Key Concepts & Formulae at a Glance
- Distance formula:PQ = √[(x₂−x₁)²+(y₂−y₁)²] for P(x₁,y₁) and Q(x₂,y₂) — derived from the Pythagoras theorem.
- Distance from the origin:OP = √(x²+y²)
- Testing collinearity:Three points are collinear if the sum of the two shorter distances equals the longest distance.
- Section formula:P dividing AB in ratio m₁:m₂ has coordinates (m₁x₂+m₂x₁m₁+m₂, m₁y₂+m₂y₁m₁+m₂)
- Midpoint formula:Special case of the section formula with ratio 1:1 — (x₁+x₂2, y₁+y₂2)
- Ratio form k:1:Coordinates become (kx₂+x₁k+1, ky₂+y₁k+1) — useful when finding the ratio a given point divides a segment in.
Exercise 7.1
10 Questions
1Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (−5, 7), (−1, 3)
(iii) (a, b), (−a, −b)
(i) d = √[(4−2)²+(1−3)²] = √[4+4] = √8 = 2√2
(ii) d = √[(−1+5)²+(3−7)²] = √[16+16] = √32 = 4√2
(iii) d = √[(−a−a)²+(−b−b)²] = √[4a²+4b²] = 2√(a²+b²)
(ii) d = √[(−1+5)²+(3−7)²] = √[16+16] = √32 = 4√2
(iii) d = √[(−a−a)²+(−b−b)²] = √[4a²+4b²] = 2√(a²+b²)
(i) 2√2 units (ii) 4√2 units (iii) 2√(a²+b²) units
2Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
d = √[(36−0)²+(15−0)²] = √[1296+225] = √1521 = 39
Town A is at the origin and town B is 36 km east and 15 km north, i.e. at (36, 15) — exactly the two points above. So the same calculation gives the distance between the towns.
Town A is at the origin and town B is 36 km east and 15 km north, i.e. at (36, 15) — exactly the two points above. So the same calculation gives the distance between the towns.
Distance = 39 units, so towns A and B are 39 km apart.
3Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
The three points, plotted — visibly not on one straight line.
Let P(1,5), Q(2,3), R(−2,−11).
PQ = √[(2−1)²+(3−5)²] = √[1+4] = √5
QR = √[(−2−2)²+(−11−3)²] = √[16+196] = √212 = 2√53
PR = √[(−2−1)²+(−11−5)²] = √[9+256] = √265
For collinearity, PQ + QR should equal PR. √5 + 2√53 ≈ 2.24 + 14.56 = 16.80, but √265 ≈ 16.28 — not equal.
PQ = √[(2−1)²+(3−5)²] = √[1+4] = √5
QR = √[(−2−2)²+(−11−3)²] = √[16+196] = √212 = 2√53
PR = √[(−2−1)²+(−11−5)²] = √[9+256] = √265
For collinearity, PQ + QR should equal PR. √5 + 2√53 ≈ 2.24 + 14.56 = 16.80, but √265 ≈ 16.28 — not equal.
The points are not collinear.
4Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Two equal sides (both √37) meeting at the apex — isosceles.
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
AB = BC = √37
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
AB = BC = √37
Yes, the triangle is isosceles (two equal sides, AB = BC = √37).
5In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and Champa asks, "Don't you think ABCD is a square?" Chameli disagrees. Using the distance formula, find who is correct.
ABCD with all four sides equal and both diagonals equal — a square.
From the figure, A(3,4), B(6,7), C(9,4), D(6,1).
AB = √[(6−3)²+(7−4)²] = √[9+9] = 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
AC = √[(9−3)²+(4−4)²] = √36 = 6
BD = √[(6−6)²+(1−7)²] = √36 = 6
All four sides are equal (3√2) and both diagonals are equal (6) — a quadrilateral with equal sides and equal diagonals is a square.
AB = √[(6−3)²+(7−4)²] = √[9+9] = 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
AC = √[(9−3)²+(4−4)²] = √36 = 6
BD = √[(6−6)²+(1−7)²] = √36 = 6
All four sides are equal (3√2) and both diagonals are equal (6) — a quadrilateral with equal sides and equal diagonals is a square.
Champa is correct — ABCD is a square.
6Name the type of quadrilateral formed, if any, by the following points, and give reasons:
(i) (−1, −2), (1, 0), (−1, 2), (−3, 0)
(ii) (−3, 5), (3, 1), (0, 3), (−1, −4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i)
All sides 2√2, both diagonals 4 — a square.
Let A(−1,−2), B(1,0), C(−1,2), D(−3,0).
AB = BC = CD = DA = √[2²+2²] = 2√2
AC = √[0+16] = 4, BD = √[16+0] = 4
All sides equal, both diagonals equal.
AB = BC = CD = DA = √[2²+2²] = 2√2
AC = √[0+16] = 4, BD = √[16+0] = 4
All sides equal, both diagonals equal.
A square.
(ii)
(-3,5), (0,3) and (3,1) lie on one straight line — no quadrilateral is formed.
Let A(−3,5), B(3,1), C(0,3), D(−1,−4).
Checking the area of ΔABC using these three points gives exactly 0 — A, C and B lie on a single straight line.
Checking the area of ΔABC using these three points gives exactly 0 — A, C and B lie on a single straight line.
No quadrilateral is formed — three of the four points (A, C and B) are collinear.
(iii)
Opposite sides equal (√10 and 3√2) but diagonals unequal — a parallelogram.
Let A(4,5), B(7,6), C(4,3), D(1,2).
AB = √[9+1] = √10, BC = √[9+9] = 3√2, CD = √[9+1] = √10, DA = √[9+9] = 3√2
AC = √[0+4] = 2, BD = √[36+16] = 2√13
Opposite sides are equal (AB=CD, BC=DA) but the diagonals are unequal — not a rhombus or rectangle.
AB = √[9+1] = √10, BC = √[9+9] = 3√2, CD = √[9+1] = √10, DA = √[9+9] = 3√2
AC = √[0+4] = 2, BD = √[36+16] = 2√13
Opposite sides are equal (AB=CD, BC=DA) but the diagonals are unequal — not a rhombus or rectangle.
A parallelogram (general, not a rhombus or rectangle).
7Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
A point on the x-axis has the form (x, 0). Let P(x, 0) be equidistant from A(2,−5) and B(−2,9).
PA² = PB²
(x−2)²+(0+5)² = (x+2)²+(0−9)²
x²−4x+4+25 = x²+4x+4+81
−8x = 56
x = −7
PA² = PB²
(x−2)²+(0+5)² = (x+2)²+(0−9)²
x²−4x+4+25 = x²+4x+4+81
−8x = 56
x = −7
The required point is (−7, 0).
8Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
PQ² = (10−2)²+(y+3)² = 100
64+(y+3)² = 100
(y+3)² = 36
y+3 = ±6
y = 3 or y = −9
64+(y+3)² = 100
(y+3)² = 36
y+3 = ±6
y = 3 or y = −9
y = 3 or y = −9
9If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.
QP² = QR²
(0−5)²+(1+3)² = (0−x)²+(1−6)²
25+16 = x²+25
x² = 16
x = ±4
For x = 4: QR = √[(0−4)²+(1−6)²] = √[16+25] = √41; PR = √[(5−4)²+(−3−6)²] = √[1+81] = √82
For x = −4: QR = √[(0+4)²+(1−6)²] = √[16+25] = √41; PR = √[(5+4)²+(−3−6)²] = √[81+81] = 9√2
(0−5)²+(1+3)² = (0−x)²+(1−6)²
25+16 = x²+25
x² = 16
x = ±4
For x = 4: QR = √[(0−4)²+(1−6)²] = √[16+25] = √41; PR = √[(5−4)²+(−3−6)²] = √[1+81] = √82
For x = −4: QR = √[(0+4)²+(1−6)²] = √[16+25] = √41; PR = √[(5+4)²+(−3−6)²] = √[81+81] = 9√2
x = 4 or x = −4. In both cases QR = √41; PR = √82 (when x=4) or 9√2 (when x=−4).
10Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (−3, 4).
Let P(x,y) be equidistant from A(3,6) and B(−3,4).
(x−3)²+(y−6)² = (x+3)²+(y−4)²
x²−6x+9+y²−12y+36 = x²+6x+9+y²−8y+16
−12x−4y+20 = 0
3x+y−5 = 0 (dividing by −4)
(x−3)²+(y−6)² = (x+3)²+(y−4)²
x²−6x+9+y²−12y+36 = x²+6x+9+y²−8y+16
−12x−4y+20 = 0
3x+y−5 = 0 (dividing by −4)
3x + y − 5 = 0 (equivalently, y = 5−3x)
Exercise 7.2
10 Questions
1Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Using the section formula with A(−1,7), B(4,−3), m₁:m₂=2:3:
x = [2(4)+3(−1)]/(2+3) = [8−3]/5 = 1
y = [2(−3)+3(7)]/(2+3) = [−6+21]/5 = 3
x = [2(4)+3(−1)]/(2+3) = [8−3]/5 = 1
y = [2(−3)+3(7)]/(2+3) = [−6+21]/5 = 3
The required point is (1, 3).
2Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Let P and Q trisect AB (A(4,−1), B(−2,−3)), i.e. AP=PQ=QB.
P divides AB in the ratio 1:2:
x = [1(−2)+2(4)]/3 = 63 = 2, y = [1(−3)+2(−1)]/3 = −53
Q divides AB in the ratio 2:1:
x = [2(−2)+1(4)]/3 = 03 = 0, y = [2(−3)+1(−1)]/3 = −73
P divides AB in the ratio 1:2:
x = [1(−2)+2(4)]/3 = 63 = 2, y = [1(−3)+2(−1)]/3 = −53
Q divides AB in the ratio 2:1:
x = [2(−2)+1(4)]/3 = 03 = 0, y = [2(−3)+1(−1)]/3 = −73
The points of trisection are (2, −53) and (0, −73).
3In a rectangular school ground ABCD, flower pots are placed 1 m apart along AD. Niharika runs 14 of the distance AD on the 2nd line and posts a green flag. Preet runs 15 of the distance AD on the 8th line and posts a red flag. Find the distance between the flags, and where Rashmi should post a blue flag exactly halfway between them.
Green flag on line 2 (14 of AD=25), red flag on line 8 (15 of AD=20), blue flag at their midpoint.
AD = 100 m (100 flower pots, 1 m apart).
Niharika's green flag: on line 2 (x=2), at 14 of AD = 25 m up — so green flag is at (2, 25).
Preet's red flag: on line 8 (x=8), at 15 of AD = 20 m up — so red flag is at (8, 20).
Distance between flags = √[(8−2)²+(20−25)²] = √[36+25] = √61
Blue flag = midpoint of green and red = (2+82, 25+202) = (5, 22.5)
Niharika's green flag: on line 2 (x=2), at 14 of AD = 25 m up — so green flag is at (2, 25).
Preet's red flag: on line 8 (x=8), at 15 of AD = 20 m up — so red flag is at (8, 20).
Distance between flags = √[(8−2)²+(20−25)²] = √[36+25] = √61
Blue flag = midpoint of green and red = (2+82, 25+202) = (5, 22.5)
The flags are √61 m apart. Rashmi should post the blue flag at (5, 22.5) — on the 5th line, 22.5 m up.
4Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Let (−1,6) divide A(−3,10), B(6,−8) in ratio m₁:m₂.
−1 = [6m₁−3m₂]/(m₁+m₂)
−m₁−m₂ = 6m₁−3m₂
2m₂ = 7m₁
m₁:m₂ = 2:7
−1 = [6m₁−3m₂]/(m₁+m₂)
−m₁−m₂ = 6m₁−3m₂
2m₂ = 7m₁
m₁:m₂ = 2:7
The ratio is 2 : 7.
5Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the ratio be k:1. The y-coordinate of the dividing point is [k(5)+1(−5)]/(k+1). On the x-axis, y=0:
5k−5 = 0
k = 1
So the ratio is 1:1 (the x-axis bisects AB).
x-coordinate = [1(−4)+1(1)]/2 = −32
5k−5 = 0
k = 1
So the ratio is 1:1 (the x-axis bisects AB).
x-coordinate = [1(−4)+1(1)]/2 = −32
The x-axis divides AB in the ratio 1 : 1, at the point (−32, 0).
6If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
The diagonals of a parallelogram bisect each other, so the midpoint of the diagonal joining (1,2) and (x,6) equals the midpoint of the diagonal joining (4,y) and (3,5).
x-coordinate: 1+x2 = 4+32 ⇒ 1+x = 7 ⇒ x = 6
y-coordinate: 2+62 = y+52 ⇒ 8 = y+5 ⇒ y = 3
x-coordinate: 1+x2 = 4+32 ⇒ 1+x = 7 ⇒ x = 6
y-coordinate: 2+62 = y+52 ⇒ 8 = y+5 ⇒ y = 3
x = 6, y = 3
7Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
A, the centre, and B are collinear with the centre as the midpoint of AB.
The centre of a circle is the midpoint of any diameter, so the centre (2,−3) is the midpoint of A and B(1,4).
2 = Aₓ+12 ⇒ Aₓ = 3
−3 = Aỿ+42 ⇒ Aỿ = −10
2 = Aₓ+12 ⇒ Aₓ = 3
−3 = Aỿ+42 ⇒ Aỿ = −10
A = (3, −10)
8If A and B are (−2, −2) and (2, −4) respectively, find the coordinates of P such that AP = 37 AB and P lies on line segment AB.
AP = 37 AB means P divides AB in the ratio AP:PB = 3:4.
x = [3(2)+4(−2)]/(3+4) = [6−8]/7 = −27
y = [3(−4)+4(−2)]/(3+4) = [−12−8]/7 = −207
x = [3(2)+4(−2)]/(3+4) = [6−8]/7 = −27
y = [3(−4)+4(−2)]/(3+4) = [−12−8]/7 = −207
P = (−27, −207)
9Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
A and B divided into four equal parts by P1, P2 (midpoint) and P3.
Let P₁, P₂, P₃ divide AB into four equal parts, so they divide it in ratios 1:3, 2:2 (=1:1) and 3:1.
P₁ (ratio 1:3): x=[1(2)+3(−2)]/4=−1, y=[1(8)+3(2)]/4=72 → (−1, 72)
P₂ (ratio 1:1, the midpoint): x=−2+22=0, y=2+82=5 → (0, 5)
P₃ (ratio 3:1): x=[3(2)+1(−2)]/4=1, y=[3(8)+1(2)]/4=132 → (1, 132)
P₁ (ratio 1:3): x=[1(2)+3(−2)]/4=−1, y=[1(8)+3(2)]/4=72 → (−1, 72)
P₂ (ratio 1:1, the midpoint): x=−2+22=0, y=2+82=5 → (0, 5)
P₃ (ratio 3:1): x=[3(2)+1(−2)]/4=1, y=[3(8)+1(2)]/4=132 → (1, 132)
The three points are (−1, 72), (0, 5) and (1, 132).
10Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1), taken in order. [Hint: Area of a rhombus = ½ × product of its diagonals]
Rhombus with diagonals 4√2 and 6√2 drawn in — area = ½ × product of diagonals.
Let A(3,0), B(4,5), C(−1,4), D(−2,−1). The diagonals are AC and BD.
AC = √[(−1−3)²+(4−0)²] = √[16+16] = 4√2
BD = √[(−2−4)²+(−1−5)²] = √[36+36] = 6√2
Area = ½ × AC × BD = ½ × 4√2 × 6√2 = ½ × 48
AC = √[(−1−3)²+(4−0)²] = √[16+16] = 4√2
BD = √[(−2−4)²+(−1−5)²] = √[36+36] = 6√2
Area = ½ × AC × BD = ½ × 4√2 × 6√2 = ½ × 48
Area of the rhombus = 24 square units.
Frequently Asked Questions
The distance between P(x₁,y₁) and Q(x₂,y₂) is √[(x₂−x₁)²+(y₂−y₁)²]. It comes directly from the Pythagoras theorem — the horizontal and vertical gaps between the points form the two legs of a right triangle, and the distance is the hypotenuse.
Find the three pairwise distances. If the two shorter distances add up to exactly the longest one, the points lie on a single straight line (there's no genuine triangle). If the sum is even slightly more than the longest distance, the points form a proper triangle instead — see Exercise 7.1, Q3.
The midpoint formula is just the section formula with the ratio fixed at 1:1. The general section formula lets you find the point dividing a segment in any ratio m₁:m₂, while the midpoint formula only ever gives the exact centre.
Set up the section formula with an unknown ratio (commonly written k:1), equate the x-coordinate (or y-coordinate) to the known value, and solve for k. It's often quicker to use just one coordinate first and then verify with the other — as in Exercise 7.2, Q4 and Q5.
Yes — a printable Formula Card covering all 7 units, including Coordinate Geometry, is available for ₹99.
Continue to the Next Chapter
Chapter 8 (Introduction to Trigonometry) moves from coordinates into ratios and angles.