**Coordinate geometry**- In the beginning, a short description of different

**coordinates**are given with reference to what students have already learned in class IX.

- The distance of a point from the
**y****-axis**is called its-*y***coordinate**or**abscissa.** - The distance of a point from the
is called its*x*-axis**x-****coordinate**or**ordinate**.

- How to find the
**distance between the two points**whose**coordinates**are given - To
**find the area of the****triangles**formed by**three given points**. - How to find the
**coordinates**of the points which divide a**line segment**joining two given points in a given**ratio.**

**distance formula.**Distance formula is well explained with the help of diagrams followed by important points to be noted at the end. Solved examples are given to make the concept more understandable to students. Different variety of questions are given in the first exercise i.e 7.1 from the topic

**distance formula.**

Moving further,

**section formula**is given and a special case called the

**mid-point formula**is also explained which is important from the examination point of view.

- In this special case, it is discussed what happens when the
**mid poin**t of a**line segment**divides the**line segment**in the**ratio 1 : 1.**An exercise of 10 questions is given after the explanation of the**section formula.** - In the last section, the
**area of a triangle**is discussed.

**area of triangles and quadrilaterals**whose

**vertices**are given. An optional exercise is given which is not from the examination point of view. After that, the chapter is concluded in 5 points for a quick revision.

#### Page No 161:

#### Question 1:

Find the distance between the following pairs of points:

(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (*a*, *b*), (− *a*, − *b*)

#### Answer:

(i) Distance between the two points is given by

(ii) Distance between is given by

(iii) Distance between is given by

##### Video Solution for real numbers (Page: 161 , Q.No.: 1)

NCERT Solution for Class 10 math - real numbers 161 , Question 1

#### Page No 161:

#### Question 2:

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

#### Answer:

Distance between points

Yes, we can find the distance between the given towns A and B.

Assume town A at origin point (0, 0).

Therefore, town B will be at point (36, 15) with respect to town A.

And hence, as calculated above, the distance between town A and B will be

39 km.

##### Video Solution for real numbers (Page: 161 , Q.No.: 2)

NCERT Solution for Class 10 math - real numbers 161 , Question 2

#### Page No 161:

#### Question 3:

Determine if the points (1, 5), (2, 3) and (− 2, − 11) are collinear.

#### Answer:

Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.

Let

Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

##### Video Solution for real numbers (Page: 161 , Q.No.: 3)

NCERT Solution for Class 10 math - real numbers 161 , Question 3

#### Page No 161:

#### Question 4:

Check whether (5, − 2), (6, 4) and (7, − 2) are the vertices of an isosceles triangle.

#### Answer:

Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given triangle respectively.

As two sides are equal in length, therefore, ABCis an isosceles triangle.

##### Video Solution for real numbers (Page: 161 , Q.No.: 4)

NCERT Solution for Class 10 math - real numbers 161 , Question 4

#### Page No 161:

#### Question 5:

In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.

Using distance formula, find which of them is correct.

#### Answer:

It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.

$\mathrm{CD}=\sqrt{{\left(9-6\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{{\left(3\right)}^{2}+{\left(3\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.

Therefore, ABCD is a square and hence, Champa was correct

##### Video Solution for real numbers (Page: 161 , Q.No.: 5)

NCERT Solution for Class 10 math - real numbers 161 , Question 5

#### Page No 161:

#### Question 6:

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (− 1, − 2), (1, 0), (− 1, 2), (− 3, 0)

(ii) (− 3, 5), (3, 1), (0, 3), (− 1, − 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

#### Answer:

(i) Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii)Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

##### Video Solution for real numbers (Page: 161 , Q.No.: 6)

NCERT Solution for Class 10 math - real numbers 161 , Question 6

#### Page No 161:

#### Question 7:

Find the point on the *x*-axis which is equidistant from (2, − 5) and (− 2, 9).

#### Answer:

We have to find a point on *x*-axis. Therefore, its *y*-coordinate will be 0.

Let the point on *x*-axis be.

By the given condition, these distances are equal in measure.

Therefore, the point is (− 7, 0).

##### Video Solution for real numbers (Page: 161 , Q.No.: 7)

NCERT Solution for Class 10 math - real numbers 161 , Question 7

#### Page No 161:

#### Question 8:

Find the values of *y* for which the distance between the points P (2, − 3) and Q (10, *y*) is 10 units.

#### Answer:

It is given that the distance between (2, −3) and (10, *y*) is 10.

##### Video Solution for real numbers (Page: 161 , Q.No.: 8)

NCERT Solution for Class 10 math - real numbers 161 , Question 8

#### Page No 162:

#### Question 9:

If Q (0, 1) is equidistant from P (5, − 3) and R (*x*, 6), find the values of *x*. Also find the distance QR and PR.

#### Answer:

Therefore, point R is (4, 6) or (−4, 6).

When point R is (4, 6),

When point R is (−4, 6),

##### Video Solution for real numbers (Page: 162 , Q.No.: 9)

NCERT Solution for Class 10 math - real numbers 162 , Question 9

#### Page No 162:

#### Question 10:

Find a relation between *x* and *y* such that the point (*x*, *y*) is equidistant from the point (3, 6) and (− 3, 4).

#### Answer:

Point (*x*, *y*) is equidistant from (3, 6) and (−3, 4).

##### Video Solution for real numbers (Page: 162 , Q.No.: 10)

NCERT Solution for Class 10 math - real numbers 162 , Question 10

#### Page No 167:

#### Question 1:

Find the coordinates of the point which divides the join of (− 1, 7) and (4, − 3) in the ratio 2:3.

#### Answer:

Let P(*x*, *y*) be the required point. Using the section formula, we obtain

Therefore, the point is (1, 3).

##### Video Solution for real numbers (Page: 167 , Q.No.: 1)

NCERT Solution for Class 10 math - real numbers 167 , Question 1

#### Page No 167:

#### Question 2:

Find the coordinates of the points of trisection of the line segment joining (4, − 1) and (− 2, − 3).

#### Answer:

Let P (*x*_{1}, *y*_{1}) and Q (*x*_{2}, *y*_{2}) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

Point Q divides AB internally in the ratio 2:1.

##### Video Solution for real numbers (Page: 167 , Q.No.: 2)

NCERT Solution for Class 10 math - real numbers 167 , Question 2

#### Page No 167:

#### Question 3:

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs the distance AD on the 2^{nd }line and posts a green flag. Preet runs the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

#### Answer:

It can be observed that Niharika posted the green flag at of the distance AD i.e., m from the starting point of 2^{nd} line. Therefore, the coordinates of this point G is (2, 25).

Similarly, Preet posted red flag at of the distance AD i.e., m from the starting point of 8^{th} line. Therefore, the coordinates of this point R are (8, 20).

Distance between these flags by using distance formula = GR

=

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (*x*, *y*).

Therefore, Rashmi should post her blue flag at 22.5m on 5^{th} line.

##### Video Solution for real numbers (Page: 167 , Q.No.: 3)

NCERT Solution for Class 10 math - real numbers 167 , Question 3

#### Page No 167:

#### Question 4:

Find the ratio in which the line segment joining the points (− 3, 10) and (6, − 8) is divided by (− 1, 6).

#### Answer:

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be* k *: 1.

##### Video Solution for real numbers (Page: 167 , Q.No.: 4)

NCERT Solution for Class 10 math - real numbers 167 , Question 4

#### Page No 167:

#### Question 5:

Find the ratio in which the line segment joining A (1, − 5) and B (− 4, 5) is divided by the *x*-axis. Also find the coordinates of the point of division.

#### Answer:

Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by *x-*axisbe.

Therefore, the coordinates of the point of division is .

We know that *y*-coordinate of any point on *x*-axis is 0.

Therefore, *x*-axis divides it in the ratio 1:1.

Division point =

##### Video Solution for real numbers (Page: 167 , Q.No.: 5)

NCERT Solution for Class 10 math - real numbers 167 , Question 5

#### Page No 167:

#### Question 6:

If (1, 2), (4, *y*), (*x*, 6) and (3, 5) are the vertices of a parallelogram taken in order, find *x* and *y*.

#### Answer:

Let (1, 2), (4, *y*), (*x*, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are

If O is the mid-point of BD, then the coordinates of O are

Since both the coordinates are of the same point O,

##### Video Solution for real numbers (Page: 167 , Q.No.: 6)

NCERT Solution for Class 10 math - real numbers 167 , Question 6

#### Page No 167:

#### Question 7:

Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, − 3) and B is (1, 4)

#### Answer:

Let the coordinates of point A be (*x*, *y*).

Mid-point of AB is (2, −3), which is the center of the circle.

##### Video Solution for real numbers (Page: 167 , Q.No.: 7)

NCERT Solution for Class 10 math - real numbers 167 , Question 7

#### Page No 167:

#### Question 8:

If A and B are (− 2, − 2) and (2, − 4), respectively, find the coordinates of P such that and P lies on the line segment AB.

#### Answer:

The coordinates of point A and B are (−2, −2) and (2, −4) respectively.

Since ,

Therefore, AP: PB = 3:4

Point P divides the line segment AB in the ratio 3:4.

##### Video Solution for real numbers (Page: 167 , Q.No.: 8)

NCERT Solution for Class 10 math - real numbers 167 , Question 8

#### Page No 167:

#### Question 9:

Find the coordinates of the points which divide the line segment joining A (− 2, 2) and B (2, 8) into four equal parts.

#### Answer:

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

##### Video Solution for real numbers (Page: 167 , Q.No.: 9)

NCERT Solution for Class 10 math - real numbers 167 , Question 9

#### Page No 167:

#### Question 10:

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (− 1, 4) and (− 2, −1) taken in order. [**Hint:** Area of a rhombus = (product of its diagonals)]

#### Answer:

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.

##### Video Solution for real numbers (Page: 167 , Q.No.: 10)

NCERT Solution for Class 10 math - real numbers 167 , Question 10

#### Page No 170:

#### Question 1:

Find the area of the triangle whose vertices are:

(i) (2, 3), (− 1, 0), (2, − 4) (ii) (− 5, − 1), (3, − 5), (5, 2)

#### Answer:

(i) Area of a triangle is given by

(ii)

##### Video Solution for real numbers (Page: 170 , Q.No.: 1)

NCERT Solution for Class 10 math - real numbers 170 , Question 1

#### Page No 170:

#### Question 2:

In each of the following find the value of ‘*k*’, for which the points are collinear.

(i) (7, − 2), (5, 1), (3, − *k*) (ii) (8, 1), (*k*, − 4), (2, − 5)

#### Answer:

(i) For collinear points, area of triangle formed by them is zero.

Therefore, for points (7, −2) (5, 1), and (3, *k*), area = 0

(ii) For collinear points, area of triangle formed by them is zero.

Therefore, for points (8, 1), (*k*, −4), and (2, −5), area = 0

##### Video Solution for real numbers (Page: 170 , Q.No.: 2)

NCERT Solution for Class 10 math - real numbers 170 , Question 2

#### Page No 170:

#### Question 3:

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, − 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

#### Answer:

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

##### Video Solution for real numbers (Page: 170 , Q.No.: 3)

NCERT Solution for Class 10 math - real numbers 170 , Question 3

#### Page No 170:

#### Question 4:

Find the area of the quadrilateral whose vertices, taken in order, are (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3)

#### Answer:

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

##### Video Solution for real numbers (Page: 170 , Q.No.: 4)

NCERT Solution for Class 10 math - real numbers 170 , Question 4

#### Page No 170:

#### Question 5:

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2)

#### Answer:

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

However, area cannot be negative. Therefore, area of ΔADC is 3 square units.

Clearly, median AD has divided ΔABC in two triangles of equal areas.

##### Video Solution for real numbers (Page: 170 , Q.No.: 5)

NCERT Solution for Class 10 math - real numbers 170 , Question 5

#### Page No 171:

#### Question 1:

Determine the ratio in which the line 2*x* + *y* − 4 = 0 divides the line segment joining the points A(2, − 2) and B(3, 7)

#### Answer:

Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio *k *: 1.

Coordinates of the point of division

This point also lies on 2*x* + *y* − 4 = 0

Therefore, the ratio in which the line 2*x* + *y* − 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.

##### Video Solution for real numbers (Page: 171 , Q.No.: 1)

NCERT Solution for Class 10 math - real numbers 171 , Question 1

#### Page No 171:

#### Question 2:

Find a relation between *x* and *y* if the points (*x*, *y*), (1, 2) and (7, 0) are collinear.

#### Answer:

If the given points are collinear, then the area of triangle formed by these points will be 0.

This is the required relation between* x *and *y*.

##### Video Solution for real numbers (Page: 171 , Q.No.: 2)

NCERT Solution for Class 10 math - real numbers 171 , Question 2

#### Page No 171:

#### Question 3:

Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).

#### Answer:

Let O (*x*, *y*) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

On adding equation (1) and (2), we obtain

10*y* = −20

*y* = −2

From equation (1), we obtain

3*x* − 2 = 7

3*x* = 9

*x* = 3

Therefore, the centre of the circle is (3, −2).

##### Video Solution for real numbers (Page: 171 , Q.No.: 3)

NCERT Solution for Class 10 math - real numbers 171 , Question 3

#### Page No 171:

#### Question 4:

The two opposite vertices of a square are (− 1, 2) and (3, 2). Find the coordinates of the other two vertices.

#### Answer:

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (*x*, *y*), (*x*_{1}, *y*_{1}) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

∴ AB = BC

We know that in a square, all interior angles are of 90°.

In ΔABC,

AB^{2} + BC^{2} = AC^{2}

⇒ 4 + *y*^{2} + 4 − 4*y* + 4 + *y*^{2} − 4*y* + 4 =16

⇒ 2*y*^{2} + 16 − 8 *y* =16

⇒ 2*y*^{2} − 8 *y* = 0

⇒ *y* (*y *− 4) = 0

⇒ *y* = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.

⇒ *y* + *y*_{1} = 4

If *y* = 0,

*y*_{1} = 4

If *y* = 4,

*y*_{1} = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

##### Video Solution for real numbers (Page: 171 , Q.No.: 4)

NCERT Solution for Class 10 math - real numbers 171 , Question 4

#### Page No 171:

#### Question 5:

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of Δ PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

#### Answer:

(i) Taking A as origin, we will take AD as *x*-axis and AB as *y*-axis. It can be observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

(ii) Taking C as origin, CB as *x*-axis, and CD as *y*-axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

It can be observed that the area of the triangle is same in both the cases.

##### Video Solution for real numbers (Page: 171 , Q.No.: 5)

NCERT Solution for Class 10 math - real numbers 171 , Question 5

#### Page No 171:

#### Question 6:

The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that. Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to

ratio of areas of two similar triangles)

#### Answer:

Given that,

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.

**Alternatively,**

We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.

Therefore, ratio between the areas of ΔADE and ΔABC =

##### Video Solution for real numbers (Page: 171 , Q.No.: 6)

NCERT Solution for Class 10 math - real numbers 171 , Question 6

#### Page No 171:

#### Question 7:

Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP: PD = 2:1

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR: RF = 2:1.

(iv) What do you observe?

(v) If A(*x*_{1}, *y*_{1}), B(*x*_{2}, *y*_{2}), and C(*x*_{3}, *y*_{3}) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.

#### Answer:

(i) Median AD of the triangle will divide the side BC in two equal parts.

Therefore, D is the mid-point of side BC.

(ii) Point P divides the side AD in a ratio 2:1.

(iii) Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC.

Point Q divides the side BE in a ratio 2:1.

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB.

Point R divides the side CF in a ratio 2:1.

(iv) It can be observed that the coordinates of point P, Q, R are the same.

Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.

(v) Consider a triangle, ΔABC, having its vertices as A(*x*_{1}, *y*_{1}), B(*x*_{2}, *y*_{2}), and C(*x*_{3},

*y*_{3}).

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC.

Let the centroid of this triangle be O.

Point O divides the side AD in a ratio 2:1.

##### Video Solution for real numbers (Page: 171 , Q.No.: 7)

NCERT Solution for Class 10 math - real numbers 171 , Question 7

#### Page No 172:

#### Question 8:

ABCD is a rectangle formed by the points A (− 1, − 1), B (− 1, 4), C (5, 4) and D (5, − 1). P, Q, R and S are the mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS is a square? a rectangle? or a rhombus? Justify your answer.

#### Answer:

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.

##### Video Solution for real numbers (Page: 172 , Q.No.: 8)

NCERT Solution for Class 10 math - real numbers 172 , Question 8

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