Coordinate Geometry — Flashcards
Karnataka Board · Class 10 · Mathematics
20 flashcards for Coordinate Geometry (Karnataka Board Class 10 Mathematics) to test yourself on key terms and facts.
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Find the distance between points A(3, 4) and B(7, 1).
Answer
Step 1: Apply distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Step 2: Substitute values: d = √[(7-3)² + (1-4)²] Step 3: Simplify: d = √[4² + (-3)²] = √[16 + 9] Step 4: Calculate: d = √25 = 5 units Answer…
Prove that points P(1, 7), Q(4, 2), R(-1, -1), and S(-4, 4) form a square.
Answer
Step 1: Find all side lengths using distance formula PQ = √[(4-1)² + (2-7)²] = √[9+25] = √34 QR = √[(-1-4)² + (-1-2)²] = √[25+9] = √34 RS = √[(-4-(-1))² + (4-(-1))²] = √[9+25] = √34 SP = √[(1-(-4))² +…
Find the coordinates of point P that divides the line segment joining A(2, 3) and B(8, 9) in ratio 2:1 internally.
Answer
Step 1: Apply section formula: P(x,y) = [(m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂)] Step 2: Here m₁:m₂ = 2:1, A(2,3), B(8,9) Step 3: x-coordinate: x = (2×8 + 1×2)/(2+1) = (16+2)/3 = 18/3 = 6 Step 4…
When do you use the section formula and what does each part mean?
Answer
Use when: Finding coordinates of a point dividing a line segment in a given ratio Formula: P(x,y) = [(m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂)] Where: m₁:m₂ is the division ratio, (x₁,y₁) and (x₂,y…
Find the midpoint of the line segment joining A(-3, 5) and B(7, -1).
Answer
Step 1: Apply midpoint formula (special case of section formula with ratio 1:1) Midpoint = [(x₁+x₂)/2, (y₁+y₂)/2] Step 2: Substitute A(-3,5) and B(7,-1) Midpoint = [(-3+7)/2, (5+(-1))/2] Step 3: Simpl…
In what ratio does point P(1, 4) divide the line segment joining A(-2, 1) and B(7, 10)?
Answer
Step 1: Let ratio be k:1. Use section formula P(1,4) = [(k×7 + 1×(-2))/(k+1), (k×10 + 1×1)/(k+1)] Step 2: From x-coordinate: 1 = (7k-2)/(k+1) Step 3: Cross multiply: 1(k+1) = 7k-2 k+1 = 7k-2 3 = 6k k …
Check if points A(1, 5), B(2, 3), and C(-2, -11) are collinear.
Answer
Step 1: Find distances AB, BC, and AC AB = √[(2-1)² + (3-5)²] = √[1+4] = √5 BC = √[(-2-2)² + (-11-3)²] = √[16+196] = √212 = 2√53 AC = √[(-2-1)² + (-11-5)²] = √[9+256] = √265 Step 2: Check if AB + BC =…
Find a point on the y-axis equidistant from A(6, 5) and B(-4, 3).
Answer
Step 1: Point on y-axis has form P(0, y) Step 2: Set PA = PB (equidistant condition) √[(0-6)² + (y-5)²] = √[(0-(-4))² + (y-3)²] Step 3: Square both sides to eliminate square roots 36 + (y-5)² = 16 + (…
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Sources & Official References
- Karnataka SSLC — kseeb.kar.nic.in
- Dept of Pre-University Education, Karnataka
- National Education Policy 2020 — education.gov.in
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