Ex 7.1 Q8 - Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:-Given that,
*Distance between the points P(2, -3) and Q(10, y) is 10 units
∴ PQ = 10
➙ Using distance formula,
D² = (x₂ - x₁)² + (y₂ - y₁)²
∴ PQ² = (10 - 2)² + (y + 3)²
∴ (10)² = (8)² + (y + 3)²
∴ 100 = 64 + (y + 3)²
∴ (y + 3)² = 100 - 64
∴ (y + 3)² = 36
*Taking square root on both sides,
∴ y + 3 = ±6
Now,
y + 3 = 6 OR y + 3 = -6
∴ y = 6 - 3 ∴ y = -6 - 3
∴ y = 3 ∴ y = -9
Hence,
The values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units are 3 or -9.
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Try This..
✒ The coordinates of one point are (x, 7) and the other is (2, –1). Distance = 10 units. Find x.
✒ If the distance between A(3, y) and B(6, –3) is 5 units, find the value of y.
✒ Find the distance between the points (4, 5) and (1, 1).
❌ Common Mistakes
Forgetting to square negatives:▸ Remember, (–3)² = 9, not –9!Wrong sign while simplifying:▸ (y – (–3)) = y + 3, not y – 3.Missing ± while taking square roots:▸ Always consider both positive and negative roots.Not squaring both sides correctly:▸ Squaring removes the square root. Don’t forget it!
📝 Related Questions:-
Queries Solved:-
Class 10 Ex 7.1
Ex 7.1 Q8 Class 10
Class 10 Ex 7.1 Q8
Class 10 Chap 7 Ex 7.1 Q8
Class 10 Coordinate Geometry
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