Coordinate Geometry: Case Study PYQs with Answers

(4, 3)

The diagonals of a rectangle bisect each other.

Their intersection is the midpoint of AC.

Midpoint = ($\frac{\text{(}\text{1 + 7}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{1 + 5}\text{)/(}}{\text{2}\text{)}}$) = (4, 3).

2√13 units

By the distance formula, AC = √[(7 − 1)² + (5 − 1)²].

AC = √(36 + 16) = √52 = 2√13 units.

24 square units

AB = 7 − 1 = 6 units.

BC = 5 − 1 = 4 units.

Area of rectangle ABCD = AB × BC = 6 × 4 = 24 square units.

3 : √13

AB = 6 units and AC = 2√13 units.

AB : AC = 6 : 2√13 = 3 : √13.

AB = CD = √26 units

AB = √[(4 − 3)² + (4 − (−1))²] = √(1 + 25) = √26 units.

CD = √[(−2 − (−1))² + (−2 − 3)²] = √(1 + 25) = √26 units.

BC = AD = √26 units

BC = √[(−1 − 4)² + (3 − 4)²] = √(25 + 1) = √26 units.

AD = √[(−2 − 3)² + (−2 − (−1))²] = √(25 + 1) = √26 units.

AC = 4√2 units; BD = 6√2 units

AC = √[(−1 − 3)² + (3 − (−1))²] = √32 = 4√2 units.

BD = √[(−2 − 4)² + (−2 − 4)²] = √72 = 6√2 units.

All four sides are equal and the diagonals are unequal, so ABCD is a rhombus.

Both midpoints are (1, 1).

Midpoint of AC = ($\frac{\text{(}\text{3 + (−1)}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{−1 + 3}\text{)/(}}{\text{2}\text{)}}$) = (1, 1).

Midpoint of BD = ($\frac{\text{(}\text{4 + (−2)}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{4 + (−2)}\text{)/(}}{\text{2}\text{)}}$) = (1, 1).

R(200, 400), S(−200, 400)

PQ = 200 − (−200) = 400 units.

The square lies above PQ, so its upper vertices have y-coordinate 400.

R = (200, 400) and S = (−200, 400).

1,60,000 square units

Side PQ = 400 units.

Area of square PQRS = 400² = 1,60,000 square units.

400√2 units

PR = √[(200 − (−200))² + (400 − 0)²].

PR = √(400² + 400²) = 400√2 units.

K = 1

From the figure, C = (−600, 0) and S = (−200, 400).

Using the section formula on CA, 400 = $\frac{\text{(}\text{800K + 0}\text{)/(}}{\text{K + 1}\text{)}}$.

400K + 400 = 800K.

K = 1.

3√2 units

AB = √[(6 − 3)² + (7 − 4)²].

AB = √18 = 3√2 units.

3√2 units

CD = √[(6 − 9)² + (1 − 4)²].

CD = √18 = 3√2 units.

ABCD is a parallelogram.

AB = CD = 3√2 units.

BC = √[(9 − 6)² + (4 − 7)²] = 3√2 units.

AD = √[(6 − 3)² + (1 − 4)²] = 3√2 units.

Both pairs of opposite sides are equal, so ABCD is a parallelogram.

Both midpoints are (6, 4).

Midpoint of AC = ($\frac{\text{(}\text{3 + 9}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{4 + 4}\text{)/(}}{\text{2}\text{)}}$) = (6, 4).

Midpoint of BD = ($\frac{\text{(}\text{6 + 6}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{7 + 1}\text{)/(}}{\text{2}\text{)}}$) = (6, 4).

A(1, 9), B(5, 13)

Read the horizontal coordinate first, then the vertical coordinate.

A = (1, 9) and B = (5, 13).

(11, 11)

C = (9, 13) and D = (13, 9).

Midpoint of CD = ($\frac{\text{(}\text{9 + 13}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{13 + 9}\text{)/(}}{\text{2}\text{)}}$) = (11, 11).

4√5 units

M = (5, 11) and Q = (9, 3).

MQ = √[(9 − 5)² + (3 − 11)²].

MQ = √80 = 4√5 units.

(6, 11)

M = (5, 11) and N = (9, 11).

By the section formula, the point is ($\frac{\text{(}\text{1 × 9 + 3 × 5}\text{)/(}}{\text{4}\text{)}}$, $\frac{\text{(}\text{1 × 11 + 3 × 11}\text{)/(}}{\text{4}\text{)}}$).

Therefore, the point is (6, 11).

C(5, 4)

Read the point C from the grid: x = 5 and y = 4.

C = (5, 4).

6√2 units

O = (0, 0) and B = (6, 6).

OB = √[(6 − 0)² + (6 − 0)²] = √72 = 6√2 units.

D(1, 5)

Let D = (x, y).

The diagonals of a parallelogram bisect each other.

Midpoint of AC = ($\frac{\text{(}\text{2 + 5}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{7 + 4}\text{)/(}}{\text{2}\text{)}}$) = (7/2, 11/2).

Midpoint of BD = ($\frac{\text{(}\text{6 + x}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{6 + y}\text{)/(}}{\text{2}\text{)}}$).

6 + x = 7, so x = 1.

6 + y = 11, so y = 5.

D = (1, 5).

3√2 units

A = (2, 7) and C = (5, 4).

AC = √[(5 − 2)² + (4 − 7)²].

AC = √18 = 3√2 units.

(−1, 2)

F = (−3, 0) and G = (1, 4).

Midpoint of FG = ($\frac{\text{(}\text{−3 + 1}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{0 + 4}\text{)/(}}{\text{2}\text{)}}$) = (−1, 2).

2√13 units

A = (3, 4) and C = (−1, −2).

AC = √[(−1 − 3)² + (−2 − 4)²].

AC = √52 = 2√13 units.

(3, 7/2)

A = (3, 4) and B = (3, 2).

By the section formula, the point is ($\frac{\text{(}\text{1 × 3 + 3 × 3}\text{)/(}}{\text{4}\text{)}}$, $\frac{\text{(}\text{1 × 2 + 3 × 4}\text{)/(}}{\text{4}\text{)}}$).

Therefore, the point is (3, 7/2).

D(−2, −5)

Read point D from the grid: x = −2 and y = −5.

D = (−2, −5).

P(4, 6), Q(3, 2), R(6, 5)

Count horizontal and vertical grid intervals from A.

P = (4, 6), Q = (3, 2) and R = (6, 5).

PQ = √17 m; QR = 3√2 m

PQ = √[(3 − 4)² + (2 − 6)²] = √17 m.

QR = √[(6 − 3)² + (5 − 2)²] = √18 = 3√2 m.

(16/3, 16/3)

By the section formula, the point is ($\frac{\text{(}\text{2 × 6 + 1 × 4}\text{)/(}}{\text{3}\text{)}}$, $\frac{\text{(}\text{2 × 5 + 1 × 6}\text{)/(}}{\text{3}\text{)}}$).

Therefore, the point is (16/3, 16/3).

No; △PQR is scalene.

PQ² = 17 and QR² = 18.

PR² = (6 − 4)² + (5 − 6)² = 5.

All three side lengths differ, so △PQR is not isosceles.

200 units × 150 units

AB = 100 − (−100) = 200 units.

BC = 150 − 0 = 150 units.

250 units

AC = √[(100 − (−100))² + (150 − 0)²].

AC = √62500 = 250 units.

4x + 3y = 225

PA² = PC².

(x + 100)² + y² = (x − 100)² + (y − 150)².

400x + 300y = 22500.

4x + 3y = 225.

Q(0, 75)

Let Q = (0, y).

QA² = QD².

100² + y² = 100² + (y − 150)².

300y = 22500.

y = 75, so Q = (0, 75).

6√5 units

AB = √[(−3 − 9)² + (−1 − 5)²].

AB = √180 = 6√5 units.

E(7, 0)

E is the midpoint of AC.

E = ($\frac{\text{(}\text{9 + 5}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{5 + (−5)}\text{)/(}}{\text{2}\text{)}}$) = (7, 0).

8√2 units

D is the midpoint of BC.

D = ($\frac{\text{(}\text{−3 + 5}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{−1 + (−5)}\text{)/(}}{\text{2}\text{)}}$) = (1, −3).

AD = √[(1 − 9)² + (−3 − 5)²].

AD = √128 = 8√2 units.

a = 1; b = 5

B, P, Q and A occur in that order with equal gaps.

P is the midpoint of BQ, so a = $\frac{\text{(}\text{−1 + 3}\text{)/(}}{\text{2}\text{)}}$ = 1.

Q is the midpoint of PA, so b = $\frac{\text{(}\text{1 + 9}\text{)/(}}{\text{2}\text{)}}$ = 5.

C(30, 35)

C is the midpoint of diameter AB.

C = ($\frac{\text{(}\text{10 + 50}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{20 + 50}\text{)/(}}{\text{2}\text{)}}$) = (30, 35).

25 units

CA = √[(30 − 10)² + (35 − 20)²].

CA = √625 = 25 units.

P(70/3, 30)

AP : PB = 1 : 2.

By the section formula, P = ($\frac{\text{(}\text{1 × 50 + 2 × 10}\text{)/(}}{\text{3}\text{)}}$, $\frac{\text{(}\text{1 × 50 + 2 × 20}\text{)/(}}{\text{3}\text{)}}$).

P = (70/3, 30).

100/3 units

AB = 2 × 25 = 50 units.

AQ = (2/3)AB = $\frac{\text{(}\text{2 × 50}\text{)/(}}{\text{3}\text{)}}$ = 100/3 units.

E(25, 30)

The diagonals of a rectangle bisect each other.

E = ($\frac{\text{(}\text{10 + 40}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{10 + 50}\text{)/(}}{\text{2}\text{)}}$) = (25, 30).

40 units

B and C have the same x-coordinate.

BC = 50 − 10 = 40 units.

D(10, 50); BD = 50 units

For this rectangle, D has A’s x-coordinate and C’s y-coordinate.

D = (10, 50).

BD = √[(10 − 40)² + (50 − 10)²].

BD = √2500 = 50 units.

80 units

AE = √[(25 − 10)² + (30 − 10)²] = 25 units.

EB = √[(40 − 25)² + (10 − 30)²] = 25 units.

BA = 40 − 10 = 30 units.

Total distance = AE + EB + BA = 25 + 25 + 30 = 80 units.

O(12, 3)

O is the midpoint of AB.

O = ($\frac{\text{(}\text{2 + 22}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{3 + 3}\text{)/(}}{\text{2}\text{)}}$) = (12, 3).

10 units

AB = 22 − 2 = 20 units.

Radius = AB/2 = 10 units.

C(12, −7), for the pictured semicircle.

OC = 10 units.

(12 − 12)² + (y − 3)² = 100.

y − 3 = ±10, giving y = 13 or y = −7.

The figure places C below AB, so y < 3.

Therefore, C = (12, −7).

Source note: the 2025 Basic 430/3/1 Q38(iii)(a) marking scheme, page 20, lists both (12,13) and (12,−7). Only (12,−7) lies on the semicircle shown.

P(7, 3)

AP : PB = 1 : 3.

By the section formula, P = ($\frac{\text{(}\text{1 × 22 + 3 × 2}\text{)/(}}{\text{4}\text{)}}$, $\frac{\text{(}\text{1 × 3 + 3 × 3}\text{)/(}}{\text{4}\text{)}}$).

P = (7, 3).

10 units

OP = √[(8 − 0)² + (6 − 0)²].

OP = √100 = 10 units.

(0, 14/3)

On the y-axis, x = 0.

2(0) + 9y = 42.

y = 42/9 = 14/3.

The intersection is (0, 14/3).

OR : RP = 3 : 2; y = 18/5

Let OR : RP = k : 1.

4.8 = $\frac{\text{(}\text{8k + 0}\text{)/(}}{\text{k + 1}\text{)}}$.

4.8k + 4.8 = 8k.

k = 3/2, so OR : RP = 3 : 2.

y = $\frac{\text{(}\text{3 × 6 + 2 × 0}\text{)/(}}{\text{5}\text{)}}$ = 18/5.

PQ/OS = 2/3

PQ = √[(12 − 8)² + (2 − 6)²] = 4√2 units.

OS = √[(−6 − 0)² + (6 − 0)²] = 6√2 units.

$\frac{\text{(}\text{PQ}\text{)/(}}{\text{OS}\text{)}}$ = $\frac{\text{(}\text{4√2}\text{)/(}}{\text{6√2}\text{)}}$ = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$.

√65 units

A = (−5, 9) and D = (−6, 1).

AD = √[(−6 − (−5))² + (1 − 9)²] = √65 units.

(−11/2, 5) and (−3/2, 6)

Midpoint of AD = ($\frac{\text{(}\text{−5 + (−6)}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{9 + 1}\text{)/(}}{\text{2}\text{)}}$) = (−11/2, 5).

In △ADB, the line through this midpoint parallel to DB meets AB at its midpoint.

Midpoint of AB = ($\frac{\text{(}\text{−5 + 2}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{9 + 3}\text{)/(}}{\text{2}\text{)}}$) = (−3/2, 6).

M(0, 5/2)

D = (−6, 1) and B = (2, 3).

By the section formula, M = ($\frac{\text{(}\text{3 × 2 + 1 × (−6)}\text{)/(}}{\text{4}\text{)}}$, $\frac{\text{(}\text{3 × 3 + 1 × 1}\text{)/(}}{\text{4}\text{)}}$).

M = (0, 5/2).

C(1, −5)

Let C = (x, y). Since P is the midpoint of AC,

$\frac{\text{(}\text{−5 + x}\text{)/(}}{\text{2}\text{)}}$ = −2 and $\frac{\text{(}\text{9 + y}\text{)/(}}{\text{2}\text{)}}$ = 2.

x = 1 and y = −5.

C = (1, −5).

4√5 units

For a = 1, O = (1, 7).

Radius OA = √[(1 − (−3))² + (7 − (−1))²].

OA = √80 = 4√5 units.

a = 1/2

The x-coordinate of a point on the y-axis is 0.

2a − 1 = 0.

a = 1/2.

a = 2 or a = −4

OA² = 100.

(2a − 1 + 3)² + (7 + 1)² = 100.

(2a + 2)² = 36.

2a + 2 = 6 or 2a + 2 = −6.

a = 2 or a = −4.

10√2 units

OA = OB = 10 units and ∠AOB = 90°.

By Pythagoras theorem, AB² = OA² + OB².

AB = √(100 + 100) = 10√2 units.

AB = AD = 5 units

AB = 7 − 2 = 5 units.

AD = 8 − 3 = 5 units.

√29 − 7 units

BC = √[(9 − 7)² + (3 − 8)²] = √29 units.

CD = 9 − 2 = 7 units.

BC − CD = √29 − 7 units.

This is a signed difference, so a negative value is possible.

M(31/5, 3)

D = (2, 3) and C = (9, 3).

M = ($\frac{\text{(}\text{3 × 9 + 2 × 2}\text{)/(}}{\text{5}\text{)}}$, $\frac{\text{(}\text{3 × 3 + 2 × 3}\text{)/(}}{\text{5}\text{)}}$).

M = (31/5, 3).

N(24/5, 6)

N = ($\frac{\text{(}\text{2 × 9 + 3 × 2}\text{)/(}}{\text{5}\text{)}}$, $\frac{\text{(}\text{2 × 3 + 3 × 8}\text{)/(}}{\text{5}\text{)}}$).

N = (24/5, 6).

2√65 units

One-way distance CA = √[(−4 − 4)² + (2 − 3)²] = √65 units.

Return distance = 2 × √65 = 2√65 units.

1 : 1

Let the ratio be k : 1. The crossing has y-coordinate 0.

0 = $\frac{\text{(}\text{k(−2) + 1 × 2}\text{)/(}}{\text{k + 1}\text{)}}$.

−2k + 2 = 0, so k = 1.

The ratio is 1 : 1.

No; BC = 7 units and BD = √85 units.

BC = |3 − (−4)| = 7 units.

BD = √[(−5 − 4)² + (−2 − (−4))²] = √85 units.

Since 7 ≠ √85, Birla Mandir is not equidistant from the two places.

A, O and B are not collinear.

If O lay on AB, its zero x-coordinate would place it internally between A and B.

Let AO : OB = k : 1.

From the x-coordinate, 0 = $\frac{\text{(}\text{4k − 4}\text{)/(}}{\text{k + 1}\text{)}}$, giving k = 1.

From the y-coordinate, 0 = $\frac{\text{(}\text{−4k + 2}\text{)/(}}{\text{k + 1}\text{)}}$, giving k = 1/2.

A single division point cannot have two different ratios.

Therefore, A, O and B are not collinear.

AC = BD = 6 units

AC = 7 − 1 = 6 units.

BD = 9 − 3 = 6 units.

Both midpoints are (4, 6).

Midpoint of AC = ($\frac{\text{(}\text{1 + 7}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{6 + 6}\text{)/(}}{\text{2}\text{)}}$) = (4, 6).

Midpoint of BD = ($\frac{\text{(}\text{4 + 4}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{9 + 3}\text{)/(}}{\text{2}\text{)}}$) = (4, 6).

ABCD is a square.

The diagonals have the same midpoint, so ABCD is a parallelogram.

AB = √[(4 − 1)² + (9 − 6)²] = 3√2 units.

BC = √[(7 − 4)² + (6 − 9)²] = 3√2 units.

A parallelogram with equal adjacent sides is a rhombus.

Its diagonals AC and BD are equal, so it is also a rectangle.

Therefore, ABCD is a square.

(4, 6)

The diagonals of the square meet at (4, 6).

The distances from this point to A, B, C and D are each 3 units.

Ravi should sit at (4, 6).

2√13 units

AP = √[(7 − 1)² + (3 − 7)²].

AP = √52 = 2√13 units.

AR : RP = 1 : 1

Midpoint of AP = ($\frac{\text{(}\text{1 + 7}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{7 + 3}\text{)/(}}{\text{2}\text{)}}$) = (4, 5).

This is R, so R divides AP in the ratio 1 : 1.

S(2/3, 0)

Let S = (x, 0).

AS² = PS².

(x − 1)² + 7² = (x − 7)² + 3².

−2x + 50 = −14x + 58.

12x = 8, so x = 2/3.

S = (2/3, 0).

L(0, −17/4)

Let L = (0, y).

RL² = PL².

4² + (y − 5)² = 7² + (y − 3)².

−10y + 41 = −6y + 58.

−4y = 17, so y = −17/4.

L = (0, −17/4).

150√34 km

L = (5, 10) and B = (0, 7).

LB = √[(0 − 5)² + (7 − 10)²] = √34 grid units.

Using the paper’s scale, distance = 150√34 km.

K(2, 41/5)

LK : KB = 3 : 2.

Using L(5,10) first and B(0,7) second,

K = ($\frac{\text{(}\text{3 × 0 + 2 × 5}\text{)/(}}{\text{5}\text{)}}$, $\frac{\text{(}\text{3 × 7 + 2 × 10}\text{)/(}}{\text{5}\text{)}}$).

K = (2, 41/5).

Source note: the 2022–23 Basic official sample Q37(II) marking scheme, page 8, prints (3,41/5). Its x-coordinate uses the opposite weights. The point (2,41/5) satisfies LK : KB = 3 : 2.

Isosceles triangle

LN = √[(2 − 5)² + (6 − 10)²] = 5 grid units.

LP = √[(8 − 5)² + (6 − 10)²] = 5 grid units.

NP = 8 − 2 = 6 grid units.

LN = LP, so △LNP is isosceles.

Source note: the 2022–23 Basic official sample Q37(III) marking scheme, page 8, prints NP = 4. The stated coordinates give NP = 8 − 2 = 6; the isosceles classification is unchanged.

(0, 25/8)

Let the point be T(0, y).

TL² = TP².

5² + (10 − y)² = 8² + (6 − y)².

125 − 20y = 100 − 12y.

8y = 25, so y = 25/8.

T = (0, 25/8).

√58 units

B = (1, 2) and F = (−2, 9).

BF = √[(−2 − 1)² + (9 − 2)²] = √58 units.

Z(−1/2, 11/2)

WXOP is a rectangle, so its diagonals bisect each other.

Z is the midpoint of WO.

Z = ($\frac{\text{(}\text{−6 + 5}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{2 + 9}\text{)/(}}{\text{2}\text{)}}$) = (−1/2, 11/2).

(0, 57/10)

Let T = (0, y).

TA² = TG².

2² + (y − 2)² = 4² + (y − 7)².

8 − 4y = 65 − 14y.

10y = 57, so T = (0, 57/10).

10 square units

AF = 9 − 2 = 7 units.

GH = 7 − 4 = 3 units.

AF and GH are parallel vertical sides, 2 units apart.

Area = (1/2)(AF + GH) × height.

Area = (1/2)(7 + 3) × 2 = 10 square units.

2 units

A and B have the same x-coordinate.

AB = 3 − 1 = 2 units.

2 units

B and C have the same y-coordinate.

BC = 4 − 2 = 2 units.

BO = √5 > BA = BC = 2 units

BO = √[(2 − 0)² + (1 − 0)²] = √5 units.

BA = BC = 2 units.

Since √5 > 2, the school is farther from Shagun’s house than either Alia’s house or the library.

△ABC is an isosceles right triangle at B.

AB = BC = 2 units.

AC² = (4 − 2)² + (1 − 3)² = 8.

AB² + BC² = 4 + 4 = 8 = AC².

By the converse of Pythagoras theorem, ∠ABC = 90°.

Since AB = BC, the triangle is also isosceles.

D(3, 5)

Let D = (x, y).

The diagonals AC and BD have the same midpoint.

1 + 6 = 4 + x, so x = 3.

2 + 6 = 3 + y, so y = 5.

D = (3, 5).

Yes; K is the midpoint of GH.

Midpoint of GH = ($\frac{\text{(}\text{−3 + 3}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{5 + 1}\text{)/(}}{\text{2}\text{)}}$) = (0, 3).

This is K, so G, K and H lie on the same straight line.

CI = CJ = √41 units; C is not the midpoint of IJ.

CJ = √[(5 − 0)² + (−3 − 1)²] = √41 units.

CI = √[(−4 − 0)² + (6 − 1)²] = √41 units.

Therefore, I and J are equidistant from C.

Midpoint of IJ = ($\frac{\text{(}\text{−4 + 5}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{6 + (−3)}\text{)/(}}{\text{2}\text{)}}$) = (1/2, 3/2).

This differs from C(0,1), so C is not the midpoint.

E(3, −10)

For these distinct collinear positions, B is the midpoint of AE.

$\frac{\text{(}\text{1 + a}\text{)/(}}{\text{2}\text{)}}$ = 2, so a = 3.

$\frac{\text{(}\text{4 + b}\text{)/(}}{\text{2}\text{)}}$ = −3, so b = −10.

E = (3, −10).

4√2 units

BC = √[(2 − 6)² + (6 − 2)²].

BC = √32 = 4√2 units.

(4, 4)

Midpoint of BC = ($\frac{\text{(}\text{6 + 2}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{2 + 6}\text{)/(}}{\text{2}\text{)}}$).

The club is at (4, 4).

Society A; √41 units

OA² = 4² + 5² = 41.

OB² = 6² + 2² = 40.

OC² = 2² + 6² = 40.

OA is greatest, so society A is farthest, at √41 units.

C is nearer; AB = √13 units and AC = √5 units.

AB = √[(6 − 4)² + (2 − 5)²] = √13 units.

AC = √[(2 − 4)² + (6 − 5)²] = √5 units.

Since √5 < √13, C is nearer to A.

2√10 units

PR = √[(8 − 2)² + (3 − 5)²].

PR = √40 = 2√10 units.

No; the midpoint is (5, 4).

Midpoint of PR = ($\frac{\text{(}\text{2 + 8}\text{)/(}}{\text{2}\text{)}}$, $\frac{\text{(}\text{5 + 3}\text{)/(}}{\text{2}\text{)}}$) = (5, 4).

Q = (4, 4), so Q is not the midpoint of PR.

(3/4, 0)

Let T = (x, 0).

TP² = TQ².

(x − 2)² + 5² = (x − 4)² + 4².

−4x + 29 = −8x + 32.

4x = 3, so x = 3/4.

T = (3/4, 0).

S(14/5, 23/5)

PS : SQ = 2 : 3.

S = ($\frac{\text{(}\text{2 × 4 + 3 × 2}\text{)/(}}{\text{5}\text{)}}$, $\frac{\text{(}\text{2 × 4 + 3 × 5}\text{)/(}}{\text{5}\text{)}}$).

S = (14/5, 23/5).