Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 7 Co Ordinate Geometry are provided here with simple step-by-step explanations. These solutions for Co Ordinate Geometry are extremely popular among Class 9 students for Maths Co Ordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 93:

#### Question 1:

• A($-$3, 2), • B($-$5, $-$2), • K(3.5, 1.5), • D(2, 10),

#### Answer:

The *x* co-ordinate of A(−3, 2) is negative and its *y* coordinate is positive. Therefore, point A(−3, 2) is in the second quadrant.

The *x* co-ordinate of B(−5, −2) is negative and its *y* coordinate is negative. Therefore, point B(−5, −2) is in the third quadrant.

The *x* co-ordinate of K(3.5, 1.5) is positive and its *y* coordinate is positive. Therefore, point K(3.5, 1.5) is in the first quadrant.

The *x* co-ordinate of D(2, 10) is positive and its *y* coordinate is positive. Therefore, point D(2, 10) is in the first quadrant.

The

*x*co-ordinate of E(37, 35) is positive and its

*y*coordinate is positive. Therefore, point E(37, 35) is in the first quadrant.

The

*x*co-ordinate of F(15, −18) is positive and its

*y*coordinate is negative. Therefore, point F(15, −18) is in the fourth quadrant.

The

*x*co-ordinate of G(3, −7) is positive and its

*y*coordinate is negative. Therefore, point G(3, −7) is in the fourth quadrant.

The

*x*co-ordinate of H(0, −5) is zero. Therefore, point H(0, −5) is on the Y-axis.

The

*y*co-ordinate of M(12, 0) is zero. Therefore, point M(12, 0) is on the X-axis.

The

*x*co-ordinate of N(0, 9) is zero. Therefore, point N(0, 9) is on the Y-axis.

The

*x*co-ordinate of P(0, 2.5) is zero. Therefore, point P(0, 2.5) is on the Y-axis.

The

*x*co-ordinate of Q(−7, −3) is negative and its

*y*coordinate is negative. Therefore, point Q(−7, −3) is in the third quadrant.

#### Page No 93:

#### Question 2:

*x*co-ordinate is positive, and the

*y*co-ordinate is negative.

*x*co-ordinate is negative and

*y*co-ordinate is positive.

#### Answer:

(i) The *x* co-ordinate and *y* co-ordinate of a point are both positive in the **first** quadrant.

(ii) The *x* co-ordinate and *y* co-ordinate of a point are both negative in the **third** quadrant.

(iii) The *x* co-ordinate of a point is positive and *y* co-ordinate of a point is negative in the **fourth** quadrant.

(iv) The *x* co-ordinate of a point is negative and *y* co-ordinate of a point is positive in the **second** quadrant.

#### Page No 93:

#### Question 3:

L($-$2, 4), M(5, 6), N($-$3, $-$4), P(2, $-$3), Q(6, $-$5), S(7, 0), T(0,$-$5)

#### Answer:

The given points are L(−2, 4), M(5, 6), N(−3, −4), P(2, −3), Q(6, −5), S(7, 0) and T(0, −5).

These point can be plotted on the co-ordinate system as follows:

#### Page No 97:

#### Question 1:

On a graph paper plot the points A (3,0), B(3,3), C(0,3). Join A, B and B, C. What is the figure formed?

#### Answer:

The given points are A(3, 0), B(3, 3) and C(0, 3). These points can be plotted on the co-ordinate plane as follows:

The *x* co-ordinate of point is its distance from the Y-axis and *y* co-ordinate of point is its distance from the X-axis.

Here, OA = AB = BC = OC = 3 units

Therefore, the figure formed is a square.

#### Page No 97:

#### Question 2:

Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.

#### Answer:

The equation of the line parallel to Y-axis at a distance of 7 units from it to its left is *x* = −7.

#### Page No 97:

#### Question 3:

Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.

#### Answer:

The equation of the line parallel to X-axis at a distance of 5 units from it and below the X-axis is *y* = −5.

#### Page No 97:

#### Question 4:

The point Q($-$3, $-$2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.

#### Answer:

The *x* co-ordinate of a point is its distance from the Y-axis and *y* co-ordinate is its distance from the X-axis.

The point of intersection of the line parallel to Y-axis at a distance 3 units from it to its left and the line parallel to X-axis at a distance 2 units below it is (−3, −2).

The equation of the line parallel to Y-axis at a distance 3 units from it to its left is *x* = −3.

#### Page No 98:

#### Question 5:

X-axis and line *x* = $-$4 are parallel lines. What is the distance between them?

#### Answer:

* Disclaimer*: The question is incorrect. The question should be "Y-axis and line

*x*= $-$4 are parallel lines. What is the distance between them?

**or**X-axis and line

*y*= $-$4 are parallel lines. What is the distance between them?"

If the question is "Y-axis and line

*x*= $-$4 are parallel lines. What is the distance between them?" then the answer is as follows.

*x*= −4 is the equation of the line parallel to the Y-axis at a distance of 4 units and to the left of Y-axis.

Thus, the distance between them is 4 units.

**OR**

If the question is "X-axis and line

*y*= $-$4 are parallel lines. What is the distance between them?" then the answer is as follows.

*y*= −4 is the equation of the line parallel to the X-axis at a distance of 4 units and below the X-axis.

Thus, the distance between them is 4 units.

#### Page No 98:

#### Question 6:

*x*= 3 (ii)

*y*$-$2 = 0 (iii)

*x*+ 6 = 0 (iv)

*y*= $-$5

#### Answer:

(i)

*x* = 3 is the equation of the line parallel to Y-axis at a distance of 3 units and to the right of Y-axis.

Thus, the graph of the line *x* = 3 is parallel to Y-axis.

(ii)

*y* − 2 = 0

⇒ *y* = 2

*y* = 2 is the equation of the line parallel to X-axis at a distance of 2 units and above the X-axis.

Thus, the graph of the line *y* − 2 = 0 is parallel to X-axis.

(iii)

*x* + 6 = 0

⇒ *x* = −6

*x* = −6 is the equation of the line parallel to Y-axis at a distance of 6 units from it to its left.

Thus, the graph of the line *x* + 6 = 0 is parallel to Y-axis.

(iv)

*y* = −5 is the equation of the line parallel to X-axis at a distance of 5 units and below the X-axis.

Thus, the graph of the line *y* = −5 is parallel to X-axis.

#### Page No 98:

#### Question 7:

On a graph paper, plot the points A(2, 3), B(6,$-$1) and C(0, 5). If those points are collinear then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.

#### Answer:

The given points are A(2, 3), B(6, −1) and C(0, 5).

The line which includes these points is shown below:

It can be seen from the graph that the line intersects the X-axis at (5, 0) and the Y-axis at (0, 5).

#### Page No 98:

#### Question 8:

*x*+ 4 = 0,

*y*$-$1 = 0, 2

*x*+ 3 = 0, 3

*y*$-$ 15 = 0

#### Answer:

*x* + 4 = 0 ⇒ *x* = −4

*x* = −4 is the equation of the line parallel to Y-axis at a distance of 4 units from it to its left.

*y* − 1 = 0 ⇒ *y* = 1

*y* = 1 is the equation of the line parallel to X-axis at a distance of 1 unit and above the X-axis.

2*x* + 3 = 0 ⇒ *x* = $-\frac{3}{2}$

*x* = $-\frac{3}{2}$ is the equation of the line parallel to Y-axis at a distance of $\frac{3}{2}$ units from it to its left.

3*y* − 15 = 0 ⇒ *y* = $\frac{15}{3}$ = 5

*y* = 5 is the equation of the line parallel to X-axis at a distance of 5 units and above the X-axis.

It can be seen from the figure that the co-ordinates of the points of intersection are $\left(-\frac{3}{2},1\right)$, $\left(-\frac{3}{2},5\right)$, (−4, 5) and (−4, 1).

#### Page No 98:

#### Question 9:

(i)

*x*+

*y*= 2 (ii) 3

*x*$-$

*y*= 0 (iii) 2

*x*+

*y*= 1

#### Answer:

(i)

The equation of the given line is *x* + *y* = 2.

*x* + *y* = 2

⇒ *y* = 2 − *x* .....(1)

Putting *x* = 0 in (1), we get

*y* = 2 − 0 = 2

Putting *x* = 2 in (1), we get

*y* = 2 − 2 = 0

Putting *x* = −1 in (1), we get

*y* = 2 − (−1) = 2 + 1 = 3

Putting *x* = 5 in (1), we get

*y* = 2 − 5 = −3

These values can be represented in the table in the form of ordered pairs as follows:

x |
0 | 2 | −1 | 5 |

y |
2 | 0 | 3 | −3 |

(x, y) |
(0, 2) | (2, 0) | (−1, 3) | (5, −3) |

Plot these points on the graph paper.

The line is the graph of the equation

*x*+

*y*= 2.

(ii)

The equation of the given line is 3

*x*−

*y*= 0.

3

*x*−

*y*= 0

⇒

*y*= 3

*x*.....(1)

Putting

*x*= 0 in (1), we get

*y*= 3 × 0 = 0

Putting

*x*= 1 in (1), we get

*y*= 3 × 1 = 3

Putting

*x*= −1 in (1), we get

*y*= 3 × (−1) = −3

Putting

*x*= 2 in (1), we get

*y*= 3 × 2 = 6

These values can be represented in the table in the form of ordered pairs as follows:

x |
0 | 1 | −1 | 2 |

y |
0 | 3 | −3 | 6 |

(x, y) |
(0, 0) | (3, 0) | (−1, −3) | (2, 6) |

Plot these points on the graph paper.

The line is the graph of the equation 3

*x*−

*y*= 0.

(iii)

The equation of the given line is 2

*x*+

*y*= 1.

2

*x*+

*y*= 1

⇒

*y*= 1 − 2

*x*.....(1)

Putting

*x*= 0 in (1), we get

*y*= 1 − 2 × 0 = 1 − 0 = 1

Putting

*x*= 1 in (1), we get

*y*= 1 − 2 × 1 = 1 − 2 = −1

Putting

*x*= −1 in (1), we get

*y*= 1 − 2 × (−1) = 1 + 2 = 3

Putting

*x*= 2 in (1), we get

*y*= 1 − 2 × 2 = 1 − 4 = −3

These values can be represented in the table in the form of ordered pairs as follows:

x |
0 | 1 | −1 | 2 |

y |
1 | −1 | 3 | −3 |

(x, y) |
(0, 1) | (1, −1) | (−1, 3) | (2, −3) |

Plot these points on the graph paper.

The line is the graph of the equation 2

*x*+

*y*= 1.

#### Page No 98:

#### Question 1:

*b*,

*b*) (B) (

*o*,

*b*) (C) (

*a*,

*o*) (D) (

*a*,

*a*)

*y*=

*x*is of the form .....

*a*,

*a*) (B) (

*o*,

*a*) (C) (

*a*,

*o*) (D) (

*a*, $-$

*a*)

*x*= 0 (B)

*y*= 0 (C)

*x*+

*y*= 0 (D)

*x*=

*y*

(vi) Which of the points P ($-$1,1), Q (3,$-$4), R(1,$-$1), S ($-$2,$-$3), T ($-$4,4) lie in the fourth quadrant ?

#### Answer:

(i)

The *y* co-ordinate of every point on the X-axis is 0. Thus, the co-ordinates of a point on the X-axis is (*a*, 0).

Hence, the correct answer is option (C).

(ii)

Putting *x* = *a* in *y* = *x*, we get

*y* = *a*

Thus, any point on the line *y* = *x* is of the form (*a*, *a*).

Hence, the correct answer is option (A).

(iii)

The *y* co-ordinate of every point on the X-axis is 0. Therefore, the equation of the X-axis is *y* = 0.

Hence, the correct answer is option (B).

(iv)

The *x* co-ordinate of (−4, −3) is negative and its *y* co-ordinate is negative. Therefore, the point (−4, −3) lies in the third quadrant.

Hence, the correct answer is option (C).

(v)

The *y* co-ordinate of all the points (−5, 5), (6, 5), (−3, 5) and (0, 5) is 5. All these points lies on the line *y* = 5, which is parallel to the X-axis.

Thus, the line which includes the points (−5, 5), (6, 5), (−3, 5) and (0, 5) is parallel to the X-axis.

Hence, the correct answer is option (C).

(vi)

The point whose *x* co-ordinate is positive and *y* co-ordinate is negative lie in the fourth quadrant.

Thus, the points Q(3, −4) and R(1, −1) lie in the fourth quadrant.

Hence, the correct answer is option (B).

#### Page No 99:

#### Question 2:

(i) Write the co-ordinates of the points Q and R.

(ii) Write the co-ordinates of the points T and M.

(iii) Which point lies in the third quadrant ?

(iv) Which are the points whose

*x*and

*y*co-ordinates are equal?

#### Answer:

(i) The *x *co-ordinate of a point is its distance from the Y-axis and *y* co-ordinate of a point is its distance from the X-axis.

The co-ordinates of point Q are (−2, 2) and the co-ordinates of point R are (4, −1).

(ii) The *x* co-ordinate of every point on the Y-axis is 0 and the *y* co-ordinate of every point on the X-axis is 0.

The co-ordinates of point T are (0, −1) and the co-ordinates of point M are (3, 0).

(iii) The point whose *x* co-ordinate is negative and *y* co-ordinate is negative lies in the third quadrant.

Thus, the point S(−3, −2) lies in the third quadrant.

(iv) The co-ordinates of point O are (0, 0).

Thus, O is the point whose *x* and *y* co-ordinates are equal.

#### Page No 99:

#### Question 3:

(i) (5,$-$3) (ii) ($-$7, $-$12) (iii) ($-$23, 4) (iv) ($-$9, 5) (v) (0, $-$3) (vi) ($-$6, 0)

#### Answer:

(i) The *x* co-ordinate of the point (5, −3) is positive and its *y* co-ordinate is negative. Therefore, the point (5, −3) lie in the fourth quadrant.

(ii) The *x* co-ordinate of the point (−7, −12) is negative and its *y* co-ordinate is negative. Therefore, the point (−7, −12) lie in the third quadrant.

(iii) The *x* co-ordinate of the point (−23, 4) is negative and its *y* co-ordinate is positive. Therefore, the point (−23, 4) lie in the second quadrant.

(iv) The *x* co-ordinate of the point (−9, 5) is negative and its *y* co-ordinate is positive. Therefore, the point (−9, 5) lie in the second quadrant.

(v) The *x* co-ordinate of the point (0, −3) is zero. Therefore, the point (0, −3) lie on the Y-axis.

(vi) The *y* co-ordinate of the point (−6, 0) is zero. Therefore, the point (−6, 0) lie on the X-axis.

#### Page No 99:

#### Question 4:

A(1, 3), B($-$3, $-$1), C(1, $-$4), D($-$2, 3), E(0, $-$8), F(1, 0)

#### Answer:

The given points are A(1, 3), B(−3, −1), C(1, −4), D(−2, 3), E(0, −8) and F(1, 0).

These points can be plotted on the co-ordinate system as follows:

#### Page No 99:

#### Question 5:

*x*co-ordinates of the points L and M?

#### Answer:

(i) The line LM is parallel to the Y-axis and passing through the point (3, 0).

The equation of the line LM is *x* = 3 and at a distance of 3 units to the right of the Y-axis.

Thus, the distance of line LM from the Y-axis is 3 units.

(ii) The *x* co-ordinate of a point is its distance from the Y-axis and *y* co-ordinate is its distance from the X-axis.

Thus, the co-ordinates of the points P, Q and R are (3, 2), (3, −1) and (3, 0), respectively.

(iii)

*x* co-ordinate of point L = 3

*x* co-ordinate of point M = 3

∴ Difference between the *x* co-ordinates of the points L and M = 3 − 3 = 0

Thus, the difference between the *x* co-ordinates of the points L and M is 0.

#### Page No 99:

#### Question 6:

How many lines are there which are parallel to X-axis and having a distance 5 units?

#### Answer:

The equations of the lines parallel to X-axis and at a distance 5 units from it are *y* = 5 and *y* = −5.

Thus, there are two lines which are parallel to X-axis and having a distance 5 units from it.

#### Page No 99:

#### Question 7:

If ‘*a*’ is a real number, what is the distance between the Y-axis and the line *x* = *a* ?

#### Answer:

The line *x* = *a* is parallel to the Y-axis passing through (*a*, 0).

If *a* > 0, then the line will be a distance of *a* units to the right of the Y-axis.

If *a* < 0, then the line will be a distance of *a* units to the left of the Y-axis.

Thus, the distance between the Y-axis and the line *x* = *a* is |*a*| units.

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