Exercise 7.3 — Area of Triangle

Area of the triangle and collinearity of points.

Lesson Notes PDF

1 / —

Loading PDF…

Exercise 7.3 — Area of a Triangle Using Coordinates

Exercise 7.3 from Chapter 7, Coordinate Geometry, of Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) teaches you how to find the area of a triangle directly from the coordinates of its three vertices — without ever needing to draw the triangle or measure anything. The exercise also introduces Heron's formula as an alternative route to the same answer, and shows how the very same area formula doubles up as a quick test for collinearity — checking whether three points lie on a single straight line.

Across nine problems, you will calculate triangle areas from given vertices, find an unknown coordinate that makes three points collinear, discover a neat property of the triangle formed by joining midpoints, split a quadrilateral into two triangles to find its area, and verify the coordinate method against Heron's formula.

Area by Coordinates Collinearity Test Heron's Formula Quadrilateral Area

💡 Foundation fact: For a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃), the area is ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Every problem in this exercise builds on this single formula — and on the fact that this expression equals exactly zero when the three points are collinear.

The Three Tools in This Exercise

Before working through the problems, it helps to have all three formulas side by side. Each one answers a slightly different question, and recognising which one a problem is asking for is half the battle.

1. Area from coordinates — when you know all three vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
This comes from drawing perpendiculars from each vertex to the x-axis, which splits the region into three trapeziums — adding and subtracting their areas in the right combination leaves exactly this expression.

2. Heron's formula — when you know only the three side lengths a, b, c:
Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) / 2
Here "s" is the semi-perimeter — literally half the perimeter of the triangle. If the vertices are given instead of the sides, simply use the distance formula on each pair of vertices first.

3. Collinearity condition — three points A, B, C are collinear (lie on one line) exactly when:
x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0
This is simply the area formula with the area set to zero — geometrically, three collinear points can never enclose any area, since they don't form a "real" triangle at all.

📌 Why the absolute value? Depending on whether you list the vertices clockwise or anticlockwise, the bracketed expression can come out negative. Since a physical area can never be negative, we always take the absolute value before multiplying by ½.

Question 1 — Find the Area of the Triangle Whose Vertices Are Given

Each part below is a direct application of the coordinate formula. The pattern is always the same: label the vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) in the order given, substitute carefully — watching every sign — and simplify.

Question 1(i)

Vertices: (2, 3), (−1, 0), (2, −4)

△ABC plotted from its three vertices

Let A(x₁,y₁) = (2, 3), B(x₂,y₂) = (−1, 0), C(x₃,y₃) = (2, −4) Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = ½ |2(0−(−4)) + (−1)(−4−3) + 2(3−0)| = ½ |2(4) + (−1)(−7) + 2(3)| = ½ |8 + 7 + 6| = ½ × 21 = 10.5 sq. units

Question 1(ii)

Vertices: (−5, −1), (3, −5), (5, 2)

△ABC plotted from its three vertices

Let A = (−5, −1), B = (3, −5), C = (5, 2) Area = ½ |(−5)(−5−2) + 3(2−(−1)) + 5(−1−(−5))| = ½ |(−5)(−7) + 3(3) + 5(4)| = ½ |35 + 9 + 20| = ½ × 64 = 32 sq. units

Question 1(iii)

Vertices: (0, 0), (3, 0), (0, 2)

A right triangle sitting at the origin

Let A = (0, 0), B = (3, 0), C = (0, 2) Area = ½ |0(0−2) + 3(2−0) + 0(0−0)| = ½ |0 + 6 + 0| = ½ × 6 = 3 sq. units (Quick check: this triangle has legs of length 3 and 2 along the axes, so ½ × base × height = ½ × 3 × 2 = 3 — matches!)

Question 2 — Find the Value of 'K' for Which the Points Are Collinear

Three points are collinear when the "triangle" they form has zero area. So for each part, we set the area expression equal to zero and solve the resulting linear equation for k.

Question 2(i)

Points: (7, −2), (5, 1), (3, k)

All three points fall on one line when k = 4

Setting area = 0: x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) = 0 7(1−k) + 5(k−(−2)) + 3(−2−1) = 0 7 − 7k + 5k + 10 − 9 = 0 −2k + 8 = 0 ⟹ k = 4

Question 2(ii)

Points: (8, 1), (k, −4), (2, −5)

All three points fall on one line when k = 3

8(−4−(−5)) + k(−5−1) + 2(1−(−4)) = 0 8(1) − 6k + 2(5) = 0 8 − 6k + 10 = 0 18 − 6k = 0 ⟹ k = 3

Question 2(iii)

Points: (k, k), (2, 3), (4, −1)

All three points fall on one line when k = 7/3

k(3−(−1)) + 2(−1−k) + 4(k−3) = 0 4k − 2 − 2k + 4k − 12 = 0 6k − 14 = 0 k = 14/6 = 7/3

⚠️ Most common error: Losing a minus sign while expanding brackets like k(−5−1) or 2(−1−k). Write each bracket out fully before combining like terms — rushing this step is where most marks are lost in Question 2.

Question 3 — Area of the Triangle Formed by the Midpoints

Given a triangle with vertices (0, −1), (2, 1) and (0, 3), this question asks for the area of the smaller triangle formed by joining the midpoints of its three sides — and then asks how that smaller area compares to the original.

△ABC (solid blue) with medial triangle △DEF (dashed orange)

Solution

Step 1 — Area of the original triangle ABC

A(0,−1), B(2,1), C(0,3) Area = ½ |0(1−3) + 2(3−(−1)) + 0(−1−1)| = ½ |0 + 8 + 0| = ½ × 8 = 4 sq. units

Step 2 — Find the midpoints D, E, F

D = midpoint of BC = ((2+0)/2, (1+3)/2) = (1, 2) E = midpoint of AC = ((0+0)/2, (−1+3)/2) = (0, 1) F = midpoint of AB = ((0+2)/2, (−1+1)/2) = (1, 0)

Step 3 — Area of the medial triangle DEF

Area = ½ |1(1−0) + 0(0−2) + 1(2−1)| = ½ |1 + 0 + 1| = ½ × 2 = 1 sq. unit

✅ Conclusion: Area of △DEF : Area of △ABC = 1 : 4.

💡 General rule worth remembering: The triangle formed by joining the midpoints of any triangle's sides — called the medial triangle — always has exactly ¼ of the original triangle's area, no matter what shape the original triangle is. This 1 : 4 ratio is a standard board-exam fact, not a coincidence of this particular example.

Question 4 — Area of a Quadrilateral Using a Diagonal Split

The coordinate formula only works directly on triangles — so to find the area of a quadrilateral, we draw one diagonal to split it into two triangles, find each triangle's area separately, and add the results. The vertices here, taken in order, are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).

Diagonal AC splits ABCD into △ABC and △ACD

Solution

Split using diagonal AC

Area of △ABC = ½ |(−4)(−5−(−2)) + (−3)(−2−(−2)) + 3(−2−(−5))| = ½ |(−4)(−3) + (−3)(0) + 3(3)| = ½ |12 + 0 + 9| = 21/2 Area of △ACD = ½ |(−4)(−2−3) + 3(3−(−2)) + 2(−2−(−2))| = ½ |(−4)(−5) + 3(5) + 2(0)| = ½ |20 + 15 + 0| = 35/2 Area of ABCD = Area(△ABC) + Area(△ACD) = 21/2 + 35/2 = 56/2 = 28 sq. units

📌 Why "taken in order" matters: The vertices of a quadrilateral must be listed in the order you'd meet them walking around the boundary (either clockwise or anticlockwise). Picking a diagonal that joins two non-adjacent vertices in this cyclic order — like AC here — guarantees the two triangles share only that diagonal as an edge and don't overlap.

Question 5 — Area Using Heron's Formula

This question deliberately asks for Heron's formula instead of the coordinate formula, using the triangle with vertices (2, 3), (6, 3) and (2, 6). The first job is to convert the three vertices into three side lengths using the distance formula.

A 3–4–5 right triangle, with sides a, b, c labelled

Solution

Step 1 — Find the three side lengths

a = BC = √[(2−6)² + (6−3)²] = √[16 + 9] = √25 = 5 b = AC = √[(2−2)² + (6−3)²] = √[0 + 9] = √9 = 3 c = AB = √[(2−6)² + (3−3)²] = √[16 + 0] = √16 = 4

Step 2 — Semi-perimeter and Heron's formula

s = (a + b + c) / 2 = (5 + 3 + 4) / 2 = 12/2 = 6 Area = √[s(s−a)(s−b)(s−c)] = √[6(6−5)(6−3)(6−4)] = √[6 × 1 × 3 × 2] = √36 = 6 sq. units

✅ Cross-check: Sides 3, 4, 5 satisfy 3² + 4² = 5², so this is a right triangle with legs 3 and 4. Using the simple right-triangle formula: ½ × 3 × 4 = 6 sq. units — exactly matching Heron's formula. This kind of cross-check is a great way to confidently verify your answer in the exam.

Common Mistakes to Avoid

Dropping a negative sign: Expressions like x₃(y₁ − y₂) involve double subtraction — when y₂ is itself negative, you're subtracting a negative number. Write out every bracket fully rather than simplifying mentally.
Forgetting the absolute value: The raw bracketed expression can be negative if vertices are listed clockwise instead of anticlockwise. Area itself is never negative — always take the modulus before reporting your final answer.
Mixing up the two formulas: Use the coordinate formula when you're given (x, y) vertices directly, and Heron's formula when you're given (or have just calculated) the three side lengths. Question 5 specifically tests whether you reach for Heron's formula even though you could solve it with coordinates too.
Splitting a quadrilateral with the wrong pair of vertices: A diagonal must join two vertices that are not adjacent in the listed order. Joining adjacent vertices does not split the quadrilateral into two triangles at all.
Arithmetic slips inside the square root: In Heron's formula, a single sign or subtraction error inside the brackets propagates straight into the final square root. Compute s, then each of (s−a), (s−b), (s−c) separately and double-check before multiplying.

⛔ High-risk exam trap: In collinearity problems (Question 2), students often substitute into the area formula but then forget to actually solve for k — they simplify down to an equation like "−2k + 8 = 0" and stop, without isolating k. Always finish the algebra and state the numeric value of k clearly.

Quick Reference — All Answers at a Glance

Question	What's Being Found	Answer
Q1(i)	Area of triangle (2,3), (−1,0), (2,−4)	10.5 sq. units
Q1(ii)	Area of triangle (−5,−1), (3,−5), (5,2)	32 sq. units
Q1(iii)	Area of triangle (0,0), (3,0), (0,2)	3 sq. units
Q2(i)	Value of k for collinear points (7,−2),(5,1),(3,k)	k = 4
Q2(ii)	Value of k for collinear points (8,1),(k,−4),(2,−5)	k = 3
Q2(iii)	Value of k for collinear points (k,k),(2,3),(4,−1)	k = 7/3
Q3	Medial triangle area & ratio to original	1 sq. unit; ratio 1 : 4
Q4	Area of quadrilateral (−4,−2),(−3,−5),(3,−2),(2,3)	28 sq. units
Q5	Area via Heron's formula, triangle (2,3),(6,3),(2,6)	6 sq. units

Frequently Asked Questions

Why do we take the absolute value in the area-of-a-triangle formula?

Because the raw expression inside the bars can come out negative or positive depending on whether the vertices happen to be listed in clockwise or anticlockwise order. Since a physical area can never be negative, the modulus (absolute value) is applied before multiplying by ½.

Can the same formula be used to test if points are collinear?

Yes — that's exactly what Question 2 in this exercise relies on. Three points are collinear precisely when the area "enclosed" by them is zero, so setting the bracketed expression equal to zero and solving gives the condition for collinearity.

What is the difference between the coordinate formula and Heron's formula?

The coordinate formula needs the (x, y) positions of all three vertices. Heron's formula instead needs the three side lengths. If you're given vertices but asked specifically for Heron's formula (as in Question 5), first use the distance formula to convert each pair of vertices into a side length.

How do you find the area of a quadrilateral using coordinates?

Draw one diagonal connecting two non-adjacent vertices, which splits the quadrilateral into two triangles. Find the area of each triangle with the coordinate formula, then add the two areas together, as shown in Question 4.

Does the order in which I list the vertices change the final answer?

No — swapping the order can flip the sign of the expression inside the bars, but after taking the absolute value, the final area is identical. The one place order genuinely matters is the quadrilateral-splitting method, where vertices should be taken in their natural cyclic (walk-around-the-shape) order so the diagonal produces two non-overlapping triangles.

What This Exercise Prepares You For

The coordinate skills sharpened in Exercise 7.3 — careful substitution, sign-handling, and comfort with the distance formula — carry directly into Exercise 7.4 (Slope of a Line), the next exercise in this chapter, which uses a very similarly structured formula to measure how steeply a line rises or falls. The collinearity test from Question 2 also reappears whenever later problems ask you to verify whether three given points genuinely form a triangle or instead lie on one straight line.

More broadly, the technique of splitting an irregular shape into triangles (Question 4) is a strategy you'll reuse whenever coordinate geometry asks for the area of any polygon with more than three sides, and the medial-triangle result from Question 3 occasionally resurfaces in similarity and ratio-based problems in later chapters.

📐 Board Exam Tip (Telangana & AP): The area-of-a-triangle formula is one of the highest-yield formulas in the entire coordinate geometry chapter — examiners frequently combine it with collinearity checks (Question 2 style) or quadrilateral-splitting (Question 4 style) in 4-mark questions. Memorise the formula in the exact order x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) so you can write it from memory without hesitation.