💡 Print paper or share solutions as PDF via WhatsApp to students.
CHAPTER 1 — THE USE OF COORDINATES
Class IX Mathematics — Coordinate Geometry
Preeti Kushwah Classes — Unit Test (Set 4: Expert)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Draw neat figures wherever required. Use graph paper for plotting questions.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Draw neat figures wherever required. Use graph paper for plotting questions.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
The point dividing the segment joining (1, −2) and (4, 7) in the ratio 2 : 1 internally is ___.
Q2.1
If the midpoint of (a, b) and (3, 5) is (2, 3), find the values of a and b.
Q3.1
The area of the triangle with vertices (0, 0), (6, 0), (0, 8) is ___ sq units.
Q4.1
If the three points (1, 1), (k, 2), (5, 5) are collinear, find the value of k.
Q5.1
The point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3 is ___.
Q6.1
True or False: The points (1, 5), (2, 3), (−2, −1) are collinear.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Find the trisection points of the line segment joining A(2, −2) and B(−7, 4).
Q8.2
Find the coordinates of the point P that divides the segment joining A(1, −1) and B(4, 5) in the ratio 2 : 1 internally.
Q9.2
Show that the points A(−2, −1), B(1, 0), C(4, 3), D(1, 2) form a parallelogram. (Hint: Show that diagonals bisect each other.)
Q10.2
Find the area of the triangle with vertices A(1, 2), B(3, 5), C(−1, 6).
Q11.2
Check whether the points (1, 1), (2, 3), (4, 7) are collinear.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Find the area of the triangle formed by the midpoints of the sides of the triangle with vertices A(0, 0), B(6, 0), C(0, 8).
Q13.3
Find the ratio in which the point (−3, p) divides the segment joining (−5, −4) and (−2, 3). Also find the value of p.
Q14.3
Show that the points A(−1, 0), B(3, 1), C(2, 2), D(−2, 1) form a parallelogram. Find the area of the parallelogram.
Q15.3
The midpoints of the sides of a triangle are P(2, 1), Q(5, 3), R(3, 5). Find the coordinates of the vertices of the triangle.
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
A park is shaped as a rhombus with vertices A(−6, 0), B(0, 8), C(6, 0), D(0, −8).
(a) Plot the rhombus and verify that all sides are equal. [1]
(b) Find the length of both diagonals AC and BD. [1]
(c) Find the area of the park. [1]
(d) A statue is placed at the point dividing AB in the ratio 1 : 3 from A. Find the coordinates of the statue. [1]
(e) The boundary path from A to C via B is how much longer than the direct path from A to C? [1]
(a) Plot the rhombus and verify that all sides are equal. [1]
(b) Find the length of both diagonals AC and BD. [1]
(c) Find the area of the park. [1]
(d) A statue is placed at the point dividing AB in the ratio 1 : 3 from A. Find the coordinates of the statue. [1]
(e) The boundary path from A to C via B is how much longer than the direct path from A to C? [1]
Q17.5
The vertices of a triangle are A(−3, 2), B(5, 4), C(1, −2).
(a) Find the lengths of all three sides. [1]
(b) Check whether the triangle is right-angled. [1]
(c) Find the midpoints of each side. [1]
(d) Find the centroid of the triangle (the point where medians intersect). [1]
(e) Verify that the centroid lies on each median by showing that the point dividing each median from vertex to opposite midpoint in ratio 2 : 1 gives the same point. [1]
(a) Find the lengths of all three sides. [1]
(b) Check whether the triangle is right-angled. [1]
(c) Find the midpoints of each side. [1]
(d) Find the centroid of the triangle (the point where medians intersect). [1]
(e) Verify that the centroid lies on each median by showing that the point dividing each median from vertex to opposite midpoint in ratio 2 : 1 gives the same point. [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ Find the area of the triangle formed by the points (a, b + c), (b, c + a), (c, a + b). What conclusion can you draw?
Preeti Kushwah Classes
Scan for more notes & papers
www.preetikushwahclasses.com
Scan for more notes & papers
www.preetikushwahclasses.com
Answer Key & Detailed Solutions
Q1. [1 Mark]
Using section formula (ratio 2 : 1):
x = (2×4 + 1×1)/(2+1) = 9/3 = 3.
y = (2×7 + 1×(−2))/(2+1) = 12/3 = 4.
Point = (3, 4).
x = (2×4 + 1×1)/(2+1) = 9/3 = 3.
y = (2×7 + 1×(−2))/(2+1) = 12/3 = 4.
Point = (3, 4).
Q2. [1 Mark]
(a + 3)/2 = 2 → a = 1.
(b + 5)/2 = 3 → b = 1.
So a = 1, b = 1.
(b + 5)/2 = 3 → b = 1.
So a = 1, b = 1.
Q3. [1 Mark]
Area = ½ × 6 × 8 = 24 sq units.
Q4. [1 Mark]
For collinear points, area of triangle = 0:
½|1(2 − 5) + k(5 − 1) + 5(1 − 2)| = 0
½|−3 + 4k − 5| = 0
4k − 8 = 0 → k = 2.
½|1(2 − 5) + k(5 − 1) + 5(1 − 2)| = 0
½|−3 + 4k − 5| = 0
4k − 8 = 0 → k = 2.
Q5. [1 Mark]
Using section formula (ratio 2 : 3):
x = (2×4 + 3×(−1))/(2+3) = (8−3)/5 = 1.
y = (2×(−3) + 3×7)/(2+3) = (−6+21)/5 = 3.
Point = (1, 3).
x = (2×4 + 3×(−1))/(2+3) = (8−3)/5 = 1.
y = (2×(−3) + 3×7)/(2+3) = (−6+21)/5 = 3.
Point = (1, 3).
Q6. [1 Mark]
False.
Area = ½|1(3−(−1)) + 2(−1−5) + (−2)(5−3)|
= ½|4 − 12 − 4| = ½|−12| = 6 ≠ 0.
Since area is not zero, the points are NOT collinear.
Area = ½|1(3−(−1)) + 2(−1−5) + (−2)(5−3)|
= ½|4 − 12 − 4| = ½|−12| = 6 ≠ 0.
Since area is not zero, the points are NOT collinear.
Q7. [2 Marks]
Trisection divides AB in ratios 1 : 2 and 2 : 1.
Point P (ratio 1 : 2): [1 mark]
x = (1×(−7) + 2×2)/3 = (−7+4)/3 = −1.
y = (1×4 + 2×(−2))/3 = (4−4)/3 = 0.
P = (−1, 0).
Point Q (ratio 2 : 1): [1 mark]
x = (2×(−7) + 1×2)/3 = (−14+2)/3 = −4.
y = (2×4 + 1×(−2))/3 = (8−2)/3 = 2.
Q = (−4, 2).
Point P (ratio 1 : 2): [1 mark]
x = (1×(−7) + 2×2)/3 = (−7+4)/3 = −1.
y = (1×4 + 2×(−2))/3 = (4−4)/3 = 0.
P = (−1, 0).
Point Q (ratio 2 : 1): [1 mark]
x = (2×(−7) + 1×2)/3 = (−14+2)/3 = −4.
y = (2×4 + 1×(−2))/3 = (8−2)/3 = 2.
Q = (−4, 2).
Q8. [2 Marks]
Ratio 2 : 1 from A(1, −1) to B(4, 5):
x = (2×4 + 1×1)/3 = 9/3 = 3.
y = (2×5 + 1×(−1))/3 = 9/3 = 3.
P = (3, 3). [1 mark for formula, 1 mark for answer]
x = (2×4 + 1×1)/3 = 9/3 = 3.
y = (2×5 + 1×(−1))/3 = 9/3 = 3.
P = (3, 3). [1 mark for formula, 1 mark for answer]
Q9. [2 Marks]
If diagonals bisect each other, it’s a parallelogram.
Midpoint of diagonal AC: ((−2+4)/2, (−1+3)/2) = (1, 1). [1 mark]
Midpoint of diagonal BD: ((1+1)/2, (0+2)/2) = (1, 1). [½ mark]
Both midpoints are (1, 1). Diagonals bisect each other.
Hence ABCD is a parallelogram. [½ mark]
Midpoint of diagonal AC: ((−2+4)/2, (−1+3)/2) = (1, 1). [1 mark]
Midpoint of diagonal BD: ((1+1)/2, (0+2)/2) = (1, 1). [½ mark]
Both midpoints are (1, 1). Diagonals bisect each other.
Hence ABCD is a parallelogram. [½ mark]
Q10. [2 Marks]
Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
= ½|1(5 − 6) + 3(6 − 2) + (−1)(2 − 5)|
= ½|−1 + 12 + 3|
= ½ × 14 = 7 sq units. [1 mark for formula, 1 mark for calculation]
= ½|1(5 − 6) + 3(6 − 2) + (−1)(2 − 5)|
= ½|−1 + 12 + 3|
= ½ × 14 = 7 sq units. [1 mark for formula, 1 mark for calculation]
Q11. [2 Marks]
Area = ½|1(3 − 7) + 2(7 − 1) + 4(1 − 3)|
= ½|−4 + 12 − 8|
= ½|0| = 0. [1 mark]
Since area = 0, the three points are collinear. [1 mark]
= ½|−4 + 12 − 8|
= ½|0| = 0. [1 mark]
Since area = 0, the three points are collinear. [1 mark]
Q12. [3 Marks]
Midpoints of sides:
M₁ (midpoint of AB) = ((0+6)/2, (0+0)/2) = (3, 0). [½ mark]
M₂ (midpoint of BC) = ((6+0)/2, (0+8)/2) = (3, 4). [½ mark]
M₃ (midpoint of AC) = ((0+0)/2, (0+8)/2) = (0, 4). [½ mark]
Area of medial triangle = ½|3(4 − 4) + 3(4 − 0) + 0(0 − 4)|
= ½|0 + 12 + 0| = 6 sq units. [1 mark]
Note: Original triangle area = ½ × 6 × 8 = 24. Medial triangle = 24/4 = 6. ✓ [½ mark]
M₁ (midpoint of AB) = ((0+6)/2, (0+0)/2) = (3, 0). [½ mark]
M₂ (midpoint of BC) = ((6+0)/2, (0+8)/2) = (3, 4). [½ mark]
M₃ (midpoint of AC) = ((0+0)/2, (0+8)/2) = (0, 4). [½ mark]
Area of medial triangle = ½|3(4 − 4) + 3(4 − 0) + 0(0 − 4)|
= ½|0 + 12 + 0| = 6 sq units. [1 mark]
Note: Original triangle area = ½ × 6 × 8 = 24. Medial triangle = 24/4 = 6. ✓ [½ mark]
Q13. [3 Marks]
Let the ratio be k : 1.
Using section formula for x-coordinate:
−3 = (k×(−2) + 1×(−5))/(k + 1)
−3(k + 1) = −2k − 5
−3k − 3 = −2k − 5
−k = −2 → k = 2. [1 mark]
Ratio = 2 : 1. [½ mark]
Finding p:
p = (2×3 + 1×(−4))/(2 + 1) = (6 − 4)/3 = 2/3. [1 mark]
p = 2/3. [½ mark]
Using section formula for x-coordinate:
−3 = (k×(−2) + 1×(−5))/(k + 1)
−3(k + 1) = −2k − 5
−3k − 3 = −2k − 5
−k = −2 → k = 2. [1 mark]
Ratio = 2 : 1. [½ mark]
Finding p:
p = (2×3 + 1×(−4))/(2 + 1) = (6 − 4)/3 = 2/3. [1 mark]
p = 2/3. [½ mark]
Q14. [3 Marks]
Midpoint of AC = ((−1+2)/2, (0+2)/2) = (1/2, 1) = (0.5, 1). [½ mark]
Midpoint of BD = ((3+(−2))/2, (1+1)/2) = (1/2, 1) = (0.5, 1). [½ mark]
Diagonals bisect each other → parallelogram. [½ mark]
Area = ½|xA(yB − yD) + xB(yC − yA) + xC(yD − yB) + xD(yA − yC)|
= ½|(−1)(1 − 1) + 3(2 − 0) + 2(1 − 1) + (−2)(0 − 2)|
= ½|0 + 6 + 0 + 4| = ½ × 10 = 5 sq units. [1½ marks]
Midpoint of BD = ((3+(−2))/2, (1+1)/2) = (1/2, 1) = (0.5, 1). [½ mark]
Diagonals bisect each other → parallelogram. [½ mark]
Area = ½|xA(yB − yD) + xB(yC − yA) + xC(yD − yB) + xD(yA − yC)|
= ½|(−1)(1 − 1) + 3(2 − 0) + 2(1 − 1) + (−2)(0 − 2)|
= ½|0 + 6 + 0 + 4| = ½ × 10 = 5 sq units. [1½ marks]
Q15. [3 Marks]
Let the vertices be A, B, C such that P = midpoint of AB, Q = midpoint of BC, R = midpoint of AC.
A = P + R − Q = (2+3−5, 1+5−3) = (0, 3). [1 mark]
B = P + Q − R = (2+5−3, 1+3−5) = (4, −1). [1 mark]
C = Q + R − P = (5+3−2, 3+5−1) = (6, 7). [½ mark]
Verification:
Mid of AB = ((0+4)/2, (3+(−1))/2) = (2, 1) = P ✓
Mid of BC = ((4+6)/2, (−1+7)/2) = (5, 3) = Q ✓
Mid of AC = ((0+6)/2, (3+7)/2) = (3, 5) = R ✓ [½ mark]
A = P + R − Q = (2+3−5, 1+5−3) = (0, 3). [1 mark]
B = P + Q − R = (2+5−3, 1+3−5) = (4, −1). [1 mark]
C = Q + R − P = (5+3−2, 3+5−1) = (6, 7). [½ mark]
Verification:
Mid of AB = ((0+4)/2, (3+(−1))/2) = (2, 1) = P ✓
Mid of BC = ((4+6)/2, (−1+7)/2) = (5, 3) = Q ✓
Mid of AC = ((0+6)/2, (3+7)/2) = (3, 5) = R ✓ [½ mark]
Q16. [5 Marks]
(a) [1 mark]
AB = √((0−(−6))² + (8−0)²) = √(36+64) = √100 = 10.
BC = √((6−0)² + (0−8)²) = √(36+64) = 10.
CD = √((0−6)² + (−8−0)²) = √(36+64) = 10.
DA = √((−6−0)² + (0−(−8))²) = √(36+64) = 10.
All sides = 10. Rhombus confirmed.
(b) [1 mark]
AC = |6−(−6)| = 12 units.
BD = |8−(−8)| = 16 units.
(c) [1 mark]
Area = ½ × d₁ × d₂ = ½ × 12 × 16 = 96 sq units.
(d) [1 mark]
Ratio 1 : 3 from A(−6, 0) to B(0, 8):
x = (1×0 + 3×(−6))/(1+3) = −18/4 = −4.5.
y = (1×8 + 3×0)/(1+3) = 8/4 = 2.
Statue at (−4.5, 2).
(e) [1 mark]
Boundary A→B→C = AB + BC = 10 + 10 = 20 units.
Direct A→C = AC = 12 units.
Difference = 20 − 12 = 8 units longer.
AB = √((0−(−6))² + (8−0)²) = √(36+64) = √100 = 10.
BC = √((6−0)² + (0−8)²) = √(36+64) = 10.
CD = √((0−6)² + (−8−0)²) = √(36+64) = 10.
DA = √((−6−0)² + (0−(−8))²) = √(36+64) = 10.
All sides = 10. Rhombus confirmed.
(b) [1 mark]
AC = |6−(−6)| = 12 units.
BD = |8−(−8)| = 16 units.
(c) [1 mark]
Area = ½ × d₁ × d₂ = ½ × 12 × 16 = 96 sq units.
(d) [1 mark]
Ratio 1 : 3 from A(−6, 0) to B(0, 8):
x = (1×0 + 3×(−6))/(1+3) = −18/4 = −4.5.
y = (1×8 + 3×0)/(1+3) = 8/4 = 2.
Statue at (−4.5, 2).
(e) [1 mark]
Boundary A→B→C = AB + BC = 10 + 10 = 20 units.
Direct A→C = AC = 12 units.
Difference = 20 − 12 = 8 units longer.
Q17. [5 Marks]
(a) [1 mark]
AB = √((5−(−3))² + (4−2)²) = √(64+4) = √68 = 2√17.
BC = √((1−5)² + (−2−4)²) = √(16+36) = √52 = 2√13.
AC = √((1−(−3))² + (−2−2)²) = √(16+16) = √32 = 4√2.
(b) [1 mark]
AB² = 68, BC² = 52, AC² = 32.
Check: 52 + 32 = 84 ≠ 68. 68 + 32 = 100 ≠ 52. 68 + 52 = 120 ≠ 32.
No pair satisfies Pythagoras. Triangle is NOT right-angled.
(c) [1 mark]
Mid AB = ((−3+5)/2, (2+4)/2) = (1, 3).
Mid BC = ((5+1)/2, (4+(−2))/2) = (3, 1).
Mid AC = ((−3+1)/2, (2+(−2))/2) = (−1, 0).
(d) [1 mark]
Centroid = ((−3+5+1)/3, (2+4+(−2))/3) = (3/3, 4/3) = (1, 4/3).
(e) [1 mark]
Median from A(−3, 2) to midpoint of BC (3, 1):
Point dividing in 2 : 1 from A = ((2×3+1×(−3))/3, (2×1+1×2)/3) = (3/3, 4/3) = (1, 4/3) ✓
Median from B(5, 4) to midpoint of AC (−1, 0):
Point dividing in 2 : 1 from B = ((2×(−1)+1×5)/3, (2×0+1×4)/3) = (3/3, 4/3) = (1, 4/3) ✓
Median from C(1, −2) to midpoint of AB (1, 3):
Point dividing in 2 : 1 from C = ((2×1+1×1)/3, (2×3+1×(−2))/3) = (3/3, 4/3) = (1, 4/3) ✓
All three medians pass through (1, 4/3).
AB = √((5−(−3))² + (4−2)²) = √(64+4) = √68 = 2√17.
BC = √((1−5)² + (−2−4)²) = √(16+36) = √52 = 2√13.
AC = √((1−(−3))² + (−2−2)²) = √(16+16) = √32 = 4√2.
(b) [1 mark]
AB² = 68, BC² = 52, AC² = 32.
Check: 52 + 32 = 84 ≠ 68. 68 + 32 = 100 ≠ 52. 68 + 52 = 120 ≠ 32.
No pair satisfies Pythagoras. Triangle is NOT right-angled.
(c) [1 mark]
Mid AB = ((−3+5)/2, (2+4)/2) = (1, 3).
Mid BC = ((5+1)/2, (4+(−2))/2) = (3, 1).
Mid AC = ((−3+1)/2, (2+(−2))/2) = (−1, 0).
(d) [1 mark]
Centroid = ((−3+5+1)/3, (2+4+(−2))/3) = (3/3, 4/3) = (1, 4/3).
(e) [1 mark]
Median from A(−3, 2) to midpoint of BC (3, 1):
Point dividing in 2 : 1 from A = ((2×3+1×(−3))/3, (2×1+1×2)/3) = (3/3, 4/3) = (1, 4/3) ✓
Median from B(5, 4) to midpoint of AC (−1, 0):
Point dividing in 2 : 1 from B = ((2×(−1)+1×5)/3, (2×0+1×4)/3) = (3/3, 4/3) = (1, 4/3) ✓
Median from C(1, −2) to midpoint of AB (1, 3):
Point dividing in 2 : 1 from C = ((2×1+1×1)/3, (2×3+1×(−2))/3) = (3/3, 4/3) = (1, 4/3) ✓
All three medians pass through (1, 4/3).
Q18. Bonus [2 Marks]
Area = ½|a((c+a) − (a+b)) + b((a+b) − (b+c)) + c((b+c) − (c+a))|
= ½|a(c − b) + b(a − c) + c(b − a)|
= ½|ac − ab + ab − bc + bc − ac|
= ½|0| = 0. [1 mark]
Conclusion: The area is always zero regardless of the values of a, b, c. This means the three points are always collinear. [1 mark]
= ½|a(c − b) + b(a − c) + c(b − a)|
= ½|ac − ab + ab − bc + bc − ac|
= ½|0| = 0. [1 mark]
Conclusion: The area is always zero regardless of the values of a, b, c. This means the three points are always collinear. [1 mark]