Std 10 Mathematics Part 2 · Chapter 5

5. Co-ordinate Geometry

Solutions to Practice Sets 5.1 to 5.3 and Problem Set 5, with full working and graphs. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 5.1

1Find the distance between each of the following pairs of points.

(1) A(2, 3), B(4, 1)

AB = √(4 − 2)² + (1 − 3)² … by distance formula

= √4 + 4 = √8 = AB = 2√2

(2) P(−5, 7), Q(−1, 3)

PQ = √(−1 − (−5))² + (3 − 7)² … by distance formula

= √16 + 16 = √32 = PQ = 4√2

(3) R(0, −3), S(0, 52)

RS = √(0 − 0)² + (52 − (−3))² … by distance formula

= √0 + 1214 = √1214 = RS = 112

(4) L(5, −8), M(−7, −3)

LM = √(−7 − 5)² + (−3 − (−8))² … by distance formula

= √144 + 25 = √169 = LM = 13

(5) T(−3, 6), R(9, −10)

TR = √(9 − (−3))² + (−10 − 6)² … by distance formula

= √144 + 256 = √400 = TR = 20

(6) W(−72, 4), X(11, 4)

WX = √(11 − (−72))² + (4 − 4)² … by distance formula

= √8414 + 0 = √8414 = WX = 292

2Determine whether the points are collinear.

(1) A(1, −3), B(2, −5), C(−4, 7)

AB = √1² + (−2)² = √5

BC = √(−6)² + 12² = √180 = 6√5

AC = √(−5)² + 10² = √125 = 5√5

AB + AC = √5 + 5√5 = 6√5 = BC

∴ Points A, B, C are collinear.

(2) L(−2, 3), M(1, −3), N(5, 4)

LM = √3² + (−6)² = √45 = 3√5 ≈ 6.71

MN = √4² + 7² = √65 ≈ 8.06

LN = √7² + 1² = √50 = 5√2 ≈ 7.07

No sum of two distances is equal to the third. (6.71 + 7.07 ≠ 8.06)

∴ Points L, M, N are not collinear.

(3) R(0, 3), D(2, 1), S(3, −1)

RD = √2² + (−2)² = 2√2 ≈ 2.83

DS = √1² + (−2)² = √5 ≈ 2.24

RS = √3² + (−4)² = 5

RD + DS ≈ 5.07 ≠ RS

∴ Points R, D, S are not collinear.

(4) P(−2, 3), Q(1, 2), R(4, 1)

PQ = √3² + (−1)² = √10

QR = √3² + (−1)² = √10

PR = √6² + (−2)² = √40 = 2√10

PQ + QR = √10 + √10 = 2√10 = PR

∴ Points P, Q, R are collinear.

3Find the point on the X-axis which is equidistant from A(−3, 4) and B(1, −4).

A point on the X-axis has y co-ordinate 0. Let the point be P(x, 0).

PA = PB, so PA² = PB².

∴ (x + 3)² + (0 − 4)² = (x − 1)² + (0 + 4)²

∴ x² + 6x + 9 + 16 = x² − 2x + 1 + 16

∴ 8x = −8

∴ x = −1

∴ The required point is (−1, 0)

4Verify that points P(−2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

PQ² = (2 + 2)² + (2 − 2)² = 16

QR² = (2 − 2)² + (7 − 2)² = 25

PR² = (2 + 2)² + (7 − 2)² = 16 + 25 = 41

∴ PQ² + QR² = 16 + 25 = 41 = PR²

∴ Δ PQR is a right angled triangle, with ∠ Q = 90°. … converse of Pythagoras theorem

5Show that points P(2, −2), Q(7, 3), R(11, −1) and S(6, −6) are vertices of a parallelogram.

PQ = √5² + 5² = √50

QR = √4² + (−4)² = √32

RS = √(−5)² + (−5)² = √50

SP = √(−4)² + 4² = √32

∴ PQ = RS and QR = SP

Both pairs of opposite sides of □ PQRS are congruent.

∴ □ PQRS is a parallelogram.

6Show that points A(−4, −7), B(−1, 2), C(8, 5) and D(5, −4) are vertices of a rhombus ABCD.

AB = √3² + 9² = √90

BC = √9² + 3² = √90

CD = √(−3)² + (−9)² = √90

DA = √(−9)² + (−3)² = √90

∴ AB = BC = CD = DA

All four sides of □ ABCD are congruent.

∴ □ ABCD is a rhombus.

7Find x if distance between points L(x, 7) and M(1, 15) is 10.

LM² = (1 − x)² + (15 − 7)² … by distance formula

∴ 100 = (1 − x)² + 64

∴ (1 − x)² = 36

∴ 1 − x = ±6

∴ x = 1 − 6 = −5 or x = 1 + 6 = 7

∴ x = −5 or x = 7

8Show that the points A(1, 2), B(1, 6), C(1 + 2√3, 4) are vertices of an equilateral triangle.

AB = √(1 − 1)² + (6 − 2)² = √16 = 4

BC = √(2√3)² + (4 − 6)² = √12 + 4 = 4

AC = √(2√3)² + (4 − 2)² = √12 + 4 = 4

∴ AB = BC = AC

∴ Δ ABC is an equilateral triangle.

Practice Set 5.2

1Find the coordinates of point P if P divides the line segment joining the points A(−1, 7) and B(4, −3) in the ratio 2 : 3.

Here (x₁, y₁) = (−1, 7), (x₂, y₂) = (4, −3) and m : n = 2 : 3.

x = mx₂ + nx₁m + n = 2 × 4 + 3 × (−1)2 + 3 = 8 − 35 = 1 … by section formula

y = my₂ + ny₁m + n = 2 × (−3) + 3 × 72 + 3 = −6 + 215 = 3

∴ Co-ordinates of P are (1, 3)

2In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.

(1) P(−3, 7), Q(1, −4), a : b = 2 : 1

Here (x₁, y₁) = (−3, 7), (x₂, y₂) = (1, −4) and m : n = 2 : 1.

x = mx₂ + nx₁m + n = 2 × 1 + 1 × (−3)2 + 1 = 2 − 33 = −13 … by section formula

y = my₂ + ny₁m + n = 2 × (−4) + 1 × 72 + 1 = −8 + 73 = −13

∴ Co-ordinates of A are (−13, −13)

(2) P(−2, −5), Q(4, 3), a : b = 3 : 4

Here (x₁, y₁) = (−2, −5), (x₂, y₂) = (4, 3) and m : n = 3 : 4.

x = mx₂ + nx₁m + n = 3 × 4 + 4 × (−2)3 + 4 = 12 − 87 = 47 … by section formula

y = my₂ + ny₁m + n = 3 × 3 + 4 × (−5)3 + 4 = 9 − 207 = −117

∴ Co-ordinates of A are (47, −117)

(3) P(2, 6), Q(−4, 1), a : b = 1 : 2

Here (x₁, y₁) = (2, 6), (x₂, y₂) = (−4, 1) and m : n = 1 : 2.

x = mx₂ + nx₁m + n = 1 × (−4) + 2 × 21 + 2 = −4 + 43 = 0 … by section formula

y = my₂ + ny₁m + n = 1 × 1 + 2 × 61 + 2 = 1 + 123 = 133

∴ Co-ordinates of A are (0, 133)

3Find the ratio in which point T(−1, 6) divides the line segment joining the points P(−3, 10) and Q(6, −8).

Let T divide seg PQ in the ratio m : n.

Using the x co-ordinate, −1 = m × 6 + n × (−3)m + n … by section formula

∴ −m − n = 6m − 3n

∴ 2n = 7m

∴ mn = 27

Check with y: 2 × (−8) + 7 × 102 + 7 = 549 = 6 ✓

∴ T divides seg PQ in the ratio 2 : 7

4Point P is the centre of the circle and AB is a diameter. Find the coordinates of point B if coordinates of point A and P are (2, −3) and (−2, 0) respectively.

The centre of a circle is the midpoint of its diameter. Let B = (x, y).

−2 = 2 + x2 and 0 = −3 + y2 … midpoint formula

∴ 2 + x = −4 and −3 + y = 0

∴ x = −6 and y = 3

∴ Co-ordinates of B are (−6, 3)

5Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) and B(1, 2). Also find k.

Let P divide seg AB in the ratio m : n.

Using the y co-ordinate, 7 = m × 2 + n × 9m + n … by section formula

∴ 7m + 7n = 2m + 9n

∴ 5m = 2n

∴ mn = 25

k = 2 × 1 + 5 × 82 + 5 = 427 = 6

∴ The ratio is 2 : 5 and k = 6

6Find the coordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Midpoint = (22 + 02, 20 + 162) … midpoint formula

= (11, 18)

∴ The midpoint is (11, 18)

7Find the centroids of the triangles whose vertices are given below.

Centroid G = (x₁ + x₂ + x₃3, y₁ + y₂ + y₃3) … centroid formula

(1) (−7, 6), (2, −2), (8, 5)

G = (−7 + 2 + 83, 6 − 2 + 53) = (33, 93)

∴ Centroid is (1, 3)

(2) (3, −5), (4, 3), (11, −4)

G = (3 + 4 + 113, −5 + 3 − 43) = (183, −63)

∴ Centroid is (6, −2)

(3) (4, 7), (8, 4), (7, 11)

G = (4 + 8 + 73, 7 + 4 + 113) = (193, 223)

∴ Centroid is (193, 223)

8In Δ ABC, G(−4, −7) is the centroid. If A(−14, −19) and B(3, 5) then find the co-ordinates of C.

Let C = (x, y).

−4 = −14 + 3 + x3 and −7 = −19 + 5 + y3 … centroid formula

∴ −12 = −11 + x and −21 = −14 + y

∴ x = −1 and y = −7

∴ Co-ordinates of C are (−1, −7)

9A(h, −6), B(2, 3) and C(−6, k) are the co-ordinates of vertices of a triangle whose centroid is G(1, 5). Find h and k.

1 = h + 2 − 63 and 5 = −6 + 3 + k3 … centroid formula

∴ 3 = h − 4 and 15 = k − 3

∴ h = 7 and k = 18

10Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(−4, −8).

Let P and Q be the points of trisection, with A − P − Q − B. AP = PQ = QB.

P divides seg AB in the ratio 1 : 2.

P = (1 × (−4) + 2 × 23, 1 × (−8) + 2 × 73) = (0, 2)

Q is the midpoint of seg PB.

Q = (0 + (−4)2, 2 + (−8)2) = (−2, −3)

∴ The points of trisection are (0, 2) and (−2, −3)

11If A(−14, −10), B(6, −2) is given, find the coordinates of the points which divide segment AB into four equal parts.

Let P, Q, R divide seg AB into four equal parts, with A − P − Q − R − B.

Q is the midpoint of seg AB. Q = (−14 + 62, −10 − 22) = (−4, −6)

P is the midpoint of seg AQ. P = (−14 − 42, −10 − 62) = (−9, −8)

R is the midpoint of seg QB. R = (−4 + 62, −6 − 22) = (1, −4)

∴ The points are (−9, −8), (−4, −6) and (1, −4)

12If A(20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.

Let P, Q, R, S divide seg AB into five congruent parts, with A − P − Q − R − S − B.

P divides AB in the ratio 1 : 4, Q in 2 : 3, R in 3 : 2 and S in 4 : 1.

P = (1 × 0 + 4 × 205, 1 × 20 + 4 × 105) = (16, 12)

Q = (2 × 0 + 3 × 205, 2 × 20 + 3 × 105) = (12, 14)

R = (3 × 0 + 2 × 205, 3 × 20 + 2 × 105) = (8, 16)

S = (4 × 0 + 1 × 205, 4 × 20 + 1 × 105) = (4, 18)

∴ The points are (16, 12), (12, 14), (8, 16) and (4, 18)

Practice Set 5.3

1Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines. (1) 45° (2) 60° (3) 90°

Slope = tan θ, where θ is the angle made with the positive direction of the X-axis.

(1) Slope = tan 45° = 1

(2) Slope = tan 60° = √3

(3) tan 90° is not defined. So the slope of this line cannot be determined. (The line is parallel to the Y-axis.)

2Find the slopes of the lines passing through the given points.

Slope = y₂ − y₁x₂ − x₁

(1) A(2, 3), B(4, 7)

Slope = 7 − 34 − 2 = 42 = 2

∴ Slope = 2

(2) P(−3, 1), Q(5, −2)

Slope = −2 − 15 − (−3) = −38

∴ Slope = −38

(3) C(5, −2), D(7, 3)

Slope = 3 − (−2)7 − 5 = 52

∴ Slope = 52

(4) L(−2, −3), M(−6, −8)

Slope = −8 − (−3)−6 − (−2) = −5−4 = 54

∴ Slope = 54

(5) E(−4, −2), F(6, 3)

Slope = 3 − (−2)6 − (−4) = 510 = 12

∴ Slope = 12

(6) T(0, −3), S(0, 4)

Slope = 4 − (−3)0 − 0 = 70

Division by zero is not possible. Line TS is the Y-axis.

∴ The slope of line TS cannot be determined.

3Determine whether the following points are collinear.

(1) A(−1, −1), B(0, 1), C(1, 3)

Slope of line AB = 1 − (−1)0 − (−1) = 2

Slope of line BC = 3 − 11 − 0 = 2

∴ Slope of line AB = slope of line BC, and point B is common to both lines.

∴ Points A, B, C are collinear.

(2) D(−2, −3), E(1, 0), F(2, 1)

Slope of line DE = 0 − (−3)1 − (−2) = 1

Slope of line EF = 1 − 02 − 1 = 1

∴ Slope of line DE = slope of line EF, and point E is common to both lines.

∴ Points D, E, F are collinear.

(3) L(2, 5), M(3, 3), N(5, 1)

Slope of line LM = 3 − 53 − 2 = −2

Slope of line MN = 1 − 35 − 3 = −1

∴ Slope of line LM ≠ slope of line MN.

∴ Points L, M, N are not collinear.

(4) P(2, −5), Q(1, −3), R(−2, 3)

Slope of line PQ = −3 − (−5)1 − 2 = −2

Slope of line QR = 3 − (−3)−2 − 1 = −2

∴ Slope of line PQ = slope of line QR, and point Q is common to both lines.

∴ Points P, Q, R are collinear.

(5) R(1, −4), S(−2, 2), T(−3, 4)

Slope of line RS = 2 − (−4)−2 − 1 = −2

Slope of line ST = 4 − 2−3 − (−2) = −2

∴ Slope of line RS = slope of line ST, and point S is common to both lines.

∴ Points R, S, T are collinear.

(6) A(−4, 4), K(−2, 52), N(4, −2)

Slope of line AK = 52 − 4−2 − (−4) = −34

Slope of line KN = −2 − 524 − (−2) = −912 = −34

∴ Slope of line AK = slope of line KN, and point K is common to both lines.

∴ Points A, K, N are collinear.

4If A(1, −1), B(0, 4), C(−5, 3) are vertices of a triangle then find the slope of each side.

Slope of AB = 4 − (−1)0 − 1 = 5−1 = −5

Slope of BC = 3 − 4−5 − 0 = −1−5 = 15

Slope of AC = 3 − (−1)−5 − 1 = 4−6 = −23

5Show that A(−4, −7), B(−1, 2), C(8, 5) and D(5, −4) are the vertices of a parallelogram.

Slope of line AB = 2 − (−7)−1 − (−4) = 93 = 3 …… (I)

Slope of line BC = 5 − 28 − (−1) = 39 = 13 …… (II)

Slope of line CD = −4 − 55 − 8 = −9−3 = 3 …… (III)

Slope of line DA = −7 − (−4)−4 − 5 = −3−9 = 13 …… (IV)

∴ line AB ∥ line CD … from (I) and (III)

∴ line BC ∥ line DA … from (II) and (IV)

Both pairs of opposite sides are parallel.

∴ □ ABCD is a parallelogram.

6Find k, if R(1, −1), S(−2, k) and slope of line RS is −2.

Slope of RS = k − (−1)−2 − 1 = k + 1−3

∴ k + 1−3 = −2

∴ k + 1 = 6

∴ k = 5

7Find k, if B(k, −5), C(1, 2) and slope of the line is 7.

Slope of BC = 2 − (−5)1 − k = 71 − k

∴ 71 − k = 7

∴ 1 − k = 1

∴ k = 0

8Find k, if PQ ∥ RS and P(2, 4), Q(3, 6), R(3, 1), S(5, k).

Slope of PQ = 6 − 43 − 2 = 2

Slope of RS = k − 15 − 3 = k − 12

PQ ∥ RS, so their slopes are equal.

∴ k − 12 = 2

∴ k − 1 = 4

∴ k = 5

Problem Set 5

1Fill in the blanks using correct alternatives.

(1) Seg AB is parallel to Y-axis and coordinates of point A are (1, 3) then co-ordinates of point B can be …… — (D) (1, −3)
A segment parallel to the Y-axis has the same x co-ordinate at every point.
(2) Out of the following, point …… lies to the right of the origin on X-axis. — (D) (2, 0)
A point on the X-axis has y = 0, and right of the origin means x > 0.
(3) Distance of point (−3, 4) from the origin is …… — (C) 5
√(−3)² + 4² = √25 = 5
(4) A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is …… — (C) 1√3
Slope = tan 30° = 1√3

2Determine whether the given points are collinear.

(1) A(0, 2), B(1, −0.5), C(2, −3)

Slope of line AB = −0.5 − 21 − 0 = −2.5

Slope of line BC = −3 − (−0.5)2 − 1 = −2.5

∴ Slope of line AB = slope of line BC, and point B is common to both lines.

∴ Points A, B, C are collinear.

(2) P(1, 2), Q(2, 85), R(3, 65)

Slope of line PQ = 85 − 22 − 1 = −25

Slope of line QR = 65 − 853 − 2 = −25

∴ Slope of line PQ = slope of line QR, and point Q is common to both lines.

∴ Points P, Q, R are collinear.

(3) L(1, 2), M(5, 3), N(8, 6)

Slope of line LM = 3 − 25 − 1 = 14

Slope of line MN = 6 − 38 − 5 = 1

∴ Slope of line LM ≠ slope of line MN.

∴ Points L, M, N are not collinear.

3Find the coordinates of the midpoint of the line segment joining P(0, 6) and Q(12, 20).

Midpoint = (0 + 122, 6 + 202) … midpoint formula

= (6, 13)

∴ The midpoint is (6, 13)

4Find the ratio in which the line segment joining the points A(3, 8) and B(−9, 3) is divided by the Y-axis.

Let the Y-axis divide seg AB at point P in the ratio m : n. The x co-ordinate of P is 0.

0 = m × (−9) + n × 3m + n … by section formula

∴ 9m = 3n

∴ mn = 39 = 13

∴ The ratio is 1 : 3

The point of division is P(0, 274), since y = 1 × 3 + 3 × 84 = 274.

5Find the point on X-axis which is equidistant from P(2, −5) and Q(−2, 9).

Let the point be A(x, 0). AP = AQ, so AP² = AQ².

∴ (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)

6Find the distances between the following points. (i) A(a, 0), B(0, a) (ii) P(−6, −3), Q(−1, 9) (iii) R(−3a, a), S(a, −2a)

(i)

AB = √(0 − a)² + (a − 0)² = √2a²

∴ AB = √2a

(ii)

PQ = √(−1 + 6)² + (9 + 3)² = √25 + 144 = √169

∴ PQ = 13

(iii)

RS = √(a + 3a)² + (−2a − a)² = √16a² + 9a² = √25a²

∴ RS = 5a

7Find the coordinates of the circumcentre of a triangle whose vertices are (−3, 1), (0, −2) and (1, 3).

Let A(−3, 1), B(0, −2), C(1, 3), and let the circumcentre be O(x, y).

The circumcentre is equidistant from the vertices. OA = OB = OC.

OA² = OB²: (x + 3)² + (y − 1)² = x² + (y + 2)²

∴ 6x + 9 − 2y + 1 = 4y + 4

∴ 6x − 6y = −6

∴ x − y = −1 …… (I)

OB² = OC²: x² + (y + 2)² = (x − 1)² + (y − 3)²

∴ 4y + 4 = −2x + 1 − 6y + 9

∴ 2x + 10y = 6

∴ x + 5y = 3 …… (II)

Subtracting (I) from (II): 6y = 4, so y = 23

From (I): x = y − 1 = 23 − 1 = −13

∴ The circumcentre is (−13, 23)

8In the following examples, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle.

(1) L(6, 4), M(−5, −3), N(−6, 8)

LM = √(−11)² + (−7)² = √170 ≈ 13.04

MN = √(−1)² + 11² = √122 ≈ 11.05

LN = √(−12)² + 4² = √160 ≈ 12.65

The sum of any two sides is greater than the third side, so a triangle is formed. All three sides are unequal.

∴ Δ LMN is a scalene triangle.

(2) P(−2, −6), Q(−4, −2), R(−5, 0)

PQ = √(−2)² + 4² = √20 = 2√5

QR = √(−1)² + 2² = √5

PR = √(−3)² + 6² = √45 = 3√5

PQ + QR = 2√5 + √5 = 3√5 = PR. So P, Q, R are collinear.

∴ A triangle cannot be formed.

(3) A(√2, √2), B(−√2, −√2), C(−√6, √6)

AB² = (−2√2)² + (−2√2)² = 8 + 8 = 16, so AB = 4

BC² = (−√6 + √2)² + (√6 + √2)² = (8 − 2√12) + (8 + 2√12) = 16, so BC = 4

AC² = (−√6 − √2)² + (√6 − √2)² = (8 + 2√12) + (8 − 2√12) = 16, so AC = 4

∴ AB = BC = AC = 4

∴ Δ ABC is an equilateral triangle.

9Find k if the line passing through points P(−12, −3) and Q(4, k) has slope 12.

Slope of PQ = k − (−3)4 − (−12) = k + 316

∴ k + 316 = 12

∴ k + 3 = 8

∴ k = 5

10Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2, 4) and D(1, 7).

Slope of AB = 5 − 85 − 4 = −3

Slope of CD = 7 − 41 − 2 = 3−1 = −3

∴ Slope of AB = slope of CD

∴ Line AB ∥ line CD

11Show that points P(1, −2), Q(5, 2), R(3, −1), S(−1, −5) are the vertices of a parallelogram.

Slope of line PQ = 2 − (−2)5 − 1 = 1 …… (I)

Slope of line QR = −1 − 23 − 5 = 32 …… (II)

Slope of line RS = −5 − (−1)−1 − 3 = 1 …… (III)

Slope of line SP = −2 − (−5)1 − (−1) = 32 …… (IV)

∴ PQ ∥ RS and QR ∥ SP … from (I), (II), (III), (IV)

∴ □ PQRS is a parallelogram.

12Show that the □ PQRS formed by P(2, 1), Q(−1, 3), R(−5, −3) and S(−2, −5) is a rectangle.

Diagonals PR and QS are shown dashed.

PQ = √(−3)² + 2² = √13, RS = √3² + (−2)² = √13

QR = √(−4)² + (−6)² = √52, SP = √4² + 6² = √52

∴ PQ = RS and QR = SP, so □ PQRS is a parallelogram.

Diagonal PR = √(−7)² + (−4)² = √65, diagonal QS = √(−1)² + (−8)² = √65

∴ The diagonals are congruent.

A parallelogram with congruent diagonals is a rectangle.

∴ □ PQRS is a rectangle.

13Find the lengths of the medians of a triangle whose vertices are A(−1, 1), B(5, −3) and C(3, 5).

Let D, E, F be the midpoints of BC, CA and AB.

D = (5 + 32, −3 + 52) = (4, 1), E = (3 − 12, 5 + 12) = (1, 3), F = (−1 + 52, 1 − 32) = (2, −1)

AD = √(4 + 1)² + (1 − 1)² = √25 = 5

BE = √(1 − 5)² + (3 + 3)² = √52 = 2√13

CF = √(2 − 3)² + (−1 − 5)² = √37

∴ The medians are AD = 5, BE = 2√13 and CF = √37

14★Find the coordinates of centroid of the triangles if points D(−7, 6), E(8, 5) and F(2, −2) are the mid points of the sides of that triangle.

Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices, with D, E, F the midpoints of BC, CA, AB.

Adding the x co-ordinates of D, E, F: x₂ + x₃2 + x₃ + x₁2 + x₁ + x₂2 = −7 + 8 + 2

∴ x₁ + x₂ + x₃ = 3

Similarly, y₁ + y₂ + y₃ = 6 + 5 − 2 = 9

Centroid = (33, 93) = (1, 3)

∴ The centroid is (1, 3)

The centroid of a triangle is the same as the centroid of the triangle formed by the midpoints of its sides.

15Show that A(4, −1), B(6, 0), C(7, −2) and D(5, −3) are vertices of a square.

AB = √2² + 1² = √5, BC = √1² + (−2)² = √5

CD = √(−2)² + (−1)² = √5, DA = √(−1)² + 2² = √5

Diagonal AC = √3² + (−1)² = √10, diagonal BD = √(−1)² + (−3)² = √10

∴ All sides are congruent, so □ ABCD is a rhombus. Its diagonals are also congruent.

A rhombus with congruent diagonals is a square.

∴ □ ABCD is a square.

16Find the coordinates of circumcentre and radius of circumcircle of Δ ABC if A(7, 1), B(3, 5) and C(2, 0) are given.

Let the circumcentre be O(x, y). OA = OB = OC.

OA² = OB²: (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)²

∴ −14x + 49 − 2y + 1 = −6x + 9 − 10y + 25

∴ −8x + 8y = −16

∴ x − y = 2 …… (I)

OB² = OC²: (x − 3)² + (y − 5)² = (x − 2)² + y²

∴ −6x + 9 − 10y + 25 = −4x + 4

∴ −2x − 10y = −30

∴ x + 5y = 15 …… (II)

Subtracting (I) from (II): 6y = 13, so y = 136

From (I): x = 2 + 136 = 256

OC² = (256 − 2)² + (136 − 0)² = 16936 + 16936 = 33836

∴ Radius = OC = √3386 = 13√26

∴ Circumcentre = (256, 136) and radius = 13√26 units

17Given A(4, −3), B(8, 5). Find the coordinates of the point that divides segment AB in the ratio 3 : 1.

Here (x₁, y₁) = (4, −3), (x₂, y₂) = (8, 5) and m : n = 3 : 1.

x = mx₂ + nx₁m + n = 3 × 8 + 1 × 43 + 1 = 24 + 44 = 7 … by section formula

y = my₂ + ny₁m + n = 3 × 5 + 1 × (−3)3 + 1 = 15 − 34 = 3

∴ Co-ordinates of the point are (7, 3)

18★Find the type of the quadrilateral if points A(−4, −2), B(−3, −7), C(3, −2) and D(2, 3) are joined serially.

AB = √1² + (−5)² = √26, BC = √6² + 5² = √61

CD = √(−1)² + 5² = √26, DA = √(−6)² + (−5)² = √61

∴ AB = CD and BC = DA. Both pairs of opposite sides are congruent.

Diagonal AC = √7² + 0² = 7, diagonal BD = √5² + 10² = √125 = 5√5

Adjacent sides are not equal, and the diagonals are not equal. So it is neither a rhombus nor a rectangle.

∴ □ ABCD is a parallelogram.

19★The line segment AB is divided into five congruent parts at P, Q, R and S such that A − P − Q − R − S − B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R, B.

A − P − Q − R − S − B

R is the midpoint of seg QS. R = (12 + 42, 14 + 182) = (8, 16)

Q is the midpoint of seg PR. Let P = (x, y). Then 12 = x + 82 and 14 = y + 162. So P = (16, 12).

P is the midpoint of seg AQ. Let A = (x, y). Then 16 = x + 122 and 12 = y + 142. So A = (20, 10).

S is the midpoint of seg RB. Let B = (x, y). Then 4 = 8 + x2 and 18 = 16 + y2. So B = (0, 20).

∴ A(20, 10), P(16, 12), R(8, 16), B(0, 20)

20Find the coordinates of the centre of the circle passing through the points P(6, −6), Q(3, −7) and R(3, 3).

Let the centre be C(x, y). CQ = CR, since both are radii.

(x − 3)² + (y + 7)² = (x − 3)² + (y − 3)²

∴ 14y + 49 = −6y + 9

∴ 20y = −40

∴ y = −2

CP = CQ: (x − 6)² + (−2 + 6)² = (x − 3)² + (−2 + 7)²

∴ x² − 12x + 36 + 16 = x² − 6x + 9 + 25

∴ −6x = −18

∴ x = 3

∴ The centre is (3, −2)

Check: the radius is 5, since CR = √0² + 5² = 5 and CP = √3² + 4² = 5.

21★Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, −2) and C(3, −2).

The three possible positions of D

The diagonals of a parallelogram bisect each other. So the two ends of each diagonal have the same midpoint.

If AC is a diagonal (□ ABCD): D = (5 + 3 − 1, 6 − 2 + 2) = (7, 6)

If AB is a diagonal (□ ACBD): D = (5 + 1 − 3, 6 − 2 + 2) = (3, 6)

If BC is a diagonal (□ ABDC): D = (1 + 3 − 5, −2 − 2 − 6) = (−1, −10)

∴ The possible positions of D are (7, 6), (3, 6) and (−1, −10)

For the order □ ABCD alone, D = (7, 6). The midpoint of AC is (4, 2), and this is also the midpoint of BD.

22Find the slope of the diagonals of a quadrilateral with vertices A(1, 7), B(6, 3), C(0, −3) and D(−3, 3).

The diagonals are AC and BD.

Slope of AC = −3 − 70 − 1 = −10−1 = 10

Slope of BD = 3 − 3−3 − 6 = 0−9 = 0

∴ Slope of AC = 10 and slope of BD = 0

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