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 \(d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}\).

   Simplifying the square root, \(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\).

   Therefore, the distance is \(2\sqrt{2}\) units.

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

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

   Distance \(d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} = \sqrt{(-1 + 5)^2 + (-4)^2} = \sqrt{(4)^2 + (-4)^2}\).

   $d\; =\; \backslash sqrt\{16\; +\; 16\}\; =\; \backslash sqrt\{32\}\backslash ).$

   Simplifying the square root, \(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\).

   Therefore, the distance is \(4\sqrt{2}\) units.

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

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

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

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

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

   Therefore, the distance is \(2\sqrt{a^2 + b^2}\) units.

First, let's find the distance between the points (0, 0) and (36, 15) using the distance formula.

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

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

$d\; =\; \backslash sqrt\{1296\; +\; 225\}\; =\; \backslash sqrt\{1521\}\backslash ).$

To find the square root of 1521, we can perform prime factorization or recognize it as a perfect square. \(39 \times 39 = 1521\).

So, \(d = 39\) units.

Yes, we can now 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 the same as the distance calculated above, which is 39 km (assuming the units in the problem context are kilometers).

To determine if the points are collinear, we need to check if the sum of the distances between any two pairs of points is equal to the distance between the remaining pair. Let the points be A = (1, 5), B = (2, 3), and C = (-2, -11).

We calculate the distances AB, BC, and CA using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Distance AB:

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

Distance BC:

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

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 \(AB + BC = CA\)? \(\sqrt{5} + \sqrt{212} \neq \sqrt{265}\).

Is \(AB + CA = BC\)? \(\sqrt{5} + \sqrt{265} \neq \sqrt{212}\).

Is \(BC + CA = AB\)? \(\sqrt{212} + \sqrt{265} \neq \sqrt{5}\).

Since the sum of the lengths of any two line segments is not equal to the length of the third line segment, the points (1, 5), (2, 3), and (-2, -11) are not collinear.

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

We calculate the distances AB, BC, and CA using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Distance AB:

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

Distance BC:

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

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