CBSE Class 10 Maths Chapter 7 Coordinate Geometry NCERT Solutions

In the Cartesian coordinate system, a point P(x, y) has coordinates where 'x' represents the perpendicular distance from the y-axis and 'y' represents the perpendicular distance from the x-axis.

For the point P(2, 3), the x-coordinate is 2 and the y-coordinate is 3.

Therefore, the perpendicular distance of the point P(2, 3) from the x-axis is its y-coordinate, which is 3.

Answer: (B) 3

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:

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

Given the points are A(0, 6) and B(0, -2).

Let $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (0, -2)$.

Substituting these values into the distance formula:

$$d\; =\; \backslash sqrt\{(0\; -\; 0)^2\; +\; (-2\; -\; 6)^2\}$$

$$d\; =\; \backslash sqrt\{(0)^2\; +\; (-8)^2\}$$

$$d\; =\; \backslash sqrt\{0\; +\; 64\}$$

$$d\; =\; \backslash sqrt\{64\}$$

$d\; =\; 8$ units

Thus, the distance between points A and B is 8 units.

Answer: (B) 8

The origin has coordinates (0, 0). We use the distance formula to find the distance between the point P(-6, 8) and the origin (0, 0).

The distance formula is:

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

Let $(x_1, y_1) = (-6, 8)$ and $(x_2, y_2) = (0, 0)$.

Substituting these values:

$$d\; =\; \backslash sqrt\{(0\; -\; (-6))^2\; +\; (0\; -\; 8)^2\}$$

$$d\; =\; \backslash sqrt\{(6)^2\; +\; (-8)^2\}$$

$$d\; =\; \backslash sqrt\{36\; +\; 64\}$$

$$d\; =\; \backslash sqrt\{100\}$$

$d\; =\; 10$ units

Therefore, the distance of the point P(-6, 8) from the origin is 10 units.

Answer: (C) 10

We apply the distance formula to find the distance between the two given points.

The distance formula is:

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

Let the points be $(x_1, y_1) = (0, 5)$ and $(x_2, y_2) = (-5, 0)$.

Substituting the coordinates into the formula:

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

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

$$d\; =\; \backslash sqrt\{25\; +\; 25\}$$

$$d\; =\; \backslash sqrt\{50\}$$

To simplify $\sqrt{50}$, we find the prime factors: $50 = 2 \times 5 \times 5 = 2 \times 5^2$.

$d\; =\; \backslash sqrt\{25\; \backslash times\; 2\}\; =\; \backslash sqrt\{25\}\; \backslash times\; \backslash sqrt\{2\}\; =\; 5\backslash sqrt\{2\}$ units

Therefore, the distance between the points (0, 5) and (-5, 0) is $5\sqrt{2}$ units.

Answer: (B) $5\sqrt{2}$