Coordinate Geometry — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Coordinate Geometry" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Coordinate Geometry" — important questions with detailed answers, download PDF…
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Q1: State the distance formula. Find the distance between points A(3, 4) and B(6, 8).
Distance Formula:
The distance d between two points P(x₁, y₁) and Q(x₂, y₂) in a coordinate plane is:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
This formula is derived from the Pythagorean theorem.
Problem: Find distance between A(3, 4) and B(6, 8)
Step 1: Identify the coordinates.
Point A: x₁ = 3, y₁ = 4
Point B: x₂ = 6, y₂ = 8
Step 2: Substitute into the distance formula.
d = √[(6 - 3)² + (8 - 4)²]
Step 3: Simplify inside the square root.
d = √[(3)² + (4)²]
d = √[9 + 16]
d = √25
d = 5 units
Answer:…
Q2: Find the coordinates of the point that divides the line segment joining A(2, 3) and B(8, 9) in the ratio 2:1 internally.
Section Formula (Internal Division):
If point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m:n, then:
P = ((m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n))
Given:
A(2, 3), B(8, 9)
Ratio = 2:1 (m:n = 2:1)
Step 1: Identify the values.
x₁ = 2, y₁ = 3
x₂ = 8, y₂ = 9
m = 2, n = 1
Step 2: Apply the section formula for x-coordinate.
x = (m·x₂ + n·x₁)/(m + n)
x = (2·8 + 1·2)/(2 + 1)
x = (16 + 2)/3
x = 18/3
x = 6
Step 3: Apply the section formula for y-coordinate.
y…
Q3: Prove that the points A(0, 0), B(3, 4), and C(6, 8) are collinear.
Given: Points A(0, 0), B(3, 4), C(6, 8)
Prove: These points are collinear (lie on the same line)
Method 1: Using Slope
Step 1: Find the slope of AB.
Slope of AB = (y₂ - y₁)/(x₂ - x₁) = (4 - 0)/(3 - 0) = 4/3
Step 2: Find the slope of BC.
Slope of BC = (8 - 4)/(6 - 3) = 4/3
Step 3: Check if slopes are equal.
Slope of AB = Slope of BC = 4/3
Also, B lies on both lines, and A and C are on the same side of B.
Step 4: Conclusion.
Since the slopes are equal and B is a common point, the three point…
Q4: Find the area of the triangle with vertices A(1, 2), B(4, 6), and C(5, 1).
Area of Triangle using Coordinate Formula:
For a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃):
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Given:
A(1, 2), B(4, 6), C(5, 1)
x₁ = 1, y₁ = 2
x₂ = 4, y₂ = 6
x₃ = 5, y₃ = 1
Step 1: Substitute into the formula.
Area = (1/2)|1(6 - 1) + 4(1 - 2) + 5(2 - 6)|
Step 2: Calculate each term.
= (1/2)|1(5) + 4(-1) + 5(-4)|
= (1/2)|5 - 4 - 20|
= (1/2)|-19|
= (1/2)(19)
= 9.5 square units
Answer: Area of triangle ABC = 9.5 square units or…
Q5: The midpoint of a line segment joining A and B is M(2, 3). If A is (0, 1), find the coordinates of B.
Midpoint Formula:
If M is the midpoint of AB where A(x₁, y₁) and B(x₂, y₂), then:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Given:
M(2, 3) is the midpoint
A(0, 1)
Find: B(x₂, y₂)
Step 1: Apply the midpoint formula for the x-coordinate.
2 = (0 + x₂)/2
4 = 0 + x₂
x₂ = 4
Step 2: Apply the midpoint formula for the y-coordinate.
3 = (1 + y₂)/2
6 = 1 + y₂
y₂ = 5
Answer: B(4, 5)
Verification:
Midpoint of A(0, 1) and B(4, 5) = ((0 + 4)/2, (1 + 5)/2) = (2, 3) ✓
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