setting the triangle area to 3 gives ∣3k−3∣=6, so k=3 or k=−1.
Concept. Area =21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
Steps. For P(k,1),Q(2,4),R(1,1):
Area=21∣k(4−1)+2(1−1)+1(1−4)∣=21∣3k−3∣=3.
∣3k−3∣=6⇒3k−3=±6⇒k=3 or k=−1.
✓Final answer(C) −1,3
ANSWER: (C)
- GUJCET 2026Set x1 markMCQQ.If area of triangle is 35 sq. units with vertices (2,−6),(5,4) and (k,4), then k is ______ (A) 12 (B) −12,−2 (C) −2 (D) 12,−2
›Reveal solutionSolution
Use the triangle-area determinant and solve the resulting absolute-value equation.
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
With (2,−6),(5,4),(k,4):
=21∣2(4−4)+5(4+6)+k(−6−4)∣=21∣0+50−10k∣=35.
So ∣50−10k∣=70:
- 50−10k=70⇒k=−2
- 50−10k=−70⇒k=12
Thus k=12,−2.
✓Final answerk=12,−2
ANSWER: (D)
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.If the area of the triangle with vertices (k,0), (4,0) and (0,2) is 4 square units, then the value of k is ____.(a) 0,−8(b) 8,−8(c) only 0(d) 0,8
›Reveal solutionSolution
Use the determinant formula for area of a triangle and solve for k, keeping both sign cases from the modulus.
Area =21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣ with (k,0),(4,0),(0,2):
Area =21∣k(0−2)+4(2−0)+0(0−0)∣=21∣−2k+8∣=4⇒∣−2k+8∣=8.
−2k+8=8⇒k=0, or −2k+8=−8⇒k=8.
✓Final answerThe correct option is (d) 0,8.
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.If the area of the triangle with vertices (−2,0), (0,4), (0,k) is 4 square units, then the value of k = ____.(a) −8(b) 0,−8(c) 0,8(d) None of the given options
›Reveal solutionSolution
Use the determinant formula for the area of a triangle and solve for k, remembering the modulus can resolve two ways.
Area =21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣. With (−2,0),(0,4),(0,k):
Area =21∣(−2)(4−k)+0(k−0)+0(0−4)∣=21∣2k−8∣=4
∣2k−8∣=8⇒2k−8=8 or 2k−8=−8⇒k=8 or k=0.
✓Final answerThe correct option is (c) 0,8.
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If area of a triangle is 3 sq. units with the vertices (3,5),(2,2) and (k,2), then k= ______.(a) 0,4(b) 0,−4(c) 3,1(d) −3,1
›Reveal solutionSolution
Set up the coordinate-geometry area formula and solve the resulting absolute-value equation.
Area =21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
=21∣3(2−2)+2(2−5)+k(5−2)∣=21∣3k−6∣.
Set equal to 3: ∣3k−6∣=6⇒3k−6=±6⇒k=4 or k=0.
✓Final answerThe correct option is (a) 0,4.
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.If area of triangle is 4 sq. units with vertices (−2,0), (0,4) and (0,k), then k is ___.(a) 0(b) −8(c) 0,8(d) 0,−8
›Reveal solutionSolution
Use the determinant area formula for the three vertices and set it to 4.
Area =21x1(y2−y3)+x2(y3−y1)+x3(y1−y2) with (−2,0),(0,4),(0,k):
=21(−2)(4−k)+0+0=21⋅2∣k−4∣=∣k−4∣.
Set ∣k−4∣=4⇒k−4=±4⇒k=8 or k=0.
✓Final answer(c) 0,8.
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.If area of triangle is 35 sq. units with vertices (2,−6),(5,4) and (k,4), then k= ___(a) 1.2(b) −20(c) −12,−2(d) 12,−2
›Reveal solutionSolution
Use the determinant formula for the area of a triangle with given vertices, set it equal to 35, and solve the resulting absolute-value equation (both cases give valid k).
Area =21x1(y2−y3)+x2(y3−y1)+x3(y1−y2) with (2,−6),(5,4),(k,4):
=212(4−4)+5(4−(−6))+k(−6−4)=210+50−10k=21∣50−10k∣.
Set equal to 35: ∣50−10k∣=70, so 50−10k=70 giving k=−2, or 50−10k=−70 giving k=12.
✓Final answer(d) k=12,−2.
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