Coordinate Geometry - Exercise 7.1

Here, $(x_1, y_1) = (3, 8)$ and $(x_2, y_2) = (6, 4)$.
Using Distance Formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(6 - 3)^2 + (4 - 8)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ units}$$

Here, $(x_1, y_1) = (-6, 8)$ and $(x_2, y_2) = (-2, 4)$.
$$d = \sqrt{(-2 - (-6))^2 + (4 - 8)^2} = \sqrt{(-2 + 6)^2 + (-4)^2} = \sqrt{4^2 + 16} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\text{ units}$$

Here, $(x_1, y_1) = (x, y)$ and $(x_2, y_2) = (-x, -y)$.
$$d = \sqrt{(-x - x)^2 + (-y - y)^2} = \sqrt{(-2x)^2 + (-2y)^2} = \sqrt{4x^2 + 4y^2} = \sqrt{4(x^2 + y^2)} = 2\sqrt{x^2 + y^2}\text{ units}$$

Let origin $O(0, 0)$ and $P(12, 5)$.
The distance of any point $(x, y)$ from origin is given by $\sqrt{x^2 + y^2}$: $$d = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{12^2 + 5^2}$$ $$d = \sqrt{144 + 25} = \sqrt{169} = 13\text{ units}$$

Let the points be $A(1, 4)$, $B(3, 2)$, and $C(-3, 10)$.
$$AB = \sqrt{(3 - 1)^2 + (2 - 4)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$ $$BC = \sqrt{(-3 - 3)^2 + (10 - 2)^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ $$AC = \sqrt{(-3 - 1)^2 + (10 - 4)^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$

For three points to be collinear, the sum of the lengths of any two segments must equal the third segment.
Here: $$AB + AC = 2\sqrt{2} + 2\sqrt{13} \approx 2(1.414) + 2(3.606) = 2.828 + 7.212 = 10.04 \neq 10$$ Since $AB + AC \neq BC$, and no sum of two distances equals the third distance: The points $A, B$, and $C$ do not lie on a single straight line.

Let the vertices be $A(2, 2)$, $B(5, 6)$, and $C(8, 2)$.
$$AB = \sqrt{(5 - 2)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ $$BC = \sqrt{(8 - 5)^2 + (2 - 6)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ $$AC = \sqrt{(8 - 2)^2 + (2 - 2)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6$$

We observe that $AB = BC = 5$, while $AC = 6$.
Since two sides of $\triangle ABC$ are equal in length ($AB = BC$), the triangle is isosceles.

$$AB = \sqrt{(4 - 1)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ $$BC = \sqrt{(7 - 4)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ $$CD = \sqrt{(4 - 7)^2 + (-2 - 1)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ $$DA = \sqrt{(1 - 4)^2 + (1 - (-2))^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ All four sides are equal: $AB = BC = CD = DA = 3\sqrt{2}\text{ units}$.

$$AC = \sqrt{(7 - 1)^2 + (1 - 1)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6$$ $$BD = \sqrt{(4 - 4)^2 + (-2 - 4)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6$$ Both diagonals are equal: $AC = BD = 6\text{ units}$.

Since all four sides are equal and both diagonals are equal, quadrilateral $ABCD$ is a square. Therefore, Ram is correct.

Sides:
$AB = \sqrt{(2 - 0)^2 + (0 - 0)^2} = 2$
$BC = \sqrt{(2 - 2)^2 + (2 - 0)^2} = 2$
$CD = \sqrt{(0 - 2)^2 + (2 - 2)^2} = 2$
$DA = \sqrt{(0 - 0)^2 + (0 - 2)^2} = 2$
Diagonals:
$AC = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2}$
$BD = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2}$
All 4 sides are equal ($2$) and both diagonals are equal ($2\sqrt{2}$), hence $ABCD$ is a Square.

Distance $AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$
Distance $AC = \sqrt{(0 - (-3))^2 + (3 - 5)^2} = \sqrt{9 + 4} = \sqrt{13}$
Distance $CB = \sqrt{(3 - 0)^2 + (1 - 3)^2} = \sqrt{9 + 4} = \sqrt{13}$
Notice that: $$AC + CB = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB$$ This means points $A, C$, and $B$ are collinear (point $C$ lies on line segment $AB$). Therefore, these 4 points do not form a closed four-sided polygon.
Result: No quadrilateral is formed.

Sides:
$AB = \sqrt{(4 - 1)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$
$BC = \sqrt{(3 - 4)^2 + (5 - 2)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$
$CD = \sqrt{(0 - 3)^2 + (4 - 5)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}$
$DA = \sqrt{(1 - 0)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$
Diagonals:
$AC = \sqrt{(3 - 1)^2 + (5 - 1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20}$
$BD = \sqrt{(0 - 4)^2 + (4 - 2)^2} = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}$
Since all four sides are equal ($\sqrt{10}$) and diagonals are equal ($\sqrt{20}$), $ABCD$ is a Square.

Since the required point lies on the $x$-axis, its $y$-coordinate is $0$.
Let the point be $P(x, 0)$, and let the given points be $A(3, -4)$ and $B(-5, 6)$.
Given that $P$ is equidistant from $A$ and $B$: $$PA = PB \implies PA^2 = PB^2$$

$$(x - 3)^2 + (0 - (-4))^2 = (x - (-5))^2 + (0 - 6)^2$$ $$(x - 3)^2 + 16 = (x + 5)^2 + 36$$ $$x^2 - 6x + 9 + 16 = x^2 + 10x + 25 + 36$$ $$x^2 - 6x + 25 = x^2 + 10x + 61$$ Canceling $x^2$ from both sides: $$-6x + 25 = 10x + 61$$ $$-6x - 10x = 61 - 25$$ $$-16x = 36$$ $$x = -\frac{36}{16} = -\frac{9}{4} = -2.25$$

Distance $PQ = 10 \implies PQ^2 = 100$.
Using coordinates $P(3, -2)$ and $Q(9, y)$: $$(9 - 3)^2 + (y - (-2))^2 = 100$$ $$6^2 + (y + 2)^2 = 100$$ $$36 + (y + 2)^2 = 100$$

$$(y + 2)^2 = 100 - 36 = 64$$ Taking square root on both sides: $$y + 2 = \pm 8$$ Case 1: $y + 2 = 8 \implies y = 8 - 2 = 6$
Case 2: $y + 2 = -8 \implies y = -8 - 2 = -10$

Given $QP = QR \implies QP^2 = QR^2$.
$$(6 - 0)^2 + (-4 - 1)^2 = (x - 0)^2 + (7 - 1)^2$$ $$6^2 + (-5)^2 = x^2 + 6^2$$ $$36 + 25 = x^2 + 36$$ $$x^2 = 25 \implies x = \pm 5$$

$$QR = \sqrt{(x - 0)^2 + (7 - 1)^2} = \sqrt{(\pm 5)^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}\text{ units}$$

When $x = 5$: Point $R$ is $(5, 7)$ and $P(6, -4)$: $$PR = \sqrt{(5 - 6)^2 + (7 - (-4))^2} = \sqrt{(-1)^2 + 11^2} = \sqrt{1 + 121} = \sqrt{122}\text{ units}$$ When $x = -5$: Point $R$ is $(-5, 7)$ and $P(6, -4)$: $$PR = \sqrt{(-5 - 6)^2 + (7 - (-4))^2} = \sqrt{(-11)^2 + 11^2} = \sqrt{121 + 121} = \sqrt{242} = 11\sqrt{2}\text{ units}$$

Let $P(x, y)$, $A(4, 5)$ and $B(-4, 3)$.
Given $PA = PB \implies PA^2 = PB^2$.
$$(x - 4)^2 + (y - 5)^2 = (x - (-4))^2 + (y - 3)^2$$ $$(x - 4)^2 + (y - 5)^2 = (x + 4)^2 + (y - 3)^2$$

$$(x^2 - 8x + 16) + (y^2 - 10y + 25) = (x^2 + 8x + 16) + (y^2 - 6y + 9)$$ $$x^2 + y^2 - 8x - 10y + 41 = x^2 + y^2 + 8x - 6y + 25$$ Canceling $x^2 + y^2$ on both sides: $$-8x - 10y + 41 = 8x - 6y + 25$$ Transposing all terms to one side: $$8x + 8x - 6y + 10y + 25 - 41 = 0$$ $$16x + 4y - 16 = 0$$ Dividing the entire equation by $4$: $$4x + y - 4 = 0$$