Class 9 - Chapter 1: Orienting Yourself: The Use of Coordinates

Q1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

The x-axis and y-axis intersect at the origin.

Therefore,

O = (0, 0)

Answer: x-coordinate = 0, y-coordinate = 0.

Q2. Point W has x-coordinate equal to – 5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Point W has x-coordinate −5.

If H lies on a line through W parallel to the y-axis, then every point on that line will have the same x-coordinate.

Therefore,

H = (−5, y)

where y can be any real number.

If y > 0, H lies in Quadrant II.

If y < 0, H lies in Quadrant III.

If y = 0, H lies on the x-axis.

Answer: H = (−5, y). H may lie in Quadrant II or Quadrant III.

Q3. Consider the points R (3, 0), A (0, – 2), M (– 5, – 2) and P (– 5, 2). If they are joined in the same order, predict:

(i) Two sides of RAMP that are perpendicular to each other.

(ii) One side of RAMP that is parallel to one of the axes.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

(i) Two sides that are perpendicular

AM is horizontal because both points have y-coordinate −2.

MP is vertical because both points have x-coordinate −5.

A horizontal line and a vertical line are perpendicular.

Answer: AM ⟂ MP

(ii) One side parallel to an axis

AM is parallel to the x-axis.

MP is parallel to the y-axis.

Answer: AM is parallel to the x-axis.

(iii) Mirror-image points

A = (0, −2)

P = (−5, 2)

M = (−5, −2)

A and M do not form mirror images.

M and P have the same x-coordinate and opposite positions about the x-axis.

Therefore M and P are mirror images in the x-axis.

Answer: M(−5, −2) and P(−5, 2) are mirror images about the x-axis.

Q4. Plot point Z (5, – 6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.

One possible construction:

• I = (0,0)
• Z = (5,−6)
• N = (5,0)

IN = 5 units

ZN = 6 units

Using Pythagoras Theorem:

IZ² = IN² + ZN²

IZ² = 5² + 6²

IZ² = 25 + 36 = 61

IZ = √61 units

Q5. What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

If negative numbers did not exist, we could only use coordinates of the form:

(x, y), where x ≥ 0 and y ≥ 0

This would allow us to locate points only in the First Quadrant and on the positive parts of the axes.

Points in Quadrants II, III and IV could not be represented.

For example:

• (−3, 4) could not be located.
• (−2, −5) could not be located.
• (4, −7) could not be located.

Answer: Without negative numbers, the coordinate system would not allow us to locate all points on a two-dimensional plane.

Q6. Are the points M (– 3, – 4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

We use the slope method.

Slope of MA

m_MA = (0 − (−4)) / (0 − (−3))

= 4/3

Slope of AG

m_AG = (8 − 0) / (6 − 0)

= 8/6

= 4/3

Since the slopes are equal,

m_MA = m_AG

Therefore the three points lie on the same straight line.

Answer: M, A and G are collinear.

Q7. Use your method (from Problem 6) to check if the points R (– 5, – 1), B (– 2, – 5) and C (4, – 12) are on the same straight line. Now plot both sets of points and check your answers.

Slope of RB

m_RB = (−5 − (−1)) / (−2 − (−5))

= −4/3

Slope of BC

m_BC = (−12 − (−5)) / (4 − (−2))

= −7/6

Since

−4/3 ≠ −7/6

the slopes are not equal.

Therefore the points do not lie on the same straight line.

Answer: R, B and C are not collinear.

Q8. Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the
other in Quadrant IV.

(i) Right-Angled Isosceles Triangle

One possible set of vertices:

• O(0,0)
• A(4,0)
• B(0,4)

OA = OB = 4 units

∠AOB = 90°

Hence OAB is a right-angled isosceles triangle.

(ii) Isosceles Triangle with One Vertex in Quadrant III and Another in Quadrant IV

One possible set of vertices:

• O(0,0)
• P(−4,−3)   (Quadrant III)
• Q(4,−3)   (Quadrant IV)

Distance OP = OQ

Therefore OPQ is an isosceles triangle.

Answer: Any correct coordinates satisfying the conditions are acceptable.

Q9. The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

Image source- NCERT

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Formula for midpoint:

Midpoint = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Connection Between Coordinates

Whenever M is the midpoint of ST:

x_M = (x_S + x_T)/2

y_M = (y_S + y_T)/2

That is, each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints.

Q10. Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, – 4) and B (x, y).

Given:

M(−7, 1) is the midpoint of A(3, −4) and B(x, y).

Using the midpoint formula:

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Finding x-coordinate of B

−7 = (3 + x)/2

−14 = 3 + x

x = −17

Finding y-coordinate of B

1 = (−4 + y)/2

2 = −4 + y

y = 6

Therefore,

B = (−17, 6)

Q11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, –2).

Given:

A(4, 7) and B(16, −2)

P and Q divide AB into three equal parts.

Step 1: Difference in Coordinates

Δx = 16 − 4 = 12

Δy = −2 − 7 = −9

One-third of AB:

Δx/3 = 4

Δy/3 = −3

Point P (closer to A)

P = (4 + 4, 7 − 3)

P = (8, 4)

Point Q (closer to B)

Q = (8 + 4, 4 − 3)

Q = (12, 1)

Answer:

Q12. (i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?

(ii) Given the points D (–5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

(i) Showing that A, B and C lie on Circle K

Centre of circle:

O(0,0)

Distance OA

OA = √[(1)² + (−8)²]

= √(1 + 64)

= √65

Distance OB

OB = √[(−4)² + 7²]

= √(16 + 49)

= √65

Distance OC

OC = √[(−7)² + (−4)²]

= √(49 + 16)

= √65

Since

OA = OB = OC = √65

all three points are at the same distance from the origin.

Therefore A, B and C lie on the same circle centred at O.

Radius of Circle K = √65 units

(ii) Position of D and E Relative to Circle K

Point D(−5, 6)

OD = √[(−5)² + 6²]

= √(25 + 36)

= √61

Since

√61 < √65

D lies inside the circle.

Point E(0, 9)

OE = √(0² + 9²)

= 9

Since

9 > √65

E lies outside the circle.

Q13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.

Given midpoints:

• D(5,1)
• E(6,5)
• F(0,3)

Let:

• D = midpoint of BC
• E = midpoint of AC
• F = midpoint of AB

Using the midpoint relations:

A = E + F − D

B = D + F − E

C = D + E − F

Coordinates of A

A = (6 + 0 − 5, 5 + 3 − 1)

A = (1, 7)

Coordinates of B

B = (5 + 0 − 6, 1 + 3 − 5)

B = (−1, −1)

Coordinates of C

C = (5 + 6 − 0, 1 + 5 − 3)

C = (11, 3)

Answer:

A(1, 7), B(−1, −1), C(11, 3)

Q14.A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

(ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find:

(a) how many street intersections can be referred to as (4, 3).

(b) how many street intersections can be referred to as (3, 4)

Three vertices of a square are:

A(1, 1), B(4, 1), C(4, 4)

Step 1: Find Length AB

AB = √[(4 − 1)² + (1 − 1)²]

= √[3² + 0]

= √9

= 3 units

Step 2: Find Length BC

BC = √[(4 − 4)² + (4 − 1)²]

= √[0 + 3²]

= √9

= 3 units

Since AB = BC and they are perpendicular, A, B and C are three consecutive vertices of a square.

The fourth vertex is:

D = (1, 4)

Answer: D(1, 4)

Q15. A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:
(i) whether any part of either circle lies outside the screen.

(ii) whether the two circles intersect each other.

Given:

A(−5, 7) and B(3, −1)

Using Midpoint Formula

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

= ( (−5 + 3)/2 , (7 + (−1))/2 )

= ( −2/2 , 6/2 )

= (−1, 3)

Answer: M(−1, 3)

Q16. Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Step 1: Find Length AB

AB = √[(−1 − 2)² + (2 − 1)²]

= √[(−3)² + 1²]

= √(9 + 1)

= √10

Step 2: Find Length BC

BC = √[(−2 + 1)² + (−1 − 2)²]

= √[(−1)² + (−3)²]

= √(1 + 9)

= √10

Step 3: Find Length CD

CD = √[(1 + 2)² + (−2 + 1)²]

= √[3² + (−1)²]

= √(9 + 1)

= √10

Step 4: Find Length DA

DA = √[(2 − 1)² + (1 + 2)²]

= √[1² + 3²]

= √(1 + 9)

= √10

Step 5: Find Diagonal AC

AC = √[(−2 − 2)² + (−1 − 1)²]

= √[(−4)² + (−2)²]

= √(16 + 4)

= √20

= 2√5

Step 6: Find Diagonal BD

BD = √[(1 + 1)² + (−2 − 2)²]

= √[2² + (−4)²]

= √(4 + 16)

= √20

= 2√5

Conclusion

All four sides are equal and both diagonals are equal.

Therefore, ABCD is a square.

Area of the Square

Area = Side²

= (√10)²

= 10 square units

Answer:

ABCD is a square and its area is 10 square units.