CBSE Class 10 Maths Chapter 7 Coordinate Geometry NCERT Solutions

We use the distance formula to find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), which is given by:

$$d\; =\; \backslash sqrt\{(x\_2\; -\; x\_1)^2\; +\; (y\_2\; -\; y\_1)^2\}$$

1. For the points (2, 3) and (4, 1):

   Let \((x_1, y_1) = (2, 3)\) and \((x_2, y_2) = (4, 1)\).

   Distance \(d = \sqrt{(4 - 2)^2 + (1 - 3)^2}

   $d\; =\; \backslash sqrt\{(2)^2\; +\; (-2)^2\}$

   $d\; =\; \backslash sqrt\{4\; +\; 4\}$

   $d\; =\; \backslash sqrt\{8\}$

   Simplifying the radical, we get \(d = 2\sqrt{2}\).

2. For the points (-5, 7) and (-1, 3):

   Let \((x_1, y_1) = (-5, 7)\) and \((x_2, y_2) = (-1, 3)\).

   Distance \(d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}

   $d\; =\; \backslash sqrt\{(-1\; +\; 5)^2\; +\; (-4)^2\}$

   $d\; =\; \backslash sqrt\{(4)^2\; +\; (-4)^2\}$

   $d\; =\; \backslash sqrt\{16\; +\; 16\}$

   $d\; =\; \backslash sqrt\{32\}$

   Simplifying the radical, we get \(d = 4\sqrt{2}\).

3. For the points (a, b) and (-a, -b):

   Let \((x_1, y_1) = (a, b)\) and \((x_2, y_2) = (-a, -b)\).

   Distance \(d = \sqrt{(-a - a)^2 + (-b - b)^2}

   $d\; =\; \backslash sqrt\{(-2a)^2\; +\; (-2b)^2\}$

   $d\; =\; \backslash sqrt\{4a^2\; +\; 4b^2\}$

   $d\; =\; \backslash sqrt\{4(a^2\; +\; b^2)\}$

   Simplifying the radical, we get \(d = 2\sqrt{a^2 + b^2}\).

To find the distance between the points (0, 0) and (36, 15), we use the distance formula.

Let \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (36, 15)\).

Distance \(d = \sqrt{(36 - 0)^2 + (15 - 0)^2}

$$d\; =\; \backslash sqrt\{36^2\; +\; 15^2\}$$

$d\; =\; \backslash sqrt\{1296\; +\; 225\}$

$d\; =\; \backslash sqrt\{1521\}$

Calculating the square root, we find \(d = 39\).

Yes, we can find the distance between the two towns A and B. If we assume town A is at the origin \((0, 0)\), then town B can be represented by the coordinates (36, 15) relative to town A. Therefore, the distance between town A and town B is 39 km, as calculated above.

Let the given points be A = (1, 5), B = (2, 3), and C = (-2, -11). To determine if these points are collinear, we need to calculate the distances between each pair of points and check if the sum of the lengths of any two segments equals the length of the third segment.

Calculate the distance AB:

$AB\; =\; \backslash sqrt\{(2\; -\; 1)^2\; +\; (3\; -\; 5)^2\}\; =\; \backslash sqrt\{(1)^2\; +\; (-2)^2\}\; =\; \backslash sqrt\{1\; +\; 4\}\; =\; \backslash sqrt\{5\}$

Calculate the distance BC:

$$BC\; =\; \backslash sqrt\{(-2\; -\; 2)^2\; +\; (-11\; -\; 3)^2\}\; =\; \backslash sqrt\{(-4)^2\; +\; (-14)^2\}\; =\; \backslash sqrt\{16\; +\; 196\}\; =\; \backslash sqrt\{212\}$$

Calculate the distance CA:

$CA\; =\; \backslash sqrt\{(1\; -\; (-2))^2\; +\; (5\; -\; (-11))^2\}\; =\; \backslash sqrt\{(1\; +\; 2)^2\; +\; (5\; +\; 11)^2\}\; =\; \backslash sqrt\{(3)^2\; +\; (16)^2\}\; =\; \backslash sqrt\{9\; +\; 256\}\; =\; \backslash sqrt\{265\}$

Now, we check if the sum of any two distances equals the third:

Is \(\sqrt{5} + \sqrt{212} = \sqrt{265}\)? No.

Is \(\sqrt{5} + \sqrt{265} = \sqrt{212}\)? No.

Is \(\sqrt{212} + \sqrt{265} = \sqrt{5}\)? No.

Since the sum of the lengths of any two line segments is not equal to the length of the third line segment (i.e., \(AB + BC \neq CA\), \(AB + CA \neq BC\), and \(BC + CA \neq AB\)), the points (1, 5), (2, 3), and (-2, -11) are not collinear.

Let the given points be A = (5, -2), B = (6, 4), and C = (7, -2). To determine if these points form an isosceles triangle, we need to calculate the lengths of the sides AB, BC, and CA and check if at least two sides have equal lengths.

Calculate the distance AB:

$AB\; =\; \backslash sqrt\{(6\; -\; 5)^2\; +\; (4\; -\; (-2))^2\}\; =\; \backslash sqrt\{(1)^2\; +\; (4\; +\; 2)^2\}\; =\; \backslash sqrt\{1^2\; +\; 6^2\}\; =\; \backslash sqrt\{1\; +\; 36\}\; =\; \backslash sqrt\{37\}$

Calculate the distance BC:

$BC\; =\; \backslash sqrt\{(7\; -\; 6)^2\; +\; (-2\; -\; 4)^2\}\; =\; \backslash sqrt\{(1)^2\; +\; (-6)^2\}\; =\; \backslash sqrt\{1\; +\; 36\}\; =\; \backslash sqrt\{37\}$

Calculate the distance CA:

$CA\; =\; \backslash sqrt\{(5\; -\; 7)^2\; +\; (-2\; -\; (-2))^2\}\; =\; \backslash sqrt\{(-2)^2\; +\; (-2\; +\; 2)^2\}\; =\; \backslash sqrt\{(-2)^2\; +\; 0^2\}\; =\; \backslash sqrt\{4\; +\; 0\}\; =\; \backslash sqrt\{4\}\; =\; 2$

We observe that the length of side AB is \(\sqrt{37}\) and the length of side BC is also \(\sqrt{37}\). Since two sides of the triangle (AB and BC) have equal lengths, the points (5, -2), (6, 4), and (7, -2) form the vertices of an isosceles triangle.