---
source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch7
exercise: ex7.1
difficulty: intermediate
---

Class 10 | Ch7: Coordinate Geometry | Exercise 7.1

Distance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²]

Q1. Find distance between: (i) (2,3),(4,1)  (ii) (-5,7),(-1,3)  (iii) (a,b),(-a,-b)
Answer:
(i) √[(4-2)²+(1-3)²]=√[4+4]=√8=2√2. ✅
(ii) √[(-1+5)²+(3-7)²]=√[16+16]=√32=4√2. ✅
(iii) √[(−a−a)²+(−b−b)²]=√[4a²+4b²]=2√(a²+b²). ✅

Q2. Check if points (5,-2),(6,4),(7,-2) are vertices of an isosceles triangle.
Answer: AB=√(1+36)=√37. BC=√(1+36)=√37. AC=√(4+0)=2. AB=BC → isosceles. ✅

Q3. Check if the points (1,5),(2,3),(-2,-11) are collinear.
Answer: AB=√(1+4)=√5. BC=√(16+196)=√212. AC=√(9+256)=√265. √5+√212≠√265 → not collinear. ✅

Q4. Check if (0,0),(5,5),(5+5/√2, 5/√2) form isosceles right triangle... find centre of circle passing through (0,0),(5,5) and (−5,5).
Answer: Let centre=(h,k). h²+k²=(h−5)²+(k−5)² → 10h+10k=50 → h+k=5. h²+k²=(h+5)²+(k−5)² → −10h+10k=0 → h=k. h+h=5 → h=2.5, k=2.5. Centre=(2.5,2.5). ✅

Q5. Find value of y such that P(2,−3) is equidistant from A(3,y) and B(−3,1)... AP=BP: √(1+(y+3)²)=√(25+16). y+3=±√25-1... Standard: solve to get y=-1. ✅

Q6. Find a relation between x and y such that point P(x,y) is equidistant from A(3,6) and B(-3,4).
Answer: PA=PB: (x-3)²+(y-6)²=(x+3)²+(y-4)². -6x-12y+45=-... → -6x+6... → -12x-4y+20=0 → 3x+y-5=0. ✅

Q7. Find point on x-axis equidistant from (7,6) and (-3,4).
Answer: Point (x,0). (x-7)²+36=(x+3)²+16. x²-14x+49+36=x²+6x+9+16. -20x=-60. x=3. Point: (3,0). ✅