Most Important Questions of Orienting Yourself: The Use of Coordinates Class 9 Maths with Solutions

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MULTIPLE CHOICE QUESTIONS

1.Abscissa of all the points on the x-axis is:

  • (a) 0
  • (b) 1
  • (c) 2
  • (d) any number

2.The point at which the two coordinate axes meet is called the:

  • (a) Abscissa
  • (b) Ordinate
  • (c) Origin
  • (d) Quadrant

3.A point both of whose coordinates are negative will lie in:

  • (a) I quadrant
  • (b) II quadrant
  • (c) III quadrant
  • (d) IV quadrant

4.The points (-5, 2) and (2, -5) lie in:

  • (a) Same quadrant
  • (b) II and III quadrants respectively
  • (c) II and IV quadrants respectively
  • (d) IV and II quadrants respectively

5.The point whose ordinate is 4 and which lies on y-axis is:

  • (a) (4, 0)
  • (b) (0, 4)
  • (c) (1, 4)
  • (d) (4, 2)

6.Which of the points P(0, 3), Q(1, 0), R(0, -1), S(-5, 0), T(1, 2) do not lie on the x-axis?

  • (a) P and R only
  • (b) Q and S only
  • (c) P, R and T
  • (d) Q, S and T

7.The point which lies on y-axis at a distance of 5 units in the negative direction of y-axis is:

  • (a) (0, 5)
  • (b) (5, 0)
  • (c) (0, -5)
  • (d) (-5, 0)

8.If the coordinates of a point are (-5, 6), the perpendicular distance of the point from x-axis is:

  • (a) 5 units
  • (b) 6 units
  • (c) 7 units
  • (d) 7.5 units

9.The coordinates of the point whose ordinate is -4 and which lies at y-axis are:

  • (a) (0, -4)
  • (b) (4, 0)
  • (c) (-4, 0)
  • (d) (0, 4)

10.If the distance between the points (2, -2) and (-1, x) is 5, one value of x is:

  • (a) -2
  • (b) 2
  • (c) -1
  • (d) 1

11.The mid-point of the line segment joining the points A(-2, 8) and B(-6, -4) is:

  • (a) (-4, -6)
  • (b) (2, 6)
  • (c) (-4, 2)
  • (d) (4, 2)

12.The distance between the points A(0, 6) and B(0, -2) is:

  • (a) 6
  • (b) 8
  • (c) 4
  • (d) 2

13.The distance of the point P(-6, 8) from the origin is:

  • (a) 8
  • (b) 2√7
  • (c) 10
  • (d) 6

14.The centre of a circle is at (2, 0), if one end of a diameter is at (6, 0), then the other end is at:

  • (a) (0, 0)
  • (b) (4, 0)
  • (c) (-2, 0)
  • (d) (-6, 0)

15.The points (-4, 0), (4, 0) and (3, 0) are the vertices of a/an:

  • (a) Right triangle
  • (b) Isosceles triangle
  • (c) Equilateral triangle
  • (d) Scalene triangle

SUBJECTIVE TYPE QUESTIONS

1.Show that the points (0, 0), (4, 0), and (0, 4) are the vertices of a right-angled isosceles triangle.

2.Find the point on the y-axis which is equidistant from the points (4, -3) and (-2, 5).

3.Find the point on the x-axis which is equidistant from the points (5, 4) and (-2, 3).

4.Determine if the points (1, 1), (2, 2), and (5, 5) are collinear.

5.ABCD is a rectangle with vertices A(0, 0), B(5, 0), and C(5, 2). Find the coordinates of D and the length of a diagonal.

6.If point P(x, y) is equidistant from A(5, 1) and B(-1, 5), prove that 3x = 2y – 4.

7.If P(x, y) is equidistant from A(a+b, a-b) and B(a-b, a+b), prove that ax = by.

8.Prove that the points (0, 0), (3, 0), and (0, 4) form a right triangle and calculate its area.

9.Show that (1, 1), (4, 1), (4, 4), and (1, 4) are the vertices of a square.

10.Show that (2, -1), (5, -1), (5, 6), and (2, 6) are the vertices of a rectangle.

11.A delivery hub is at (-2, 3) and a customer is at (10, 9). Find the straight-line distance for the delivery drone.

12.Two fence posts in a rectangular garden are placed at (1, -2) and (7, 6). Find the distance between these two posts.

13.Two colonies are located at A(2, 6) and B(10, 4). A mall is to be built at a point exactly halfway between them so that both colonies have equal access.

(i) Find the coordinates of the mall.

(ii) Check whether the mall lies on the x-axis, y-axis, or in a quadrant.

(iii) Suggest why midpoint location is useful in city planning.

14.If P(a, 5) is the midpoint of the line segment joining A(-4, 7) and B(10, 3), find the value of a.

15.Show that the diagonals of a rectangle ABCD with vertices A(0, 0), B(4, 0), C(4, 3), and D(0, 3) bisect each other and are equal in length.

16.If (p, q) is the midpoint of the segment joining A(-3, 4) and B(k, 6), and p – 2q = 10, find the value of k.

17.If two adjacent vertices of a parallelogram are (1, 2) and (3, 4), and the diagonals intersect at (0, 5), find the coordinates of the other two vertices.

18.Find the length of the median AD of triangle ABC having vertices A(5, 1), B(1, 5), and C(-3, -1).

19.ABCD is a quadrilateral with vertices A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2). If P, Q, R, S are the midpoints of sides AB, BC, CD, and DA, show that PQRS is a parallelogram.

20.Riya’s house is at (1, 2) and her friend’s house is at (7, 10). They want to meet at a bus stop exactly halfway between their houses. Find the coordinates of the bus stop.

21.In a rectangular hall, two computers are located at (3, 4) and (9, 12). A WiFi router needs to be placed exactly in the middle to provide equal signal strength. Where should it be placed?

22.Two villages are located at A(-5, 3) and B(3, -1). A bridge is to be built exactly halfway between them on the straight road connecting them. Find the location of the bridge.

ASSERTION–REASON QUESTIONS

Directions: In each question below, there are two statements labelled as Assertion (A) and Reason (R). Choose the correct answer:

  • (a) Both (A) and (R) are true and (R) is the correct explanation of (A)
  • (b) Both (A) and (R) are true but (R) is not the correct explanation of (A)
  • (c) (A) is true but (R) is false
  • (d) (A) is false but (R) is true

1.Assertion (A): (-3, 4) lies to the left of the y-axis.

Reason (R): Abscissa is negative to the right of y-axis.

2.Assertion (A): (-11, -13) lies in IV quadrant.

Reason (R): Abscissa and ordinate are negative in III quadrant.

3.Assertion (A): The abscissa is 9 and ordinate is -6, the coordinates of the point are (9, -6).

Reason (R): The point (2, 3) lies in II quadrant.

4.Assertion (A): The ordinate of point (-6, 4) is 4.

Reason (R): The perpendicular distance of a point from x-axis is called its ordinate.

5.Assertion (A): Points (1, 5), (2, 3) and (-2, -11) are collinear.

Reason (R): Three points are said to be collinear when they all lie on the same line.

CASE STUDY QUESTION

A telecommunications team is setting up a local network grid. They have placed four signal relay towers, labelled S, T, A, and R, at the following coordinates:

  • S is at (2, 4)
  • T is at (2, -4)
  • A is at (-3, -4)
  • R is at (-3, 4)

The team needs to analyze the layout to ensure maximum signal coverage and structural efficiency.

Answer the following questions:

1.

Analyze the coordinates of STAR and predict the following:

(i)

What specific type of quadrilateral is formed by connecting these four points?

(ii)

Which points are reflections of each other across the y-axis?

(iii)

Which points are reflections of each other across the x-axis?

ANSWERS AND SOLUTIONS

MCQ ANSWERS

  1. (d) any number
  2. (c) Origin
  3. (c) III quadrant
  4. (c) II and IV quadrants respectively
  5. (b) (0, 4)
  6. (c) P, R and T
  7. (c) (0, -5)
  8. (b) 6 units
  9. (a) (0, -4)
  10. (b) 2
  11. (c) (-4, 2)
  12. (b) 8
  13. (c) 10
  14. (c) (-2, 0)
  15. (d) Scalene triangle

SOLUTIONS OF SUBJECTIVE QUESTIONS

1.

Points are A(0, 0), B(4, 0), C(0, 4)

AB = 4 units AC = 4 units

Also, AB is along x-axis and AC is along y-axis. Hence, ∠A = 90°.

Since AB = AC, triangle is isosceles.

Therefore, the triangle is a right-angled isosceles triangle.

2.

Let the point on y-axis be P(0, y).

Distance from P to (4, -3):

√[(0 – 4)^2 + (y + 3)^2]

Distance from P to (-2, 5):

√[(0 + 2)^2 + (y – 5)^2]

Equating both:

16 + (y + 3)^2 = 4 + (y – 5)^2

16 + y^2 + 6y + 9 = 4 + y^2 – 10y + 25

6y + 25 = -10y + 29

16y = 4

y = 1/4

Required point = (0, 1/4)

3.

Let the point on x-axis be P(x, 0).

Using equal distance condition:

(x – 5)^2 + (0 – 4)^2 = (x + 2)^2 + (0 – 3)^2

x^2 – 10x + 25 + 16 = x^2 + 4x + 4 + 9

-10x + 41 = 4x + 13

-14x = -28

x = 2

Required point = (2, 0)

4.

Slope between (1, 1) and (2, 2):

m1 = (2 – 1)/(2 – 1) = 1

Slope between (2, 2) and (5, 5):

m2 = (5 – 2)/(5 – 2) = 1

Since slopes are equal, the points are collinear.

5.

Given: A(0, 0), B(5, 0), C(5, 2)

Since ABCD is a rectangle, D = (0, 2)

Diagonal AC:

AC = √[(5 – 0)^2 + (2 – 0)^2]

= √(25 + 4)

= √29 units

6.

P(x, y) is equidistant from A(5, 1) and B(-1, 5)

(x – 5)^2 + (y – 1)^2 = (x + 1)^2 + (y – 5)^2

x^2 – 10x + 25 + y^2 – 2y + 1 = x^2 + 2x + 1 + y^2 – 10y + 25

-10x – 2y + 26 = 2x – 10y + 26

-12x + 8y = 0

3x = 2y

Hence proved.

7.

Since P(x, y) is equidistant from A(a+b, a-b) and B(a-b, a+b),

(x-a-b)^2 + (y-a+b)^2 = (x-a+b)^2 + (y-a-b)^2

On simplifying,

ax = by

Hence proved.

8.

Points are A(0, 0), B(3, 0), C(0, 4)

AB = 3 units AC = 4 units BC = 5 units

Since:

AB^2 + AC^2 = BC^2

3^2 + 4^2 = 5^2

9 + 16 = 25

Hence, triangle is right-angled.

Area = 1/2 × 3 × 4 = 6 square units.

9.

Vertices are: (1, 1), (4, 1), (4, 4), (1, 4)

Each side length = 3 units.

Adjacent sides are perpendicular.

Therefore, the figure is a square.

10.

Vertices are: (2, -1), (5, -1), (5, 6), (2, 6)

Opposite sides are equal and parallel.

Adjacent sides are perpendicular.

Hence, the figure is a rectangle.

11.

Using distance formula:

Distance = √[(10 + 2)^2 + (9 – 3)^2]

= √(12^2 + 6^2)

= √(144 + 36)

= √180

= 6√5 units

12.

Distance = √[(7 – 1)^2 + (6 + 2)^2]

= √(6^2 + 8^2)

= √(36 + 64)

= √100

= 10 units

13.

Midpoint formula:

Mall coordinates = ((2+10)/2 , (6+4)/2)

= (12/2, 10/2)

= (6, 5)

(ii) Since both coordinates are positive, it lies in the first quadrant.

(iii) Midpoint location gives equal accessibility and reduces travel distance.

14.

Using midpoint formula:

a = (-4 + 10)/2

= 6/2

= 3

Hence, a = 3.

15.

Midpoint of AC:

((0+4)/2, (0+3)/2)

= (2, 3/2)

Midpoint of BD:

((4+0)/2, (0+3)/2)

= (2, 3/2)

Thus diagonals bisect each other.

Length of AC:

√[(4-0)^2 + (3-0)^2]

= √(16 + 9)

= 5

Length of BD:

√[(0-4)^2 + (3-0)^2]

= 5

Hence diagonals are equal.

16.

Midpoint of A(-3, 4) and B(k, 6):

p = (-3 + k)/2

q = (4 + 6)/2 = 5

Given:

p – 2q = 10

(-3 + k)/2 – 10 = 10

(-3 + k)/2 = 20

-3 + k = 40

k = 43

17.

Let the other vertices be C(x1, y1) and D(x2, y2).

Midpoint of diagonals is (0, 5).

Using midpoint formula:

For diagonal joining (1, 2) and C:

(1 + x1)/2 = 0

x1 = -1

(2 + y1)/2 = 5

y1 = 8

So C = (-1, 8)

For diagonal joining (3, 4) and D:

(3 + x2)/2 = 0

x2 = -3

(4 + y2)/2 = 5

y2 = 6

So D = (-3, 6)

18.

Midpoint of BC:

D = ((1-3)/2 , (5-1)/2)

= (-1, 2)

Length AD:

= √[(-1-5)^2 + (2-1)^2]

= √(36 + 1)

= √37 units

19.

Midpoints:

P = (1, -1/2)

Q = (7/2, 3/2)

R = (0, 5/2)

S = (-5/2, 1/2)

Diagonals PR and QS bisect each other.

Hence PQRS is a parallelogram.

20.

Midpoint = ((1+7)/2 , (2+10)/2)

= (4, 6)

Required coordinates of bus stop = (4, 6)

21.

Midpoint = ((3+9)/2 , (4+12)/2)

= (6, 8)

Router should be placed at (6, 8).

22.

Midpoint = ((-5+3)/2 , (3-1)/2)

= (-2/2 , 2/2)

= (-1, 1)

Bridge should be built at (-1, 1).

ASSERTION–REASON ANSWERS

  1. (c)
  2. (d)
  3. (c)
  4. (a)
  5. (b)

CASE STUDY SOLUTION

(i)

The points form a rectangle.

(ii)

Reflection across y-axis:

  • S(2, 4) and R(-3, 4) are not reflections.
  • T(2, -4) and A(-3, -4) are not reflections.

No exact pair is reflection across y-axis.

(iii)

Reflection across x-axis:

  • S(2, 4) and T(2, -4)
  • R(-3, 4) and A(-3, -4)

These pairs are reflections across x-axis.

Students can also check all other exercise solutions of the chapter “Introduction to Linear Polynomial” for complete preparation. Visit Exercise 2.1 Solutions, Exercise 2.2 Solutions, Exercise 2.3 Solutions, Exercise 2.4 Solutions, Exercise 2.5 Solutions, and Exercise 2.6 Solutions to understand every concept step by step and score better in exams.

Watch video solutions of ” End of Chapter Exercises” on our youtube Channel SHARMA TUTORIAL