Coordinate Geometry
Madhya Pradesh Board · Class 10 · Mathematics
Flashcards for Coordinate Geometry — Madhya Pradesh Board Class 10 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Explore the full setFind the coordinates of a point on the x-axis that is 5 units from the origin.
Answer
Step 1: A point on the x-axis has the form (x, 0). Step 2: The distance from the origin is OP = √(x² + y²). Step 3: Here y = 0 and OP = 5, so √(x²) = 5. Step 4: So, x = 5 or x = -5. Answer: The point …
x-अक्ष पर एक बिंदु के निर्देशांक ज्ञात कीजिए जो मूल बिंदु से 5 इकाई दूर है।
Answer
चरण 1: x-अक्ष पर एक बिंदु का रूप (x, 0) है। चरण 2: मूल से दूरी OP = √(x² + y²) है। चरण 3: यहाँ y = 0 और OP = 5 है, इसलिए √(x²) = 5 है। चरण 4: तो, x = 5 or x = -5. उत्तर: बिंदु (5, 0) या (-5, 0) हो सकत…
Find the distance between A(4, 0) and B(6, 0).
Answer
Step 1: Both points lie on the x-axis. Step 2: Distance on the x-axis is found by subtracting x-values. Step 3: AB = 6 - 4 = 2 units. Answer: 2 units…
A(4, 0) और B(6, 0) के बीच की दूरी ज्ञात कीजिए।
Answer
चरण 1: दोनों बिंदु x-अक्ष पर स्थित हैं। चरण 2: x-अक्ष पर दूरी x-मानों को घटाकर ज्ञात की जाती है। चरण 3: AB = 6 - 4 = 2 इकाई। उत्तर: 2 इकाई…
Find the distance between C(0, 3) and D(0, 8).
Answer
Step 1: Both points lie on the y-axis. Step 2: Distance on the y-axis is found by subtracting y-values. Step 3: CD = 8 - 3 = 5 units. Answer: 5 units…
C(0, 3) और D(0, 8) के बीच की दूरी ज्ञात कीजिए।
Answer
चरण 1: दोनों बिंदु y-अक्ष पर स्थित हैं। चरण 2: y-अक्ष पर दूरी, y-मानों को घटाकर ज्ञात की जाती है। चरण 3: CD = 8 - 3 = 5 इकाई। उत्तर: 5 इकाई…
Apply the distance formula to P(4, 6) and Q(6, 8).
Answer
Step 1: Use PQ = √((x2 - x1)² + (y2 - y1)²). Step 2: PQ = √((6 - 4)² + (8 - 6)²) Step 3: PQ = √(2² + 2²) = √8 Step 4: PQ = 2√2 units Answer: 2√2 units…
पी(4, 6) और क्यू(6, 8) पर दूरी सूत्र लागू करें।
Answer
चरण 1: PQ = √((x2 - x1)² + (y2 - y1)²) का उपयोग करें। चरण 2: PQ = √((6 - 4)² + (8 - 6)²) चरण 3: PQ = √(2² + 2²) = √8 चरण 4: PQ = 2√2 इकाई उत्तर: 2√2 इकाई…
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