CBSE Class 9 Mathematics Quadrilaterals Worksheet Set 03

Choose the Correct Answer

Question 1. Which of the following has two pairs of adjacent sides equal?
(a) Parallelogram
(b) Rectangle
(c) Trapezium
(d) Kite
Answer: (d) Kite
In simple words: A kite has neighboring sides that match in length two by two, whereas parallelograms and rectangles have matching opposite sides.

Exam Tip: Do not confuse adjacent side equality with opposite side equality; kites feature two distinct pairs of equal neighboring sides.

Question 2. D, E and F are the midpoints of the sides of \( \Delta ABC \) as shown in the figure. If \( DF = 5\text{ cm} \) then \( AC = \) ______
(a) \( 2.5\text{ cm} \)
(b) \( 6\text{ cm} \)
(c) \( 10\text{ cm} \)
Answer: (c) 10 cm
By applying the Midpoint Theorem, a line segment that connects the midpoints of two sides of any triangle runs parallel to the third side and measures exactly half of its length.
Here, points \( D \) and \( F \) serve as the midpoints of sides \( AB \) and \( BC \) respectively.
Therefore:
\( DF = \frac{1}{2}AC \)

\( \implies AC = 2 \times DF = 2 \times 5\text{ cm} = 10\text{ cm} \)
In simple words: The segment joining the two midpoints is half as long as the base it faces, so double 5 cm to get 10 cm.

Exam Tip: Be mindful of which side is being determined; multiply by 2 when computing the triangle's outer side from the midpoint segment.

Question 3. Which of the following equation does not pass through the origin?
(a) \( y = 3x \)
(b) \( x + y = 5 \)
(c) \( x - y = 0 \)
(d) \( x = 4y \)
Answer: (b) \( x + y = 5 \)
A straight line crosses through the origin \( (0, 0) \) only when its coordinates satisfy the equation.
Testing \( (0, 0) \) in each given equation:
(a) \( 0 = 3(0) \implies 0 = 0 \) (passes through the origin)
(b) \( 0 + 0 = 0 \neq 5 \) (fails to satisfy, hence does not pass through the origin)
(c) \( 0 - 0 = 0 \) (passes through the origin)
(d) \( 0 = 4(0) \implies 0 = 0 \) (passes through the origin)
In simple words: An equation without an isolated constant number always passes through (0, 0), but x + y = 5 leaves a remaining 5.

Exam Tip: Any linear relationship taking the form \( ax + by = 0 \) passes through the origin; a non-zero constant term indicates it does not.

Question 4. An equation of the form \( y = k \) represents a line parallel to the x axis and it meet the Y axis at
(a) \( (0, k) \)
(b) \( (k, 0) \)
(c) \( (0, 0) \)
Answer: (a) \( (0, k) \)
Along the entire vertical Y-axis, the x-coordinate is constantly \( 0 \). Setting \( x = 0 \) in the equation \( y = k \) yields the intersection point \( (0, k) \).
In simple words: Since every spot on the vertical axis has an x-value of zero, the horizontal line y = k intersects it at (0, k).

Exam Tip: Points on the Y-axis always take the coordinate form \( (0, y) \), whereas points resting on the X-axis appear as \( (x, 0) \).

Question 5. ABCD is a \( \|^{gm} \). Diagonals meet at O. If \( BD = 10\text{ cm} \) and \( AC = 12\text{ cm} \), the value of \( x \) is
(a) 7cm
(b) 8cm
(c) 10cm
Answer: (a) 7cm
The diagonals belonging to any parallelogram cut each other directly in half.
Thus, point \( O \) is the center of diagonal \( BD \):
\( OD = \frac{1}{2}BD = \frac{10}{2} = 5\text{ cm} \)
From the provided diagram, the length of segment \( OD \) is labeled as \( x - 2 \).
Equating these values:
\( x - 2 = 5 \)

\( \implies x = 5 + 2 = 7\text{ cm} \)
In simple words: The diagonal is split into two equal 5 cm parts, so adding 2 to 5 gives x = 7 cm.

Exam Tip: Check whether the variable expression corresponds to the full diagonal or only one of the bisected halves before setting up the equation.

Question 6. ABCD is a rhombus. Diagonals AC and BD meet at O such that \( AC = 48\text{ cm} \) and \( BD = 14\text{ cm} \). Find the perimeter of Rhombus ABCD.
Answer: The diagonals of a rhombus intersect perpendicularly and bisect each other.
Consequently, inside the right-angled triangle \( \Delta AOB \):
\( OA = \frac{AC}{2} = \frac{48}{2} = 24\text{ cm} \)
\( OB = \frac{BD}{2} = \frac{14}{2} = 7\text{ cm} \)
Using the Pythagorean theorem in \( \Delta AOB \):
\( AB^2 = OA^2 + OB^2 \)
\( AB^2 = 24^2 + 7^2 = 576 + 49 = 625 \)

\( \implies AB = \sqrt{625} = 25\text{ cm} \)
Because all four sides of a rhombus share identical lengths:
\( \text{Perimeter} = 4 \times \text{side} = 4 \times 25\text{ cm} = 100\text{ cm} \)
In simple words: Halve the diagonals to get legs of 24 cm and 7 cm, use Pythagoras to find a 25 cm side, and multiply by 4 to get a 100 cm perimeter.

Exam Tip: The combination (7, 24, 25) forms a well-known Pythagorean triple; recognizing it helps calculate the side length swiftly.

Question 7. The adjacent angles of a parallelogram are in the ratio 2 : 3. Find the angles of the \( \|^{gm} \).
Answer: Any two consecutive interior angles in a parallelogram sum up to \( 180^\circ \).
Let the measures of the two adjacent angles be \( 2x \) and \( 3x \).
Then:
\( 2x + 3x = 180^\circ \)
\( 5x = 180^\circ \)

\( \implies x = \frac{180^\circ}{5} = 36^\circ \)
Now compute each angle:
\( 2x = 2 \times 36^\circ = 72^\circ \)
\( 3x = 3 \times 36^\circ = 108^\circ \)
Because opposite angles of a parallelogram are identical, the complete set of four angles is \( 72^\circ, 108^\circ, 72^\circ, 108^\circ \).
In simple words: Divide 180 degrees into 5 equal parts of 36 degrees; the two angle measures are 72 degrees and 108 degrees.

Exam Tip: Be sure to write down all four angles of the parallelogram, not just the two adjacent ones.

Question 8. One of the diagonals of a rhombus is equal to the sides. Find the angles of the rhombus.
Answer: Let \( ABCD \) denote the rhombus with side length \( a \), so that \( AB = BC = CD = DA = a \).
Suppose diagonal \( BD = a \).
In triangle \( \Delta ABD \), all three boundaries are congruent:
\( AB = AD = BD = a \)
This indicates that \( \Delta ABD \) is an equilateral triangle, meaning:
\( \angle A = 60^\circ \)
Since opposite angles of a rhombus match:
\( \angle C = \angle A = 60^\circ \)
Consecutive angles are supplementary:
\( \angle B = 180^\circ - \angle A = 180^\circ - 60^\circ = 120^\circ \)
Likewise, the opposite angle is:
\( \angle D = \angle B = 120^\circ \)
Thus, the four angles of the rhombus are \( 60^\circ, 120^\circ, 60^\circ, 120^\circ \).
In simple words: The diagonal splits the rhombus into two equilateral triangles with 60-degree corners, making the adjacent corners 120 degrees.

Exam Tip: Recognizing that a triangle with three equal sides is equilateral immediately establishes the first angle as 60°.

Question 9. D, E and F are the midpoints of the sides of \( \Delta ABC \). Prove that perimeter of \( \Delta DEF = \frac{1}{2}\text{ perimeter of }\Delta ABC \).
Answer: Let \( D, E, \) and \( F \) represent the midpoints of sides \( AB, BC, \) and \( CA \) respectively in \( \Delta ABC \).
According to the Midpoint Theorem, the segment joining the midpoints of any two sides of a triangle equals half of the opposite parallel side:
\( DE = \frac{1}{2}AC \)
\( EF = \frac{1}{2}AB \)
\( DF = \frac{1}{2}BC \)
Summing these three side lengths gives the perimeter of \( \Delta DEF \):
\( \text{Perimeter of }\Delta DEF = DE + EF + DF \)
\( = \frac{1}{2}AC + \frac{1}{2}AB + \frac{1}{2}BC \)
\( = \frac{1}{2}(AB + BC + CA) \)

\( \implies \text{Perimeter of }\Delta DEF = \frac{1}{2}(\text{Perimeter of }\Delta ABC) \)
This completes the proof.
In simple words: Every side of the inner triangle is exactly half as long as an outer side, so the entire inner boundary is half the outer boundary.

Exam Tip: Clearly state the Midpoint Theorem for all three pairs of sides before combining them into the final perimeter formula.

Question 10. AD and BE are medians of \( \Delta ABC \) and \( BE \parallel DF \). Prove that \( CF = \frac{1}{4}AC \)
Answer: Given that \( AD \) and \( BE \) are medians of \( \Delta ABC \):
1. \( D \) is the midpoint of \( BC \), meaning \( BD = DC \).
2. \( E \) is the midpoint of \( AC \), meaning \( CE = \frac{1}{2}AC \).
Now examine triangle \( \Delta CBE \):
Point \( D \) marks the midpoint of base side \( BC \), and the segment \( DF \) is parallel to side \( BE \).
By the Converse of the Midpoint Theorem, a line passing through the midpoint of one side of a triangle parallel to a second side must bisect the remaining side.
Hence, \( F \) is the midpoint of line segment \( CE \):
\( CF = \frac{1}{2}CE \)
Substituting \( CE = \frac{1}{2}AC \) into this relationship:
\( CF = \frac{1}{2}\left(\frac{1}{2}AC\right) = \frac{1}{4}AC \)
Hence proved.
In simple words: Since E is halfway along AC, and F is halfway along EC, point F sits at one-fourth of the total distance along AC.

Exam Tip: Explicitly mention that you are applying the Converse of the Midpoint Theorem specifically inside triangle \( \Delta CBE \).

Question 11. Represent Geometrically: \( 2x + 5 = 4x + 1 \)
(a) on number line
(b) in Cartesian plane.
Answer: First, isolate the variable in the equation:
\( 2x + 5 = 4x + 1 \)
\( 5 - 1 = 4x - 2x \)
\( 4 = 2x \)

\( \implies x = 2 \)

(a) On the number line:
In one dimension, the equation \( x = 2 \) is denoted by a single point located at coordinate 2 on the real number line. (b) In the Cartesian plane:
Expressed in two variables, it becomes \( 1\cdot x + 0\cdot y = 2 \). This defines a vertical line running parallel to the Y-axis at a constant distance of 2 units to its right, crossing through \( (2, 0) \). In simple words: The value x = 2 shows up as a single dot on a simple line, but becomes a full vertical line when drawn on a two-dimensional grid.

Exam Tip: Express the equation as \( x + 0\cdot y = 2 \) to justify why it traces an entire straight vertical line in two dimensions.

Question 11. Diagonals of a square are equal and bisects each other at right angles. Prove.
Answer: Let \( ABCD \) be a square whose diagonals \( AC \) and \( BD \) cross at point \( O \).

Part 1: Diagonals are equal (\( AC = BD \))
In triangles \( \Delta ABC \) and \( \Delta BAD \):
- \( AB = BA \) (shared side)
- \( \angle ABC = \angle BAD = 90^\circ \) (square corner angles)
- \( BC = AD \) (sides of a square)
By the SAS congruence rule, \( \Delta ABC \cong \Delta BAD \).

\( \implies AC = BD \) (via CPCT).

Part 2: Diagonals bisect each other
Because every square is inherently a parallelogram, its diagonals naturally bisect one another:
\( OA = OC \quad \text{and} \quad OB = OD \)

Part 3: Diagonals meet at right angles (\( AC \perp BD \))
In triangles \( \Delta AOB \) and \( \Delta COB \):
- \( OA = OC \) (diagonals bisect)
- \( AB = CB \) (equal sides of square)
- \( OB = OB \) (shared side)
By the SSS congruence criterion, \( \Delta AOB \cong \Delta COB \).

\( \implies \angle AOB = \angle COB \) (via CPCT).
Since \( \angle AOB + \angle COB = 180^\circ \) along the straight line \( AC \):
\( 2\angle AOB = 180^\circ \implies \angle AOB = 90^\circ \)
Thus, the diagonals are equal and bisect each other at right angles.
In simple words: Congruent corner triangles prove the diagonals have equal length, and congruent inner triangles prove they cross at 90 degrees.

Exam Tip: Address all three requirements methodically: diagonal equality, mutual bisection, and perpendicular intersection.

Question 12. If the diagonals of a \( \|^{gm} \) are equal, prove that it is a rectangle.
Answer: Consider a parallelogram \( ABCD \) with equal diagonals such that \( AC = BD \).
Compare triangles \( \Delta ABC \) and \( \Delta DCB \):
- \( AB = DC \) (opposite boundaries of a parallelogram)
- \( BC = CB \) (common base)
- \( AC = DB \) (given equal diagonals)
By SSS congruence, \( \Delta ABC \cong \Delta DCB \).

\( \implies \angle ABC = \angle DCB \) (by CPCT).
Since \( AB \parallel DC \) with transversal line segment \( BC \), the interior angles on the same side are supplementary:
\( \angle ABC + \angle DCB = 180^\circ \)
Substituting \( \angle DCB = \angle ABC \):
\( 2\angle ABC = 180^\circ \)

\( \implies \angle ABC = 90^\circ \)
A parallelogram containing at least one right angle is a rectangle.
Hence, \( ABCD \) is a rectangle.
In simple words: Proving the two large overlapping triangles are identical shows the bottom two angles are equal, and because they must sum to 180, each is 90 degrees.

Exam Tip: Conclude your proof by citing the formal definition: a parallelogram possessing one right interior angle is classified as a rectangle.

Question 13. Each angle of a rectangle is a right angle. Prove.
Answer: By geometric definition, a rectangle is a parallelogram containing one angle equal to \( 90^\circ \).
Let \( ABCD \) denote the rectangle with \( \angle A = 90^\circ \).
Because opposite sides are parallel (\( AD \parallel BC \)) with transversal \( AB \):
\( \angle A + \angle B = 180^\circ \) (co-interior angles)
\( 90^\circ + \angle B = 180^\circ \implies \angle B = 90^\circ \)
Opposite angles in any parallelogram are equal:
\( \angle C = \angle A = 90^\circ \)
\( \angle D = \angle B = 90^\circ \)
Therefore, \( \angle A = \angle B = \angle C = \angle D = 90^\circ \), confirming that each angle of a rectangle is a right angle.
In simple words: Parallel lines force neighboring corners to sum to 180 degrees, turning every single corner into a 90-degree right angle.

Exam Tip: Combine the co-interior angle rule with opposite angle equality to establish all four angles as 90° cleanly.

Question 14. D, E and F are the midpoints of the sides AB, BC and AC of \( \Delta ABC \). Prove that \( \square BEFD \) is a \( \|^{gm} \).
Answer: In triangle \( \Delta ABC \), points \( D \) and \( F \) are the respective midpoints of sides \( AB \) and \( AC \).
Applying the Midpoint Theorem:
\( DF \parallel BC \quad \text{and} \quad DF = \frac{1}{2}BC \)
Since \( E \) is given as the midpoint of segment \( BC \):
\( BE = \frac{1}{2}BC \)
This directly implies:
\( DF \parallel BE \quad \text{and} \quad DF = BE \)
Whenever one pair of opposite boundaries in a four-sided figure is both parallel and equal in length, that figure constitutes a parallelogram.
Hence, quadrilateral \( \square BEFD \) is a parallelogram (\( \|^{gm} \)).
In simple words: The top segment DF is parallel to the bottom segment BE and matches it in length, which confirms BEFD is a parallelogram.

Exam Tip: Showing that just one pair of opposite sides is both parallel and equal is sufficient to establish a parallelogram.

Question 15. Draw the graph of the following equation
(a) \( x - y = 0 \)
(b) \( x + y = 0 \)
(c) \( \frac{x}{2} + \frac{y}{3} = 2 \)
(d) \( 2(x + y) + 3 = 3(x - y) - 7 \)
Answer:
(a) \( x - y = 0 \implies y = x \)
Coordinate pairs: \( (0, 0), (2, 2), (-2, -2) \). This traces a straight line passing through the origin at an inclination of \( 45^\circ \).

(b) \( x + y = 0 \implies y = -x \)
Coordinate pairs: \( (0, 0), (2, -2), (-2, 2) \). This represents a straight line crossing through the origin with a negative unit slope.

(c) \( \frac{x}{2} + \frac{y}{3} = 2 \)
Multiplying through by 6:
\( 3x + 2y = 12 \implies y = \frac{12 - 3x}{2} \)
Coordinate pairs:
- When \( x = 0 \), \( y = 6 \implies (0, 6) \)
- When \( x = 4 \), \( y = 0 \implies (4, 0) \)
- When \( x = 2 \), \( y = 3 \implies (2, 3) \)

(d) \( 2(x + y) + 3 = 3(x - y) - 7 \)
Expanding and gathering like terms:
\( 2x + 2y + 3 = 3x - 3y - 7 \)
\( 3 + 7 = 3x - 2x - 3y - 2y \)
\( 10 = x - 5y \implies x - 5y = 10 \implies x = 5y + 10 \)
Coordinate pairs:
- When \( y = 0 \), \( x = 10 \implies (10, 0) \)
- When \( y = -2 \), \( x = 0 \implies (0, -2) \)
- When \( y = -1 \), \( x = 5 \implies (5, -1) \) In simple words: Find at least two coordinate pairs for each equation by picking values for x to calculate y, and draw a straight line through them.

Exam Tip: Construct a clear table of values containing at least three distinct points for each line to prevent plotting mistakes.

Question 15. The bisectors of two adjacent angles of a \( \|^{gm} \) meet at right angles. Prove.
Answer: Let \( ABCD \) be a parallelogram, and let rays \( AP \) and \( BP \) bisect the adjacent angles \( \angle A \) and \( \angle B \), meeting at point \( P \).
Since adjacent angles are supplementary:
\( \angle A + \angle B = 180^\circ \)
Dividing by 2 on both sides:
\( \frac{1}{2}\angle A + \frac{1}{2}\angle B = \frac{180^\circ}{2} = 90^\circ \)
Because \( AP \) and \( BP \) are angle bisectors:
\( \angle PAB = \frac{1}{2}\angle A \quad \text{and} \quad \angle PBA = \frac{1}{2}\angle B \)
Hence:
\( \angle PAB + \angle PBA = 90^\circ \)
Now, inside triangle \( \Delta PAB \), the sum of all angles equals \( 180^\circ \):
\( \angle PAB + \angle PBA + \angle APB = 180^\circ \)
\( 90^\circ + \angle APB = 180^\circ \)

\( \implies \angle APB = 180^\circ - 90^\circ = 90^\circ \)
This verifies that the angle bisectors meet at a right angle.
In simple words: Halving two angles that add to 180 gives 90, which leaves exactly 90 degrees for the third corner of the small triangle.

Exam Tip: Showing that \( \frac{1}{2}\angle A + \frac{1}{2}\angle B = 90^\circ \) directly sets up the angle sum property of the triangle to finish the proof.

Question 16. ABCD is a \( \|^{gm} \). Find the value of \( x \) and \( y \).
Answer: In the given parallelogram \( ABCD \), opposite angles are identical:
\( \angle A = \angle C \)
From the figure, \( \angle A = (3y - 5)^\circ \) and \( \angle C = 85^\circ \).
Therefore:
\( 3y - 5 = 85 \)
\( 3y = 90 \)

\( \implies y = 30 \)

Now observe triangle \( \Delta ABD \):
The angle at \( A \) is \( 85^\circ \), angle \( \angle ABD = 40^\circ \), and angle \( \angle ADB = (2x)^\circ \).
Using the triangle angle sum property:
\( \angle A + \angle ABD + \angle ADB = 180^\circ \)
\( 85^\circ + 40^\circ + 2x = 180^\circ \)
\( 125^\circ + 2x = 180^\circ \)
\( 2x = 180^\circ - 125^\circ = 55^\circ \)

\( \implies x = \frac{55}{2} = 27.5 \)
Thus, \( x = 27.5 \) and \( y = 30 \).
In simple words: Set opposite angles equal to find y = 30, then use the 180-degree triangle sum in triangle ABD to determine that x = 27.5.

Exam Tip: First use the opposite angles property of the parallelogram to get \( \angle A \), then apply the triangle sum property to evaluate the diagonal angle.

Question 17. ABCD is a trapezium in which \( AB \parallel CD \) and \( AD = BC \). Prove that \( \angle A = \angle B \).
Answer: Construction: Extend side \( AB \) and construct a line segment from vertex \( C \) parallel to \( DA \), meeting the extension at point \( E \).

Proof:
In quadrilateral \( ADCE \):
- \( AE \parallel CD \) (since \( AB \parallel CD \))
- \( CE \parallel DA \) (by our construction)
Therefore, \( ADCE \) forms a parallelogram.
Since opposite sides in a parallelogram match in length:
\( AD = CE \)
However, the problem states that:
\( AD = BC \)
Hence, \( BC = CE \).
Inside triangle \( \Delta BCE \), because sides \( BC \) and \( CE \) are equal, their opposite angles are also equal:
\( \angle CEB = \angle CBE \quad \text{--- (1)} \)
For parallel lines \( AD \parallel CE \) intersected by transversal line \( AE \):
\( \angle A + \angle CEB = 180^\circ \implies \angle A = 180^\circ - \angle CEB \quad \text{--- (2)} \)
Along the linear pair on line \( AE \) at vertex \( B \):
\( \angle B + \angle CBE = 180^\circ \implies \angle B = 180^\circ - \angle CBE \quad \text{--- (3)} \)
Comparing (2) and (3) while utilizing (1):
\( \angle A = \angle B \)
Hence proved.
In simple words: Drawing a parallel line forms an isosceles triangle on the side, whose equal base angles prove that the two bottom corners of the trapezium are identical.

Exam Tip: Mention the auxiliary construction \( CE \parallel DA \) clearly at the start, as it forms the entire backbone of this standard proof.

Question 18. If the point \( (2m, m+1) \) lies on the line \( 2x+3y=10 \), find the value of \( m \).
Answer: If a point lies along a given line, its coordinates must satisfy that equation.
Substitute \( x = 2m \) and \( y = m + 1 \) into \( 2x + 3y = 10 \):
\( 2(2m) + 3(m + 1) = 10 \)
\( 4m + 3m + 3 = 10 \)
\( 7m + 3 = 10 \)
\( 7m = 10 - 3 \)
\( 7m = 7 \)

\( \implies m = 1 \)
In simple words: Plug 2m in place of x and m + 1 in place of y, simplify the resulting equation, and solve to get m = 1.

Exam Tip: Be sure to distribute the multiplication through the entire binomial \( 3(m + 1) \) to avoid common sign and coefficient errors.

Question 19. ABCD is a trapezium. \( AB \parallel EF \parallel CD \). E is the midpoint of AD. Prove that: (i) F is the midpoint of BC. (ii) \( EF = \frac{1}{2}(AB + CD) \)
Answer: Construction: Draw diagonal \( AC \) cutting segment \( EF \) at point \( G \).

(i) Proof that F is the midpoint of BC:
In triangle \( \Delta ADC \):
- \( E \) is the midpoint of side \( AD \) (given).
- \( EG \parallel CD \) (since \( EF \parallel CD \)).
By the Converse of the Midpoint Theorem, a line drawn through the midpoint of one side parallel to another side bisects the third side.
Hence, \( G \) is the midpoint of diagonal \( AC \).

Now, in triangle \( \Delta CAB \):
- \( G \) is the midpoint of side \( AC \) (proven above).
- \( GF \parallel AB \) (since \( EF \parallel AB \)).
By the Converse of the Midpoint Theorem, \( F \) is the midpoint of side \( BC \).

(ii) Proof that \( EF = \frac{1}{2}(AB + CD) \):
By the Midpoint Theorem:
In \( \Delta ADC \), \( EG = \frac{1}{2}CD \)
In \( \Delta CAB \), \( GF = \frac{1}{2}AB \)
Adding these two segments together:
\( EF = EG + GF = \frac{1}{2}CD + \frac{1}{2}AB = \frac{1}{2}(AB + CD) \)
Hence proved.
In simple words: The diagonal creates two triangles where the midpoint theorem applies, proving F is in the middle and EF is the average of the two parallel bases.

Exam Tip: Introducing diagonal \( AC \) splits the quadrilateral into two distinct triangles where the Converse Midpoint Theorem applies step by step.

Question 20. Draw the graphs of \( x+3y=6 \) and \( 2x-3y=12 \) on the same graph sheet. Find the vertices and area of the triangle formed between the two lines representing the equations and the Y axis.
Answer: First, determine points for both linear equations:

For Line 1: \( x + 3y = 6 \implies x = 6 - 3y \)
- When \( y = 0 \), \( x = 6 \implies (6, 0) \)
- When \( y = 2 \), \( x = 0 \implies (0, 2) \)
- When \( y = 1 \), \( x = 3 \implies (3, 1) \)

For Line 2: \( 2x - 3y = 12 \implies 2x = 12 + 3y \)
- When \( y = 0 \), \( 2x = 12 \implies x = 6 \implies (6, 0) \)
- When \( y = -4 \), \( 2x = 0 \implies x = 0 \implies (0, -4) \)
- When \( y = -2 \), \( 2x = 6 \implies x = 3 \implies (3, -2) \) Vertices of the formed triangle:
The two lines intersect each other at \( A(6, 0) \).
The first line crosses the Y-axis at \( B(0, 2) \).
The second line crosses the Y-axis at \( C(0, -4) \).
Thus, the three triangle vertices are \( (6, 0), (0, 2), \) and \( (0, -4) \).

Area of the triangle:
Base length along the vertical Y-axis:
\( \text{Base} = |2 - (-4)| = 6\text{ units} \)
Height corresponding to the perpendicular distance to the Y-axis:
\( \text{Height} = x\text{-coordinate of point } A = 6\text{ units} \)
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 6 = 18\text{ square units} \)
In simple words: The lines cross at (6, 0) and hit the vertical axis at 2 and -4, creating a triangle with a base of 6 and height of 6, giving an area of 18 square units.

Exam Tip: Make sure to state both the three vertex coordinates and the square unit area clearly to secure full marks.

Question 21. Draw the graph of \( \frac{x}{2}+\frac{y}{3}=2 \). Find the points where the line representing the equation meets: (a) X axis. (b) Y axis.
Answer: Clear fractions by multiplying the entire equation by 6:
\( 3x + 2y = 12 \)

(a) Intercept with the X-axis:
Setting \( y = 0 \):
\( 3x + 2(0) = 12 \implies 3x = 12 \implies x = 4 \)
Hence, the line meets the X-axis at \( (4, 0) \).

(b) Intercept with the Y-axis:
Setting \( x = 0 \):
\( 3(0) + 2y = 12 \implies 2y = 12 \implies y = 6 \)
Hence, the line meets the Y-axis at \( (0, 6) \). In simple words: Putting y = 0 reveals that the line hits the horizontal axis at (4, 0), and putting x = 0 shows it hits the vertical axis at (0, 6).

Exam Tip: Expressing intercepts as full coordinate pairs \( (x, y) \) rather than standalone numbers prevents losing partial credit.