CBSE Class 10 Maths Chapter 7 Coordinate Geometry NCERT Solutions

We use the distance formula, which states that the distance between two points $(x\_1,\; y\_1)$ and $(x\_2,\; y\_2)$ 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):

   $x\_1\; =\; 2,\; y\_1\; =\; 3$ and $x\_2\; =\; 4,\; y\_2\; =\; 1$.

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

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

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

   $d\; =\; \backslash sqrt\{8\}\; =\; 2\backslash sqrt\{2\}$

   The distance between (2, 3) and (4, 1) is $2\backslash sqrt\{2\}$ units.

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

   $x\_1\; =\; -5,\; y\_1\; =\; 7$ and $x\_2\; =\; -1,\; y\_2\; =\; 3$.

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

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

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

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

   $d\; =\; \backslash sqrt\{32\}\; =\; 4\backslash sqrt\{2\}$

   The distance between $(-5,\; 7)$ and $(-1,\; 3)$ is $4\backslash sqrt\{2\}$ units.

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

   $x\_1\; =\; a,\; y\_1\; =\; b$ and $x\_2\; =\; -a,\; y\_2\; =\; -b$.

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

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

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

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

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

   The distance between $(a,\; b)$ and $(-a,\; -b)$ is $2\backslash sqrt\{a^2\; +\; b^2\}$ units.

First, we 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 $d\; =\; \backslash sqrt\{(36-0)^2\; +\; (15-0)^2\}$

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

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

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

$d\; =\; 39$

The distance between the points (0, 0) and (36, 15) is 39 units.

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 the same as the distance calculated above, which is 39 km (assuming the units are kilometers as implied by the context of towns).

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 check if the sum of the lengths of any two line segments formed by these points is equal to the length of the third line segment. We will use the distance formula $d\; =\; \backslash sqrt\{(x\_2-x\_1)^2\; +\; (y\_2-y\_1)^2\}$.

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 distance:

$AB\; +\; BC\; =\; \backslash sqrt\{5\}\; +\; \backslash sqrt\{212\}\; \backslash approx\; 2.24\; +\; 14.56\; =\; 16.80$

$CA\; =\; \backslash sqrt\{265\}\; \backslash approx\; 16.28$

Since $AB\; +\; BC\; \backslash neq\; CA$, and similarly $AB\; +\; CA\; \backslash neq\; BC$ and $BC\; +\; CA\; \backslash neq\; AB$, the points do not lie on the same straight line.

Therefore, 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 check if these points form an isosceles triangle, we need to calculate the lengths of the three sides AB, BC, and CA using the distance formula $d\; =\; \backslash sqrt\{(x\_2-x\_1)^2\; +\; (y\_2-y\_1)^2\}$ and see if any two sides are equal.

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 $AB\; =\; \backslash sqrt\{37\}$ and $BC\; =\; \backslash sqrt\{37\}$. Since two sides of the triangle (AB and BC) have equal lengths, the triangle formed by the points (5, -2), (6, 4), and (7, -2) is an isosceles triangle.