NCERT Solutions for Class 10 Math Chapter 7 - Coordinate Geometry

Answer:

(i) Distance between the two points is given by

(ii) Distance between is given by

(iii) Distance between is given by

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.

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.

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.

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

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.

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).

Answer:

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

Answer:

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

When point R is (4, 6),

When point R is (−4, 6),

Answer:

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

Answer:

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

Therefore, the point is (1, 3).

Answer:

Let P (x₁, y₁) and Q (x₂, y₂) 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.

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.

Answer:

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

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 =

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,

Answer:

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

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

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.

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.

Answer:

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

Answer:

(i) Area of a triangle is given by

(ii)

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

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

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.

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.

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 2x + y − 4 = 0

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

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.

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

10y = −20

y = −2

From equation (1), we obtain

3x − 2 = 7

3x = 9

x = 3

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

Answer:

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x₁, y₁) 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,

AB² + BC² = AC²

⇒ 4 + y² + 4 − 4y + 4 + y² − 4y + 4 =16

⇒ 2y² + 16 − 8 y =16

⇒ 2y² − 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 + y₁ = 4

If y = 0,

y₁ = 4

If y = 4,

y₁ = 0

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

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.

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 =

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(x₁, y₁), B(x₂, y₂), and C(x₃,

y₃).

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.

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.