NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1#

Using the distance formula, we find the distance $d$:

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

$$= \sqrt{4 + 4}$$

$$= \sqrt{8}$$

$$= 2\sqrt{2}$$

∴ The distance is $\mathbf{2\sqrt{2}}$ units.

Using the distance formula, we find the distance $d$:

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

$$= \sqrt{16 + 16}$$

$$= \sqrt{32}$$

$$= 4\sqrt{2}$$

∴ The distance is $\mathbf{4\sqrt{2}}$ units.

Using the distance formula, we find the distance $d$:

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

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

$$= 2\sqrt{a^2 + b^2}$$

∴ The distance is $\mathbf{2\sqrt{a^2 + b^2}}$ units.

Using the distance formula, we find the distance $d$:

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

$$= \sqrt{1296 + 225}$$

$$= \sqrt{1521}$$

$$= 39$$

∴ The distance between the points is 39 units.

Yes. Town A is at (0, 0) and town B is at (36, 15), with distances in km, so the distance between the two towns is 39 km.

Let the points be $A(1, 5)$, $B(2, 3)$ and $C(-2, -11)$.

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

$$BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{212} = 2\sqrt{53}$$

$$CA = \sqrt{(1-(-2))^2 + (5-(-11))^2} = \sqrt{265}$$

CA is the longest, so we compare $AB + BC$ with $CA$:

$$\Rightarrow \sqrt{5} + 2\sqrt{53} \neq \sqrt{265}$$

(Approximately, $16.80 \neq 16.28$.)

∴ $AB + BC \neq CA$, so the points are not collinear.

Let the points be $A(5, -2)$, $B(6, 4)$ and $C(7, -2)$.

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

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

$$CA = \sqrt{(5-7)^2 + (-2-(-2))^2} = \sqrt{4} = 2$$$$\Rightarrow AB = BC = \sqrt{37}$$

∴ Two sides are equal, so the points are the vertices of an isosceles triangle.

From the figure, the points are $A(3, 4)$, $B(6, 7)$, $C(9, 4)$ and $D(6, 1)$.

Sides:

$$AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{18} = 3\sqrt{2}$$

$$BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{18} = 3\sqrt{2}$$

$$CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{18} = 3\sqrt{2}$$

$$DA = \sqrt{(3-6)^2 + (4-1)^2} = \sqrt{18} = 3\sqrt{2}$$$$\Rightarrow AB = BC = CD = DA = 3\sqrt{2}$$

Diagonals:

$$AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{36} = 6$$

$$BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{36} = 6$$$$\Rightarrow AC = BD = 6$$

∴ All four sides are equal and the diagonals are equal, so ABCD is a square. Champa is correct.

Let the points be $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$ and $D(-3, 0)$.

Sides:

$$AB = \sqrt{(1+1)^2 + (0+2)^2} = \sqrt{8} = 2\sqrt{2}$$

$$BC = \sqrt{(-1-1)^2 + (2-0)^2} = \sqrt{8} = 2\sqrt{2}$$

$$CD = \sqrt{(-3+1)^2 + (0-2)^2} = \sqrt{8} = 2\sqrt{2}$$

$$DA = \sqrt{(-1+3)^2 + (-2-0)^2} = \sqrt{8} = 2\sqrt{2}$$$$\Rightarrow AB = BC = CD = DA = 2\sqrt{2}$$

Diagonals:

$$AC = \sqrt{(-1+1)^2 + (2+2)^2} = \sqrt{16} = 4$$

$$BD = \sqrt{(-3-1)^2 + (0-0)^2} = \sqrt{16} = 4$$$$\Rightarrow AC = BD = 4$$

∴ All four sides are equal and the diagonals are equal, so ABCD is a square.

Let the points be $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$ and $D(-1, -4)$.

Check whether A, C and B lie on one line:

$$AC = \sqrt{(0+3)^2 + (3-5)^2} = \sqrt{13}$$

$$CB = \sqrt{(3-0)^2 + (1-3)^2} = \sqrt{13}$$

$$AB = \sqrt{(3+3)^2 + (1-5)^2} = \sqrt{52} = 2\sqrt{13}$$$$\Rightarrow AC + CB = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB$$

∴ $AC + CB = AB$, so A, C and B are collinear. Three of the points lie on one line, so the points do not form a quadrilateral.

Let the points be $A(4, 5)$, $B(7, 6)$, $C(4, 3)$ and $D(1, 2)$.

Sides:

$$AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{10}$$

$$BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{18} = 3\sqrt{2}$$

$$CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{10}$$

$$DA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{18} = 3\sqrt{2}$$$$\Rightarrow AB = CD = \sqrt{10} \text{ and } BC = DA = 3\sqrt{2}$$

Diagonals:

$$AC = \sqrt{(4-4)^2 + (3-5)^2} = \sqrt{4} = 2$$

$$BD = \sqrt{(1-7)^2 + (2-6)^2} = \sqrt{52} = 2\sqrt{13}$$$$\Rightarrow AC \neq BD$$

∴ The opposite sides are equal but the diagonals are not equal, so ABCD is a parallelogram.

Let the point on the x-axis be $P(x, 0)$, and let $A(2, -5)$ and $B(-2, 9)$.

P is equidistant from A and B, so:

$$PA^2 = PB^2$$

$$\Rightarrow (x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2$$

$$\Rightarrow x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$$

$$\Rightarrow -4x + 29 = 4x + 85$$

$$\Rightarrow -8x = 56$$

$$\Rightarrow x = -7$$

∴ The required point is $\mathbf{(-7, 0)}$.

The distance PQ is 10 units, so:

$$\sqrt{(10-2)^2 + (y+3)^2} = 10$$

$$\Rightarrow 64 + (y+3)^2 = 100$$

$$\Rightarrow (y+3)^2 = 36$$

$$\Rightarrow y + 3 = \pm 6$$

$$\Rightarrow y = 3 \text{ or } y = -9$$

∴ $y = \mathbf{3}$ or $y = \mathbf{-9}$

Q is equidistant from P and R, so:

$$QP^2 = QR^2$$

$$\Rightarrow (5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2$$

$$\Rightarrow 41 = x^2 + 25$$

$$\Rightarrow x^2 = 16$$

$$\Rightarrow x = \pm 4$$

Case 1: $x = 4$, so the point is $R(4, 6)$.

$$QR = \sqrt{(4-0)^2 + (6-1)^2} = \sqrt{41}$$

$$PR = \sqrt{(4-5)^2 + (6+3)^2} = \sqrt{82}$$

Case 2: $x = -4$, so the point is $R(-4, 6)$.

$$QR = \sqrt{(-4-0)^2 + (6-1)^2} = \sqrt{41}$$

$$PR = \sqrt{(-4-5)^2 + (6+3)^2} = \sqrt{162} = 9\sqrt{2}$$

∴ $x = \mathbf{4}$ or $x = \mathbf{-4}$, $QR = \mathbf{\sqrt{41}}$, and $PR = \mathbf{\sqrt{82}}$ or $PR = \mathbf{9\sqrt{2}}$.

Let $P(x, y)$, $A(3, 6)$ and $B(-3, 4)$.

P is equidistant from A and B, so:

$$PA^2 = PB^2$$

$$\Rightarrow (x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2$$

$$\Rightarrow x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$$

$$\Rightarrow -6x - 12y + 45 = 6x - 8y + 25$$

$$\Rightarrow -12x - 4y + 20 = 0$$

$$\Rightarrow 3x + y - 5 = 0$$

∴ The required relation is $\mathbf{3x + y - 5 = 0}$.