Class 10 Maths (NCERT)
Chapter 7: Coordinate Geometry
Exercise 7.1 – Stepwise Solutions
Distance Formula (Concept)
If A(x1, y1) and B(x2, y2), then
AB = √[(x2 − x1)² + (y2 − y1)²]
1. Find the distance between the following pairs of points
(i) (2, 3) and (4, 1)
Step 1: x1=2, y1=3, x2=4, y2=1
Step 2: Distance = √[(4−2)² + (1−3)²]
= √[2² + (−2)²] = √[4 + 4] = √8 = 2√2
(ii) (−5, 7) and (−1, 3)
Distance = √[(−1+5)² + (3−7)²]
= √[4² + (−4)²] = √[16+16] = √32 = 4√2
(iii) (a, b) and (−a, −b)
Distance = √[(−a−a)² + (−b−b)²]
= √[(−2a)² + (−2b)²]
= √[4a² + 4b²] = √[4(a²+b²)]
= 2√(a² + b²)
2. Find the distance between (0,0) and (36,15)
Distance = √[(36−0)² + (15−0)²]
= √[36² + 15²] = √[1296 + 225] = √1521
= 39
3. Check if (1,5), (2,3), (−2,−11) are collinear
Step 1: Find slope of (1,5) and (2,3)
m12 = (3−5)/(2−1) = −2
Step 2: Find slope of (2,3) and (−2,−11)
m23 = (−11−3)/(−2−2) = (−14)/(−4) = 7/2
Since m12 ≠ m23, points are NOT collinear.
4. Check whether (5,−2), (6,4), (7,−2) form an isosceles triangle
Let A(5,−2), B(6,4), C(7,−2)
AB = √[(6−5)² + (4+2)²] = √[1 + 36] = √37
BC = √[(7−6)² + (−2−4)²] = √[1 + 36] = √37
AC = √[(7−5)² + (−2+2)²] = √[4 + 0] = 2
Since AB = BC, triangle is isosceles.
5. Using distance formula, check whether ABCD is a square (Fig. 7.8)
From graph, the coordinates are:
A(3,4), B(6,7), C(9,4), D(6,1)
AB = √[(6−3)² + (7−4)²] = √[9+9] = √18 = 3√2
BC = √[(9−6)² + (4−7)²] = √[9+9] = 3√2
CD = √[(6−9)² + (1−4)²] = √[9+9] = 3√2
DA = √[(3−6)² + (4−1)²] = √[9+9] = 3√2
Diagonals:
AC = √[(9−3)² + (4−4)²] = √36 = 6
BD = √[(6−6)² + (7−1)²] = √36 = 6
All sides are equal and diagonals are equal, so ABCD is a square.
Champa is correct.
6. Name the type of quadrilateral formed (if any)
(i) (−1,−2), (1,0), (−1,2), (−3,0)
Label A(−1,−2), B(1,0), C(−1,2), D(−3,0)
AB = √[(2)²+(2)²]=√8, BC=√[(−2)²+(2)²]=√8
CD=√[(−2)²+(−2)²]=√8, DA=√[(2)²+(−2)²]=√8
All sides equal. Also diagonals AC is vertical and BD is horizontal.
Therefore it is a square.
(ii) (−3,5), (3,1), (0,3), (−1,−4)
Check slopes quickly:
Slope between first two points = (1−5)/(3+3)= −4/6= −2/3
Slope between next two points = (3−1)/(0−3)=2/(−3)= −2/3
So two sides are parallel, but other pair is not parallel.
Therefore it is a trapezium.
(iii) (4,5), (7,6), (4,3), (1,2)
Check slopes:
Slope AB = (6−5)/(7−4)=1/3
Slope CD = (2−3)/(1−4)= (−1)/(−3)=1/3
So AB ∥ CD.
Slope BC = (3−6)/(4−7)= (−3)/(−3)=1
Slope DA = (5−2)/(4−1)=3/3=1
So BC ∥ DA.
Both opposite sides parallel ⇒ parallelogram.
7. Find the point on x-axis equidistant from (2,−5) and (−2,9)
Let point be P(x,0)
Distance squared from (2,−5): (x−2)² + (0+5)²
Distance squared from (−2,9): (x+2)² + (0−9)²
Equate:
(x−2)² + 25 = (x+2)² + 81
x² − 4x + 4 + 25 = x² + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56
x = −7
Required point = (−7, 0)
8. Find y if distance between P(2,−3) and Q(10,y) is 10
Distance = √[(10−2)² + (y+3)²] = 10
√[8² + (y+3)²] = 10
64 + (y+3)² = 100
(y+3)² = 36
y + 3 = ±6
y = 3 or y = −9
Answer: y = 3, −9
9. If Q(0,1) is equidistant from P(5,−3) and R(x,6), find x and distances QR and PR
QP² = (0−5)² + (1+3)² = 25 + 16 = 41
QR² = (0−x)² + (1−6)² = x² + 25
Equidistant ⇒ QP² = QR²
41 = x² + 25
x² = 16 ⇒ x = 4 or x = −4
QR = √41
Now PR:
If x = 4, R(4,6): PR² = (4−5)² + (6+3)² = 1 + 81 = 82 ⇒ PR = √82
If x = −4, R(−4,6): PR² = (−4−5)² + (6+3)² = 81 + 81 = 162 ⇒ PR = 9√2
Answers:
x = 4 ⇒ QR = √41, PR = √82
x = −4 ⇒ QR = √41, PR = 9√2
10. Find relation between x and y if (x,y) is equidistant from (3,6) and (−3,4)
Equidistant ⇒ distances squared equal:
(x−3)² + (y−6)² = (x+3)² + (y−4)²
Expand:
(x² − 6x + 9) + (y² − 12y + 36) = (x² + 6x + 9) + (y² − 8y + 16)
Cancel x², y², 9:
−6x − 12y + 36 = 6x − 8y + 16
Bring terms together:
−12x − 4y + 20 = 0
Divide by −4:
3x + y − 5 = 0
Relation: 3x + y = 5
Final Answers (Quick)
1(i) 2√2 (ii) 4√2 (iii) 2√(a²+b²)
2) 39
3) Not collinear
4) Isosceles
5) ABCD is a square (Champa correct)
6(i) Square (ii) Trapezium (iii) Parallelogram
7) (−7,0)
8) y = 3, −9
9) x = 4 or −4; QR = √41; PR = √82 or 9√2
10) 3x + y = 5