Coordinate Geometry — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Coordinate Geometry" — important questions with detailed answers, download PDF for board exam preparation.
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Coordinate Geometry — Textbook Questions Answered (8)
Q1: Plot the points A(5, 2), B(-4, 3), C(-1, -4), D(2, -6) and state which quadrant each lies in.
A point (x, y) lies in:
Quadrant I if x > 0 and y > 0
Quadrant II if x < 0 and y > 0
Quadrant III if x < 0 and y < 0
Quadrant IV if x > 0 and y < 0
A(5, 2): x = 5 > 0, y = 2 > 0 → Quadrant I
B(-4, 3): x = -4 < 0, y = 3 > 0 → Quadrant II
C(-1, -4): x = -1 < 0, y = -4 < 0 → Quadrant III
D(2, -6): x = 2 > 0, y = -6 < 0 → Quadrant IV
Q2: Find the distance between the points P(2, 3) and Q(6, 6).
Using the distance formula: d = √[(x₂ − x₁)² + (y₂ − y₁)²]
Here (x₁, y₁) = (2, 3) and (x₂, y₂) = (6, 6).
d = √[(6 − 2)² + (6 − 3)²]
d = √[4² + 3²]
d = √[16 + 9]
d = √25
d = 5 units
Q3: Find the midpoint of the line segment joining A(-3, 5) and B(7, -1).
Midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Here A(-3, 5) and B(7, -1).
M = ((-3 + 7)/2, (5 + (-1))/2)
M = (4/2, 4/2)
M = (2, 2)
The midpoint is (2, 2).
Q4: Show that the points A(1, 1), B(2, 3), and C(4, 7) are collinear.
Three points are collinear if the slope between each consecutive pair is the same.
Slope of AB = (3 − 1)/(2 − 1) = 2/1 = 2
Slope of BC = (7 − 3)/(4 − 2) = 4/2 = 2
Since slope of AB = slope of BC = 2, all three points lie on the same straight line.
Therefore, A, B, and C are collinear.
Q5: Find the area of a triangle with vertices A(2, 3), B(6, 3), and C(6, 8).
Area formula: Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Here A(2, 3), B(6, 3), C(6, 8).
Area = ½ |2(3 − 8) + 6(8 − 3) + 6(3 − 3)|
Area = ½ |2(-5) + 6(5) + 6(0)|
Area = ½ |-10 + 30 + 0|
Area = ½ × 20
Area = 10 square units
Q6: A point lies on the x-axis. If its distance from (3, 4) is 5 units, find the coordinates of the point.
A point on the x-axis has the form (x, 0).
Using the distance formula between (x, 0) and (3, 4):
5 = √[(x − 3)² + (0 − 4)²]
Square both sides:
25 = (x − 3)² + 16
(x − 3)² = 9
x − 3 = ±3
x = 6 or x = 0
Verification: distance from (6, 0) to (3, 4) = √[9 + 16] = √25 = 5 ✓
distance from (0, 0) to (3, 4) = √[9 + 16] = √25 = 5 ✓
The point is (6, 0) or (0, 0).
Q7: Find the slope of the line through P(2, 5) and Q(5, -1). Is the line rising or falling?
Slope formula: m = (y₂ − y₁)/(x₂ − x₁)
Here P(2, 5) and Q(5, -1).
m = (-1 − 5)/(5 − 2)
m = -6/3
m = -2
Since the slope is negative, the line is falling (as x increases, y decreases).
Q8: Verify that the quadrilateral with vertices A(0, 0), B(5, 0), C(5, 4), D(0, 4) is a rectangle.
A quadrilateral is a rectangle if opposite sides are equal AND the diagonals are equal.
Side lengths (distance formula):
AB = √[(5−0)² + (0−0)²] = √25 = 5
BC = √[(5−5)² + (4−0)²] = √16 = 4
CD = √[(0−5)² + (4−4)²] = √25 = 5
DA = √[(0−0)² + (0−4)²] = √16 = 4
Opposite sides: AB = CD = 5, and BC = DA = 4 ✓
Diagonals:
AC = √[(5−0)² + (4−0)²] = √41
BD = √[(0−5)² + (4−0)²] = √41
AC = BD ✓
Since opposite sides are equal and the diagonals are equal, ABCD is a rectangle.
Related
- NCERT Solutions: Coordinate Geometry — the CBSE version of this same topic.
- Constructions — more Karnataka (SSLC) Class 9 Mathematics.
- Areas of Parallelograms and Triangles — more Karnataka (SSLC) Class 9 Mathematics.
- Probability — more Karnataka (SSLC) Class 9 Mathematics.
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