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CHAPTER 1 — THE USE OF COORDINATES
Class IX Mathematics — Coordinate Geometry
Preeti Kushwah Classes — Unit Test (Set 3: Advanced)
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 distance of point P(3, 4) from the origin is ___ units.
Q2.1
The distance between points (2, 0) and (5, 0) is ___ units.
Q3.1
The area of a triangle with vertices at the origin, (4, 0), and (0, 3) is ___ sq units.
Q4.1
The midpoint of the segment joining (2, 3) and (4, 7) is ___.
Q5.1
True or False: The points A(1, 1), B(3, 3), and C(5, 5) are collinear.
Q6.1
If the distance between (a, 0) and (0, a) is 10 units, find the positive value of a.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Find the distance between the points A(3, 4) and B(−3, −4).
Q8.2
Show that the points P(0, 0), Q(3, 0), and R(0, 4) form a right-angled triangle. Find the length of the hypotenuse.
Q9.2
Find all values of y if the distance between (2, y) and (2, 3) is 5 units.
Q10.2
Find the midpoint of the line segment joining A(−3, 4) and B(5, −2).
Q11.2
The midpoint of segment AB is M(3, 5). If A = (1, 3), find the coordinates of B.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Prove that the points A(1, 7), B(4, 2), C(−1, −1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Q13.3
Three vertices of a rectangle are A(−4, 1), B(−4, 5), C(2, 5). Find the fourth vertex D. Calculate the area of the rectangle and the length of its diagonal.
Q14.3
Find the point on the x-axis which is equidistant from the points (2, −5) and (−2, 9).
Q15.3
The vertices of a quadrilateral are A(1, 1), B(6, 1), C(6, 4), D(1, 4). Show that ABCD is a rectangle. Find its area and the length of its diagonal.
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
The vertices of a triangle are A(1, 2), B(5, 2), C(3, 6).
(a) Plot the triangle. 1
(b) Find the lengths AB, BC, and AC. 1
(c) Is the triangle equilateral, isosceles, or scalene? 1
(d) Find the midpoints of all three sides. 1
(e) Find the area of the triangle. 1
(a) Plot the triangle. 1
(b) Find the lengths AB, BC, and AC. 1
(c) Is the triangle equilateral, isosceles, or scalene? 1
(d) Find the midpoints of all three sides. 1
(e) Find the area of the triangle. 1
Q17.5
A triangle has vertices P(−3, 0), Q(3, 0), R(0, 4).
(a) Show that the triangle is isosceles. 1
(b) Find the perimeter of the triangle. 1
(c) Find the area of the triangle. 1
(d) Find the midpoint M of PQ. Show that RM is perpendicular to PQ. 1
(e) Find the point that divides PR in the ratio 1 : 2 from P. 1
(a) Show that the triangle is isosceles. 1
(b) Find the perimeter of the triangle. 1
(c) Find the area of the triangle. 1
(d) Find the midpoint M of PQ. Show that RM is perpendicular to PQ. 1
(e) Find the point that divides PR in the ratio 1 : 2 from P. 1
Bonus Question (Optional) [2 Marks]
Q18.2
★ Find the coordinates of the point which divides the line segment joining (4, −3) and (8, 5) in the ratio 3 : 1 internally.
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Distance = √(3² + 4²) = √(9 + 16) = √25 = 5 units.
Q2. [1 Mark]
Since both points are on x-axis: |5 − 2| = 3 units.
Q3. [1 Mark]
Area = ½ × base × height = ½ × 4 × 3 = 6 sq units.
(Base = 4 on x-axis, height = 3 on y-axis.)
(Base = 4 on x-axis, height = 3 on y-axis.)
Q4. [1 Mark]
Midpoint = ((2+4)/2, (3+7)/2) = (3, 5).
Q5. [1 Mark]
True.
Check: AB = √((3−1)² + (3−1)²) = √(4+4) = 2√2.
BC = √((5−3)² + (5−3)²) = 2√2.
AC = √((5−1)² + (5−1)²) = 4√2.
Since AB + BC = 2√2 + 2√2 = 4√2 = AC, the points are collinear.
Check: AB = √((3−1)² + (3−1)²) = √(4+4) = 2√2.
BC = √((5−3)² + (5−3)²) = 2√2.
AC = √((5−1)² + (5−1)²) = 4√2.
Since AB + BC = 2√2 + 2√2 = 4√2 = AC, the points are collinear.
Q6. [1 Mark]
Distance = √(a² + a²) = a√2 = 10.
So a = 10/√2 = 10√2/2 = 5√2.
So a = 10/√2 = 10√2/2 = 5√2.
Q7. [2 Marks]
AB = √((−3−3)² + (−4−4)²) = √(36 + 64) = √100 = 10 units.
[1 mark for formula, 1 mark for answer]
[1 mark for formula, 1 mark for answer]
Q8. [2 Marks]
PQ = √((3−0)² + (0−0)²) = 3.
PR = √((0−0)² + (4−0)²) = 4.
QR = √((3−0)² + (0−4)²) = √(9+16) = 5. [1 mark]
Check: PQ² + PR² = 9 + 16 = 25 = QR² = 5².
By Pythagoras theorem, right angle at P. Hypotenuse QR = 5 units. [1 mark]
PR = √((0−0)² + (4−0)²) = 4.
QR = √((3−0)² + (0−4)²) = √(9+16) = 5. [1 mark]
Check: PQ² + PR² = 9 + 16 = 25 = QR² = 5².
By Pythagoras theorem, right angle at P. Hypotenuse QR = 5 units. [1 mark]
Q9. [2 Marks]
Distance = √((2−2)² + (y−3)²) = √((y−3)²) = |y − 3| = 5. [1 mark]
y − 3 = 5 → y = 8, or y − 3 = −5 → y = −2.
Values: y = 8 or y = −2. [1 mark]
y − 3 = 5 → y = 8, or y − 3 = −5 → y = −2.
Values: y = 8 or y = −2. [1 mark]
Q10. [2 Marks]
Midpoint = ((−3+5)/2, (4+(−2))/2) = (2/2, 2/2) = (1, 1).
[1 mark for formula, 1 mark for answer]
[1 mark for formula, 1 mark for answer]
Q11. [2 Marks]
Let B = (x, y). Using midpoint formula:
(1+x)/2 = 3 → x = 5.
(3+y)/2 = 5 → y = 7. [1 mark]
B = (5, 7). [1 mark]
(1+x)/2 = 3 → x = 5.
(3+y)/2 = 5 → y = 7. [1 mark]
B = (5, 7). [1 mark]
Q12. [3 Marks]
AB = √((4−1)² + (2−7)²) = √(9+25) = √34. [½ mark]
BC = √((−1−4)² + (−1−2)²) = √(25+9) = √34. [½ mark]
AC = √((−1−1)² + (−1−7)²) = √(4+64) = √68. [½ mark]
AB = BC = √34. Triangle is isosceles. [½ mark]
AB² + BC² = 34 + 34 = 68 = AC². By Pythagoras, right angle at B. [½ mark]
Area = ½ × AB × BC = ½ × √34 × √34 = ½ × 34 = 17 sq units. [½ mark]
BC = √((−1−4)² + (−1−2)²) = √(25+9) = √34. [½ mark]
AC = √((−1−1)² + (−1−7)²) = √(4+64) = √68. [½ mark]
AB = BC = √34. Triangle is isosceles. [½ mark]
AB² + BC² = 34 + 34 = 68 = AC². By Pythagoras, right angle at B. [½ mark]
Area = ½ × AB × BC = ½ × √34 × √34 = ½ × 34 = 17 sq units. [½ mark]
Q13. [3 Marks]
A(−4, 1), B(−4, 5), C(2, 5). AB is vertical (x = −4), BC is horizontal (y = 5). [½ mark]
D must have x = 2 (same as C) and y = 1 (same as A). D = (2, 1). [1 mark]
Length = |2 − (−4)| = 6. Breadth = |5 − 1| = 4.
Area = 6 × 4 = 24 sq units. [1 mark]
Diagonal = √(6² + 4²) = √(36+16) = √52 = 2√13 units. [½ mark]
D must have x = 2 (same as C) and y = 1 (same as A). D = (2, 1). [1 mark]
Length = |2 − (−4)| = 6. Breadth = |5 − 1| = 4.
Area = 6 × 4 = 24 sq units. [1 mark]
Diagonal = √(6² + 4²) = √(36+16) = √52 = 2√13 units. [½ mark]
Q14. [3 Marks]
Let the point on x-axis be P(x, 0). [½ mark]
PA = √((x−2)² + 25), PB = √((x+2)² + 81). [½ mark]
PA = PB → (x−2)² + 25 = (x+2)² + 81. [½ mark]
x² − 4x + 4 + 25 = x² + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56 → x = −7. [1 mark]
Point = (−7, 0). [½ mark]
PA = √((x−2)² + 25), PB = √((x+2)² + 81). [½ mark]
PA = PB → (x−2)² + 25 = (x+2)² + 81. [½ mark]
x² − 4x + 4 + 25 = x² + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56 → x = −7. [1 mark]
Point = (−7, 0). [½ mark]
Q15. [3 Marks]
AB = |6−1| = 5. BC = |4−1| = 3. CD = |6−1| = 5. DA = |4−1| = 3. [1 mark]
AB = CD and BC = DA (opposite sides equal). All angles are 90° (horizontal meets vertical at right angles). ABCD is a rectangle. [1 mark]
Area = 5 × 3 = 15 sq units. Diagonal = √(5² + 3²) = √(25+9) = √34 units. [1 mark]
AB = CD and BC = DA (opposite sides equal). All angles are 90° (horizontal meets vertical at right angles). ABCD is a rectangle. [1 mark]
Area = 5 × 3 = 15 sq units. Diagonal = √(5² + 3²) = √(25+9) = √34 units. [1 mark]
Q16. [5 Marks]
(a) Plot triangle with vertices at (1, 2), (5, 2), (3, 6). [1 mark]
(b) AB = |5−1| = 4.
BC = √((3−5)² + (6−2)²) = √(4+16) = √20 = 2√5.
AC = √((3−1)² + (6−2)²) = √(4+16) = √20 = 2√5. [1 mark]
(c) BC = AC = 2√5. The triangle is isosceles. [1 mark]
(d) Midpoint of AB = ((1+5)/2, (2+2)/2) = (3, 2).
Midpoint of BC = ((5+3)/2, (2+6)/2) = (4, 4).
Midpoint of AC = ((1+3)/2, (2+6)/2) = (2, 4). [1 mark]
(e) Base AB = 4 (along y = 2). Height = perpendicular distance from C(3, 6) to line y = 2 = |6−2| = 4.
Area = ½ × 4 × 4 = 8 sq units. [1 mark]
(b) AB = |5−1| = 4.
BC = √((3−5)² + (6−2)²) = √(4+16) = √20 = 2√5.
AC = √((3−1)² + (6−2)²) = √(4+16) = √20 = 2√5. [1 mark]
(c) BC = AC = 2√5. The triangle is isosceles. [1 mark]
(d) Midpoint of AB = ((1+5)/2, (2+2)/2) = (3, 2).
Midpoint of BC = ((5+3)/2, (2+6)/2) = (4, 4).
Midpoint of AC = ((1+3)/2, (2+6)/2) = (2, 4). [1 mark]
(e) Base AB = 4 (along y = 2). Height = perpendicular distance from C(3, 6) to line y = 2 = |6−2| = 4.
Area = ½ × 4 × 4 = 8 sq units. [1 mark]
Q17. [5 Marks]
(a) PR = √((0−(−3))² + (4−0)²) = √(9+16) = √25 = 5.
QR = √((0−3)² + (4−0)²) = √(9+16) = √25 = 5.
PR = QR = 5. Isosceles. [1 mark]
(b) PQ = |3−(−3)| = 6.
Perimeter = PQ + PR + QR = 6 + 5 + 5 = 16 units. [1 mark]
(c) Base PQ = 6 (on x-axis). Height = y-coordinate of R = 4.
Area = ½ × 6 × 4 = 12 sq units. [1 mark]
(d) Midpoint M of PQ = ((−3+3)/2, (0+0)/2) = (0, 0) = origin.
RM goes from R(0, 4) to M(0, 0) — this is a vertical line (along y-axis).
PQ goes from (−3, 0) to (3, 0) — this is a horizontal line (along x-axis).
Vertical ⊥ horizontal. Hence RM ⊥ PQ. [1 mark]
(e) Section formula: ratio 1:2 from P(−3, 0) to R(0, 4).
x = (1×0 + 2×(−3))/(1+2) = −6/3 = −2.
y = (1×4 + 2×0)/(1+2) = 4/3.
Point = (−2, 4/3). [1 mark]
QR = √((0−3)² + (4−0)²) = √(9+16) = √25 = 5.
PR = QR = 5. Isosceles. [1 mark]
(b) PQ = |3−(−3)| = 6.
Perimeter = PQ + PR + QR = 6 + 5 + 5 = 16 units. [1 mark]
(c) Base PQ = 6 (on x-axis). Height = y-coordinate of R = 4.
Area = ½ × 6 × 4 = 12 sq units. [1 mark]
(d) Midpoint M of PQ = ((−3+3)/2, (0+0)/2) = (0, 0) = origin.
RM goes from R(0, 4) to M(0, 0) — this is a vertical line (along y-axis).
PQ goes from (−3, 0) to (3, 0) — this is a horizontal line (along x-axis).
Vertical ⊥ horizontal. Hence RM ⊥ PQ. [1 mark]
(e) Section formula: ratio 1:2 from P(−3, 0) to R(0, 4).
x = (1×0 + 2×(−3))/(1+2) = −6/3 = −2.
y = (1×4 + 2×0)/(1+2) = 4/3.
Point = (−2, 4/3). [1 mark]
Q18. Bonus [2 Marks]
Section formula (internal division) in ratio 3:1:
x = (3×8 + 1×4)/(3+1) = (24+4)/4 = 28/4 = 7. [1 mark]
y = (3×5 + 1×(−3))/(3+1) = (15−3)/4 = 12/4 = 3. [½ mark]
Point = (7, 3). [½ mark]
x = (3×8 + 1×4)/(3+1) = (24+4)/4 = 28/4 = 7. [1 mark]
y = (3×5 + 1×(−3))/(3+1) = (15−3)/4 = 12/4 = 3. [½ mark]
Point = (7, 3). [½ mark]