Coordinate Geometry Class 10 Mathematics — Competency Based Questions

20 Questions | 80 Marks

A civil engineer is planning a new park. She wants to place a statue at a point equidistant from two entrances, A(–2, 5) and B(6, 3). She has also decided that the statue should be on a straight pathway represented by the line x – y = 5. Find the coordinates of the point where the statue should be placed. 5marks

Two friends, Ravi and Sunita, live in different parts of a city. Ravi’s house is at point R(–3, 4) and Sunita’s house is at S(5, –2). They want to meet at a café located on the line segment RS such that the distance of the café from Ravi’s house is twice the distance from Sunita’s house. Find the coordinates of the café. 3marks

Assertion: The points P(–4, 2) and Q(2, 8) are equidistant from point A(–1, 5).Reason: If a point is equidistant from two points, it lies on the perpendicular bisector of the line segment joining these two points.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 2marks

A farmer has a rectangular plot of land with vertices at P(–2, 3), Q(4, 7), R(8, 1) and S(2, -3). He wants to plant a special variety of medicinal herb at a point that is equidistant from the two diagonals of the plot. Find the coordinates of this point. 5marks

Consider a mobile tower located at point T(2, 3). A resident at point A(–1, 5) wants to know the distance from their house to the tower. Another resident at point B(6, 1) wants to know the distance from their house to the tower. Calculate these distances. 3marks

A ship is located at point S(–3, –4). A lighthouse L is at (2, 8). The captain wants to navigate the ship to a point P on the line segment SL such that the distance from the lighthouse to P is three times the distance from the ship to P. Find the coordinates of P. 3marks

The points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a quadrilateral.(i) Show that this quadrilateral is a square using the distance formula. [3 marks](ii) Calculate the distance of the origin from the point (– 4, 4). 1mark

Assertion: The point (2, 5) lies on the line segment joining (0, 1) and (4, 9).Reason: If a point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n, then x = (mx2 + nx1)/(m+n) and y = (my2 + ny1)/(m+n).(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 2marks

A school has set up a coordinate system for a treasure hunt. The first clue leads to point A(–5, 6) and the second clue to point B(3, –2). The treasure is hidden at a point P on the line segment AB such that AP : PB = 3 : 1. Find the coordinates of the treasure. 3marks

A group of students is creating a game on a graph paper. They plot four points: P(1, 2), Q(4, 5), R(7, 2), and S(4, -1).(i) Verify if PQRS forms a rhombus by calculating the lengths of its sides. [3 marks](ii) Calculate the distance between the diagonals PR and QS. 2marks

A mapping application shows a landmark at point L(–1, 3). A user wants to find a point P on the straight road connecting L to another landmark M(5, 7) such that P is exactly one-fourth of the distance from L to M. Find the coordinates of P. 3marks

Two friends are standing at points A(–6, 4) and B(–1, 9). They want to meet at a point C on the x-axis such that AC = BC. Find the coordinates of point C. 3marks

Assertion: The points (2, –2), (–1, 1) and (4, 5) are collinear.Reason: Three points are collinear if the area of the triangle formed by them is zero.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 2marks

A scientist is studying the movement of a particle. The particle starts at A(–3, 4) and moves in a straight line to B(7, –6). The scientist marks a point P on the path such that AP = (3/10) AB. Find the coordinates of P. 3marks

A community centre is planning to build a park. They want to place a musical fountain at a point P(x, y) that is equidistant from two points where a new library will be built, A(–2, 4) and B(6, –2). The fountain will also be located on a road represented by the equation y = 2x – 1. Find the coordinates of the fountain. 5marks

Two villages, V1 and V2, are located at coordinates (–4, 5) and (6, –3) respectively. The government plans to build a community centre at a point C on the line segment V1V2 such that C is closer to V2 than V1, with the ratio V1C : CV2 = 2 : 3. Find the coordinates of the community centre. 3marks

A surveyor is mapping a region. He identifies three important locations: A(1, 5), B(–3, 2), and C(–6, –1).(i) Calculate the distance between points A and B. [1 mark](ii) Calculate the distance between points B and C. [1 mark](iii) Determine if points A, B, and C are collinear. Justify your answer. 2marks

Assertion: The point (3, 5) is the mid-point of the line segment joining (1, 1) and (5, 9).Reason: The mid-point of a line segment joining (x1, y1) and (x2, y2) is given by ((x1 + x2)/2, (y1 + y2)/2).(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 2marks

A drone is flying from point D(–5, 7) to point E(1, –1). At what point P on the line segment DE will the drone be exactly one-third of the way from D to E? 3marks

A park designer wants to place a statue at a point P such that it is equidistant from points A(–4, 3) and B(2, 1). The statue is also to be placed on a straight path represented by the line y = x + 1. Find the coordinates of the point P. 5marks

Answers

Let the coordinates of the statue be P(x, y). Since P is equidistant from A(–2, 5) and B(6, 3), we have AP² = BP².(x – (–2))² + (y – 5)² = (x – 6)² + (y – 3)²(x + 2)² + (y – 5)² = (x – 6)² + (y – 3)²x² + 4x + 4 + y² – 10y + 25 = x² – 12x + 36 + y² – 6y + 94x – 10y + 29 = –12x – 6y + 4516x – 4y = 164x – y = 4 (Equation 1)

Since P(x, y) lies on the line x – y = 5, we have y = x – 5.
Substitute this into Equation 1:
4x – (x – 5) = 4
4x – x + 5 = 4
3x = –1
x = –1/3

Now find y:
y = x – 5 = –1/3 – 5 = –1/3 – 15/3 = –16/3
The coordinates of the statue are (–1/3, –16/3). [5 marks]

Let R = (–3, 4) and S = (5, –2). The point C(x, y) divides RS such that RC = 2SC. This means the ratio RC : SC = 2 : 1.Using the section formula:x = (m x₂ + n x₁) / (m + n) = (2 * 5 + 1 * (–3)) / (2 + 1) = (10 – 3) / 3 = 7/3y = (m y₂ + n y₁) / (m + n) = (2 * (–2) + 1 * 4) / (2 + 1) = (–4 + 4) / 3 = 0/3 = 0The coordinates of the café are (7/3, 0). 3marks

Assertion: Let P(–4, 2) and Q(2, 8) and A(–1, 5).Distance AP =

\sqrt{(- 1 - (- 4))^{2} + (5 - 2)^{2}} = \sqrt{(3)^{2} + (3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.Distance AQ =

\sqrt{(- 1 - 2)^{2} + (5 - 8)^{2}} = \sqrt{(- 3)^{2} + (- 3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.Since AP = AQ, point A is equidistant from P and Q. So the assertion is true.Reason: A point equidistant from two given points lies on the perpendicular bisector of the line segment joining these two points. This is a fundamental property in coordinate geometry. So the reason is true.The reason correctly explains why the assertion is true.(a) Both assertion and reason are true and reason is the correct explanation of assertion. 2marks

Vertices are P(–2, 3), Q(4, 7), R(8, 1), S(2, -3).Mid-point of diagonal PR: M₁ = ((-2+8)/2, (3+1)/2) = (6/2, 4/2) = (3, 2).Mid-point of diagonal QS: M₂ = ((4+2)/2, (7+(-3))/2) = (6/2, 4/2) = (3, 2).Since the mid-points of the diagonals coincide, the diagonals bisect each other. The point of intersection of the diagonals is (3, 2).A point equidistant from the two diagonals of a rhombus is the intersection point of the diagonals (since the diagonals are perpendicular bisectors of each other in a rhombus and also in a square). In this case, the quadrilateral is a parallelogram, and the diagonals bisect each other at (3,2). For a parallelogram, the intersection of diagonals is equidistant from the vertices, but not necessarily from the diagonals themselves. However, if the question implies a point equidistant from the lines forming the diagonals, we would need to find the equations of the diagonals and then the locus of points equidistant from two lines. Given the context of NCERT, it is most likely asking for the intersection point of the diagonals.Let's check if it's a rhombus:PQ =

\sqrt{(4 - (- 2))^{2} + (7 - 3)^{2}} = \sqrt{6^{2} + 4^{2}} = \sqrt{36 + 16} = \sqrt{52}

QR =

\sqrt{(8 - 4)^{2} + (1 - 7)^{2}} = \sqrt{4^{2} + (- 6)^{2}} = \sqrt{16 + 36} = \sqrt{52}

RS =

\sqrt{(2 - 8)^{2} + (- 3 - 1)^{2}} = \sqrt{(- 6)^{2} + (- 4)^{2}} = \sqrt{36 + 16} = \sqrt{52}

SP =

\sqrt{(- 2 - 2)^{2} + (3 - (- 3))^{2}} = \sqrt{(- 4)^{2} + 6^{2}} = \sqrt{16 + 36} = \sqrt{52}

All sides are equal, so it's a rhombus. In a rhombus, diagonals are perpendicular bisectors. The intersection point of diagonals is equidistant from the endpoints of diagonals. If the question means equidistant from the lines containing the diagonals, the point of intersection of diagonals is the center.The coordinates of the point where the statue should be placed is the intersection of the diagonals, which is (3, 2). 5marks

Point T(2, 3), A(–1, 5), B(6, 1).Distance AT =

\sqrt{(- 1 - 2)^{2} + (5 - 3)^{2}} = \sqrt{(- 3)^{2} + (2)^{2}} = \sqrt{9 + 4} = \sqrt{13}

units.Distance BT =

\sqrt{(6 - 2)^{2} + (1 - 3)^{2}} = \sqrt{(4)^{2} + (- 2)^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}

units. 3marks

Let S = (–3, –4) and L = (2, 8). Point P(x, y) divides SL such that SL is divided in the ratio 3:1 externally is not implied by "distance of the lighthouse to P is three times the distance from the ship to P". This implies SP : PL = 1 : 3 internally.Using the section formula for internal division with ratio 1:3:x = (1 * 2 + 3 * (–3)) / (1 + 3) = (2 – 9) / 4 = –7/4y = (1 * 8 + 3 * (–4)) / (1 + 3) = (8 – 12) / 4 = –4/4 = –1The coordinates of P are (–7/4, –1). 3marks

Vertices are A(1, 7), B(4, 2), C(–1, –1), D(– 4, 4).(i) AB =

\sqrt{(4 - 1)^{2} + (2 - 7)^{2}} = \sqrt{3^{2} + (- 5)^{2}} = \sqrt{9 + 25} = \sqrt{34}

.BC =

\sqrt{(- 1 - 4)^{2} + (- 1 - 2)^{2}} = \sqrt{(- 5)^{2} + (- 3)^{2}} = \sqrt{25 + 9} = \sqrt{34}

.CD =

\sqrt{(- 4 - (- 1))^{2} + (4 - (- 1))^{2}} = \sqrt{(- 3)^{2} + 5^{2}} = \sqrt{9 + 25} = \sqrt{34}

.DA =

\sqrt{(1 - (- 4))^{2} + (7 - 4)^{2}} = \sqrt{5^{2} + 3^{2}} = \sqrt{25 + 9} = \sqrt{34}

.All sides are equal.AC =

\sqrt{(- 1 - 1)^{2} + (- 1 - 7)^{2}} = \sqrt{(- 2)^{2} + (- 8)^{2}} = \sqrt{4 + 64} = \sqrt{68}

.BD =

\sqrt{(- 4 - 4)^{2} + (4 - 2)^{2}} = \sqrt{(- 8)^{2} + 2^{2}} = \sqrt{64 + 4} = \sqrt{68}

.The diagonals are equal. Since all sides are equal and diagonals are equal, ABCD is a square. [3 marks](ii) Distance of origin (0, 0) from D(– 4, 4) is

\sqrt{(- 4 - 0)^{2} + (4 - 0)^{2}} = \sqrt{(- 4)^{2} + 4^{2}} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2}

units. 1mark

Assertion: Let A = (0, 1), B = (4, 9), P = (2, 5).Ratio AP : PB = (2-0) : (4-2) = 2 : 2 = 1 : 1.Ratio AP : PB = (5-1) : (9-5) = 4 : 4 = 1 : 1.Since the ratio is the same for x and y coordinates, P divides AB in the ratio 1:1. Thus, the assertion is true.Reason: The section formula is correctly stated.(a) Both assertion and reason are true and reason is the correct explanation of assertion. 2marks

Points are A(–5, 6) and B(3, –2). Point P(x, y) divides AB in the ratio 3:1 internally.x = (3 * 3 + 1 * (–5)) / (3 + 1) = (9 – 5) / 4 = 4/4 = 1.y = (3 * (–2) + 1 * 6) / (3 + 1) = (–6 + 6) / 4 = 0/4 = 0.The coordinates of the treasure are (1, 0). 3marks

Points are P(1, 2), Q(4, 5), R(7, 2), S(4, -1).(i) PQ =

\sqrt{(4 - 1)^{2} + (5 - 2)^{2}} = \sqrt{3^{2} + 3^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.QR =

\sqrt{(7 - 4)^{2} + (2 - 5)^{2}} = \sqrt{3^{2} + (- 3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.RS =

\sqrt{(4 - 7)^{2} + (- 1 - 2)^{2}} = \sqrt{(- 3)^{2} + (- 3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.SP =

\sqrt{(1 - 4)^{2} + (2 - (- 1))^{2}} = \sqrt{(- 3)^{2} + 3^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

.All sides are equal, so PQRS is a rhombus. [3 marks](ii) Diagonal PR length =

\sqrt{(7 - 1)^{2} + (2 - 2)^{2}} = \sqrt{6^{2} + 0^{2}} = 6

.Diagonal QS length =

\sqrt{(4 - 4)^{2} + (- 1 - 5)^{2}} = \sqrt{0^{2} + (- 6)^{2}} = 6

.The lengths of the diagonals are equal. 2marks

Point L(–1, 3), M(5, 7). Point P divides LM in the ratio 1:3 internally.x = (1 * 5 + 3 * (–1)) / (1 + 3) = (5 – 3) / 4 = 2/4 = 1/2.y = (1 * 7 + 3 * 3) / (1 + 3) = (7 + 9) / 4 = 16/4 = 4.The coordinates of P are (1/2, 4). 3marks

Points are A(–6, 4) and B(–1, 9). Point C(x, y) is on the x-axis, so y = 0.C is equidistant from A and B, so AC² = BC².(x – (–6))² + (0 – 4)² = (x – (–1))² + (0 – 9)²(x + 6)² + (–4)² = (x + 1)² + (–9)²x² + 12x + 36 + 16 = x² + 2x + 1 + 8112x + 52 = 2x + 8210x = 30x = 3So, the coordinates of C are (3, 0). 3marks

Points are A(–3, 4) and B(7, –6). Point P divides AB such that AP = (3/10) AB. This means AP : PB = 3 : 7.Using the section formula:x = (3 * 7 + 7 * (–3)) / (3 + 7) = (21 – 21) / 10 = 0/10 = 0.y = (3 * (–6) + 7 * 4) / (3 + 7) = (–18 + 28) / 10 = 10/10 = 1.The coordinates of P are (0, 1). 3marks

Let P(x, y) be the point. P is equidistant from A(–2, 4) and B(6, –2).AP² = BP²(x – (–2))² + (y – 4)² = (x – 6)² + (y – (–2))²(x + 2)² + (y – 4)² = (x – 6)² + (y + 2)²x² + 4x + 4 + y² – 8y + 16 = x² – 12x + 36 + y² + 4y + 44x – 8y + 20 = –12x + 4y + 4016x – 12y = 204x – 3y = 5 (Equation 1)

P also lies on the road y = 2x – 1. Substitute this into Equation 1:
4x – 3(2x – 1) = 5
4x – 6x + 3 = 5
–2x = 2
x = –1

Now find y:
y = 2x – 1 = 2(–1) – 1 = –2 – 1 = –3.
The coordinates of the fountain are (–1, –3). [5 marks]

Let V1 = (–4, 5) and V2 = (6, –3). Point C divides V1V2 such that V1C : CV2 = 2 : 3.Using the section formula:x = (2 * 6 + 3 * (–4)) / (2 + 3) = (12 – 12) / 5 = 0/5 = 0.y = (2 * (–3) + 3 * 5) / (2 + 3) = (–6 + 15) / 5 = 9/5.The coordinates of the community centre are (0, 9/5). 3marks

Points are A(1, 5), B(–3, 2), C(–6, –1).(i) AB =

\sqrt{(- 3 - 1)^{2} + (2 - 5)^{2}} = \sqrt{(- 4)^{2} + (- 3)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5

. [1 mark](ii) BC =

\sqrt{(- 6 - (- 3))^{2} + (- 1 - 2)^{2}} = \sqrt{(- 3)^{2} + (- 3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

. [1 mark](iii) AC =

\sqrt{(- 6 - 1)^{2} + (- 1 - 5)^{2}} = \sqrt{(- 7)^{2} + (- 6)^{2}} = \sqrt{49 + 36} = \sqrt{85}

.Check for collinearity: AB + BC = 5 +

3 \sqrt{2}

≈ 5 + 4.24 = 9.24. AC =

\sqrt{85}

≈ 9.22.Since AB + BC is approximately equal to AC, the points are collinear. 2marks

Assertion: Let A = (1, 1), B = (5, 9). Mid-point M = ((1+5)/2, (1+9)/2) = (6/2, 10/2) = (3, 5).The assertion is true.Reason: The mid-point formula is correctly stated.(a) Both assertion and reason are true and reason is the correct explanation of assertion. 2marks

Points are D(–5, 7) and E(1, –1). Point P divides DE such that DP = (1/3) DE. This means DP : PE = 1 : 2.Using the section formula:x = (1 * 1 + 2 * (–5)) / (1 + 2) = (1 – 10) / 3 = –9/3 = –3.y = (1 * (–1) + 2 * 7) / (1 + 2) = (–1 + 14) / 3 = 13/3.The coordinates of P are (–3, 13/3). 3marks

Let P(x, y) be the point. P is equidistant from A(–4, 3) and B(2, 1).AP² = BP²(x – (–4))² + (y – 3)² = (x – 2)² + (y – 1)²(x + 4)² + (y – 3)² = (x – 2)² + (y – 1)²x² + 8x + 16 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 18x – 6y + 25 = –4x – 2y + 512x – 4y = –203x – y = –5 (Equation 1)

P also lies on the path y = x + 1. Substitute this into Equation 1:
3x – (x + 1) = –5
3x – x – 1 = –5
2x = –4
x = –2

Now find y:
y = x + 1 = –2 + 1 = –1.
The coordinates of point P are (–2, –1). [5 marks]

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Coordinate Geometry — Competency Based Questions

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