Distance Formula Questions in English

(C) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(-3, 4)$ and $O(0, 0)$.
Substituting the values into the formula:
$OA = \sqrt{(0 - (-3))^2 + (0 - 4)^2}$
$OA = \sqrt{(3)^2 + (-4)^2}$
$OA = \sqrt{9 + 16}$
$OA = \sqrt{25}$
$OA = 5$ units.

(D) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(2, 3)$ and $B(-3, -9)$.
Here,$x_1 = 2, y_1 = 3, x_2 = -3, y_2 = -9$.
Substituting these values in the formula:
$AB = \sqrt{(-3 - 2)^2 + (-9 - 3)^2}$
$AB = \sqrt{(-5)^2 + (-12)^2}$
$AB = \sqrt{25 + 144}$
$AB = \sqrt{169}$
$AB = 13$.
Thus,the distance between the points is $13$ units.

(B) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $P(3, 2)$,$Q(7, k)$,and $d(P, Q) = 5$.
Squaring both sides,we get $PQ^2 = 5^2 = 25$.
Using the distance formula: $(7 - 3)^2 + (k - 2)^2 = 25$.
$(4)^2 + (k - 2)^2 = 25$.
$16 + (k - 2)^2 = 25$.
$(k - 2)^2 = 25 - 16 = 9$.
Taking the square root on both sides: $k - 2 = \pm 3$.
Case $1$: $k - 2 = 3 \implies k = 5$.
Case $2$: $k - 2 = -3 \implies k = -1$.
Since $5$ is one of the options provided,the correct value is $k = 5$.

(A) Let the point on the $X-$ axis be $P(x, 0)$.
Since $P$ is equidistant from $A(-1, 2)$ and $B(5, 4)$,we have $PA = PB$,which implies $PA^2 = PB^2$.
Using the distance formula,we get:
$[x - (-1)]^2 + (0 - 2)^2 = (x - 5)^2 + (0 - 4)^2$
$(x + 1)^2 + 4 = (x - 5)^2 + 16$
$x^2 + 2x + 1 + 4 = x^2 - 10x + 25 + 16$
$2x + 5 = -10x + 41$
$12x = 36$
$x = 3$
Therefore,the required point on the $X-$ axis is $(3, 0)$.

(B) Suppose the required point on the $Y-$ axis is $R(0, y)$.
Since the point is equidistant from $P(3, 2)$ and $Q(-1, -5)$,we have $PR = QR$.
Squaring both sides,we get $PR^2 = QR^2$.
Using the distance formula,$(3 - 0)^2 + (2 - y)^2 = (-1 - 0)^2 + (-5 - y)^2$.
Expanding the squares,$9 + (4 - 4y + y^2) = 1 + (25 + 10y + y^2)$.
Simplifying the equation,$13 - 4y + y^2 = 26 + 10y + y^2$.
Subtracting $y^2$ from both sides,$13 - 4y = 26 + 10y$.
Rearranging the terms,$-13 = 14y$.
Therefore,$y = -\frac{13}{14}$.
The required point on the $Y-$ axis is $\left(0, -\frac{13}{14}\right)$.

(A) To determine the type of triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
$1$. Length of $PQ$: $PQ^2 = (2-2)^2 + (5-3)^2 = 0^2 + 2^2 = 4$. So,$PQ = 2$.
$2$. Length of $QR$: $QR^2 = (2+\sqrt{3}-2)^2 + (4-5)^2 = (\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$. So,$QR = 2$.
$3$. Length of $PR$: $PR^2 = (2+\sqrt{3}-2)^2 + (4-3)^2 = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4$. So,$PR = 2$.
Since $PQ = QR = PR = 2$,all three sides are equal.
Therefore,$\Delta PQR$ is an equilateral triangle.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(4, 7)$ and $B(7, 3)$.
Substituting the values into the formula:
$AB = \sqrt{(7 - 4)^2 + (3 - 7)^2}$
$AB = \sqrt{(3)^2 + (-4)^2}$
$AB = \sqrt{9 + 16}$
$AB = \sqrt{25}$
$AB = 5$.

(D) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $P(2, -3)$ and $Q(7, 9)$.
Substituting the values into the formula:
$PQ = \sqrt{(7 - 2)^2 + (9 - (-3))^2}$
$PQ = \sqrt{(5)^2 + (12)^2}$
$PQ = \sqrt{25 + 144}$
$PQ = \sqrt{169}$
$PQ = 13$ units.

(A) Let the coordinates of point $P$ on the $X-$axis be $(x, 0)$.
Given that the distance $PA = 13$ units.
Using the distance formula,$PA^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
$13^2 = (x - 11)^2 + (0 - 12)^2$
$169 = (x - 11)^2 + (-12)^2$
$169 = (x - 11)^2 + 144$
$(x - 11)^2 = 169 - 144$
$(x - 11)^2 = 25$
Taking the square root on both sides,$x - 11 = \pm 5$.
Case $1$: $x - 11 = 5 \implies x = 16$.
Case $2$: $x - 11 = -5 \implies x = 6$.
Therefore,the coordinates of $P$ are $(16, 0)$ or $(6, 0)$.

(C) The coordinates of point $A$ are $(a \cos \theta, a \sin \theta)$ and the origin $O$ is $(0, 0)$.
Using the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the values,we get $d = \sqrt{(a \cos \theta - 0)^2 + (a \sin \theta - 0)^2}$.
$d = \sqrt{a^2 \cos^2 \theta + a^2 \sin^2 \theta}$.
$d = \sqrt{a^2(\cos^2 \theta + \sin^2 \theta)}$.
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have $d = \sqrt{a^2(1)} = \sqrt{a^2}$.
Since $a \in R^{+}$,the distance is $a$.

(B) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(\cos \theta, 0)$ and $B(0, \sin \theta)$.
Substituting the coordinates into the formula:
$d = \sqrt{(0 - \cos \theta)^2 + (\sin \theta - 0)^2}$
$d = \sqrt{(-\cos \theta)^2 + (\sin \theta)^2}$
$d = \sqrt{\cos^2 \theta + \sin^2 \theta}$
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$d = \sqrt{1} = 1$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(a+b, b-a)$ and $B(a-b, a+b)$.
$AB = \sqrt{((a-b) - (a+b))^2 + ((a+b) - (b-a))^2}$
$AB = \sqrt{(a - b - a - b)^2 + (a + b - b + a)^2}$
$AB = \sqrt{(-2b)^2 + (2a)^2}$
$AB = \sqrt{4b^2 + 4a^2}$
$AB = \sqrt{4(a^2 + b^2)}$
$AB = 2\sqrt{a^2 + b^2}$.

(C) To identify the quadrilateral,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Side $AB = \sqrt{(4 - 1)^2 + (3 - 3)^2} = \sqrt{3^2 + 0^2} = 3$.
Side $BC = \sqrt{(4 - 4)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = 2$.
Side $CD = \sqrt{(1 - 4)^2 + (5 - 5)^2} = \sqrt{(-3)^2 + 0^2} = 3$.
Side $DA = \sqrt{(1 - 1)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = 2$.
Since opposite sides are equal ($AB = CD = 3$ and $BC = DA = 2$) and adjacent sides are not equal,it is a parallelogram.
Now,check the diagonals: $AC = \sqrt{(4 - 1)^2 + (5 - 3)^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.
$BD = \sqrt{(1 - 4)^2 + (5 - 3)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.
Since the opposite sides are equal and the diagonals are equal,$\square ABCD$ is a rectangle.

(B) For the median $\overline{AD}$,$D$ is the midpoint of $\overline{BC}$,where $B(-3, 2)$ and $C(5, 0)$.
The coordinates of $D$ are calculated using the midpoint formula: $D = \left(\frac{-3+5}{2}, \frac{2+0}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)$.
Now,we find the length of the median $\overline{AD}$ using the distance formula between $A(1, 4)$ and $D(1, 1)$:
$AD = \sqrt{(1-1)^2 + (4-1)^2}$
$AD = \sqrt{0^2 + 3^2}$
$AD = \sqrt{9} = 3$.
Thus,the length of the median $\overline{AD}$ is $3$.

(A) The coordinates of the vertices are $A(0, 0)$,$B(4, 0)$,and $C(0, 3)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$ units.
$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ units.
$AC = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{3^2} = 3$ units.
Since $AB^2 + AC^2 = 4^2 + 3^2 = 16 + 9 = 25$ and $BC^2 = 5^2 = 25$,we have $AB^2 + AC^2 = BC^2$.
This satisfies the Pythagorean theorem,indicating that $\Delta ABC$ is a right-angled triangle at vertex $A$. Since $AB \neq AC$,it is not isosceles.

(D) The perpendicular distance of any point $(x, y)$ from the $X-$ axis is given by the absolute value of its $y-$ coordinate.
For the point $(5, 3)$,the $x-$ coordinate is $5$ and the $y-$ coordinate is $3$.
Therefore,the perpendicular distance from the $X-$ axis $= |y| = |3| = 3$ units.

(C) The perpendicular distance of a point $(x, y)$ from the $Y$-axis is given by the absolute value of its $x$-coordinate.
For the point $(-2, 5)$,the $x$-coordinate is $-2$.
Therefore,the perpendicular distance from the $Y$-axis is $|-2| = 2$ units.

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(1, 3)$ and $B(a, 0)$ with distance $d = 3$.
Substituting the values: $3 = \sqrt{(a - 1)^2 + (0 - 3)^2}$.
Squaring both sides: $9 = (a - 1)^2 + (-3)^2$.
$9 = (a - 1)^2 + 9$.
$(a - 1)^2 = 0$.
Therefore,$a - 1 = 0$,which gives $a = 1$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $A(-4, -3)$ and $B(6, a)$ with distance $d = 10$.
$10 = \sqrt{(6 - (-4))^2 + (a - (-3))^2}$
$10 = \sqrt{(10)^2 + (a + 3)^2}$
Squaring both sides:
$100 = 100 + (a + 3)^2$
$(a + 3)^2 = 0$
$a + 3 = 0$
$a = -3$

(B) To determine the type of triangle, we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(0 - 0)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.
$BC = \sqrt{(\sqrt{3} - 0)^2 + (1 - 2)^2} = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$AC = \sqrt{(\sqrt{3} - 0)^2 + (1 - 0)^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Since $AB = BC = AC = 2$, all three sides are equal.
Therefore, $\Delta ABC$ is an equilateral triangle.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(0,0), B(4,0)$ and $C(0,6)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |0(0 - 6) + 4(6 - 0) + 0(0 - 0)|$
Area $= \frac{1}{2} |0 + 24 + 0|$
Area $= \frac{1}{2} |24| = 12$ square units.

(C) The median $\overline{AD}$ connects vertex $A$ to the midpoint $D$ of the opposite side $\overline{BC}$.
Since $B = (0, 0)$ and $C = (6, 0)$,the coordinates of the midpoint $D$ are calculated using the midpoint formula:
$D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + 6}{2}, \frac{0 + 0}{2} \right) = (3, 0)$.
Now,we find the length of the segment $\overline{AD}$ between $A(3, 4)$ and $D(3, 0)$ using the distance formula:
$AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 3)^2 + (0 - 4)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$.
Therefore,the length of the median $\overline{AD}$ is $4$.

(D) Since $\overline{BE}$ is the median of $\Delta ABC$ to the side $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.
The coordinates of $A$ are $(9,5)$ and $C$ are $(3,3)$.
Using the midpoint formula,the coordinates of $E = \left(\frac{9+3}{2}, \frac{5+3}{2}\right) = \left(\frac{12}{2}, \frac{8}{2}\right) = (6,4)$.
The length of median $\overline{BE}$ is the distance between $B(6,7)$ and $E(6,4)$.
Using the distance formula,$BE = \sqrt{(6-6)^2 + (7-4)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.

(C) To determine the type of triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $PQ$: $PQ^2 = (3 - 6)^2 + (10 - 5)^2 = (-3)^2 + (5)^2 = 9 + 25 = 34$. So,$PQ = \sqrt{34}$.
$2$. Length of $QR$: $QR^2 = (6 - 1)^2 + (5 - 2)^2 = (5)^2 + (3)^2 = 25 + 9 = 34$. So,$QR = \sqrt{34}$.
$3$. Length of $PR$: $PR^2 = (3 - 1)^2 + (10 - 2)^2 = (2)^2 + (8)^2 = 4 + 64 = 68$. So,$PR = \sqrt{68}$.
Since $PQ = QR = \sqrt{34}$,two sides of the triangle are equal in length.
Therefore,$\Delta PQR$ is an isosceles triangle.

(B) To determine the type of $\Delta LMN$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $LM = \sqrt{(4 - 1)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.
$2$. Length of $MN = \sqrt{(4 - 4)^2 + (4 - 1)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
$3$. Length of $NL = \sqrt{(4 - 1)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$.
Since $MN = NL = 3$,the triangle is isosceles.
Also,check for a right angle using the Pythagorean theorem: $MN^2 + NL^2 = 3^2 + 3^2 = 9 + 9 = 18$,and $LM^2 = (3\sqrt{2})^2 = 18$.
Since $MN^2 + NL^2 = LM^2$,the triangle is a right-angled triangle.
Therefore,$\Delta LMN$ is an isosceles right-angled triangle.

(B) If a line segment $\overline{PQ}$ is parallel to the $X$-axis,all points on the line segment must have the same $Y$-coordinate.
Since $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ lie on this line,their $Y$-coordinates must be equal.
Therefore,$y_{1} = y_{2}$.

(A) line parallel to the $Y$-axis is a vertical line.
For any vertical line,the $x$-coordinate remains constant for all points on that line.
Since points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ lie on a line parallel to the $Y$-axis,their $x$-coordinates must be equal.
Therefore,$x_{1} = x_{2}$.

(C) The distance formula between two points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ is given by $PQ = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$.
If the line segment $\overline{PQ}$ is parallel to the $X$-axis,the $y$-coordinates of both points must be equal,i.e.,$y_{1} = y_{2}$.
Substituting $y_{1} = y_{2}$ into the distance formula:
$PQ = \sqrt{(x_{2} - x_{1})^2 + (y_{1} - y_{1})^2}$
$PQ = \sqrt{(x_{2} - x_{1})^2 + 0^2}$
$PQ = \sqrt{(x_{2} - x_{1})^2} = |x_{2} - x_{1}| = |x_{1} - x_{2}|$.
Thus,the distance is $|x_{1} - x_{2}|$.

(C) If a line segment $\overline{PQ}$ is parallel to the $Y$-axis,the $x$-coordinates of all points on the line segment must be equal.
Therefore,$x_{1} = x_{2}$.
The distance formula between two points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$.
Substituting $x_{1} = x_{2}$ into the formula:
$PQ = \sqrt{(x_{1}-x_{1})^2 + (y_{2}-y_{1})^2}$
$PQ = \sqrt{0^2 + (y_{2}-y_{1})^2}$
$PQ = \sqrt{(y_{2}-y_{1})^2} = |y_{2}-y_{1}| = |y_{1}-y_{2}|$.
Thus,the distance is $|y_{1}-y_{2}|$.

(A) The distance between two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ in a Cartesian plane is given by the distance formula.
According to the distance formula,the distance $d$ between points $A$ and $B$ is calculated as $d(A, B) = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$.
Since $(x_{2}-x_{1})^{2} = (x_{1}-x_{2})^{2}$ and $(y_{2}-y_{1})^{2} = (y_{1}-y_{2})^{2}$,the formula can also be written as $d(A, B) = \sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2}}$.
Therefore,option $A$ is the correct answer.

(B) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,$(x_1, y_1) = (7, 5)$ and $(x_2, y_2) = (2, 5)$.
Substituting the values into the formula:
$d = \sqrt{(2 - 7)^2 + (5 - 5)^2}$
$d = \sqrt{(-5)^2 + (0)^2}$
$d = \sqrt{25 + 0}$
$d = \sqrt{25}$
$d = 5$
Therefore,the distance between the two points is $5$ units.

(C) The distance of any point $P(x, y)$ from the $X$-axis is given by the absolute value of its $y$-coordinate,which is $|y|$.
Given the point $P(2, -3)$,the $x$-coordinate is $2$ and the $y$-coordinate is $-3$.
Therefore,the distance from the $X$-axis is $|-3| = 3$ units.
Thus,the correct option is $C$.

(A) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(2, 4)$ and $B(-3, 4)$.
Here,$x_1 = 2, y_1 = 4$ and $x_2 = -3, y_2 = 4$.
Substituting these values into the formula:
$d = \sqrt{(-3 - 2)^2 + (4 - 4)^2}$
$d = \sqrt{(-5)^2 + (0)^2}$
$d = \sqrt{25 + 0}$
$d = \sqrt{25} = 5$.
Therefore,the distance between the points is $5$ units.

(A) The distance between points $A(3, a)$ and $B(4, 1)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $d = \sqrt{10}$,we have $\sqrt{(4 - 3)^2 + (1 - a)^2} = \sqrt{10}$.
Squaring both sides,we get $(4 - 3)^2 + (1 - a)^2 = 10$.
$1^2 + (1 - 2a + a^2) = 10$.
$1 + 1 - 2a + a^2 = 10$.
$a^2 - 2a + 2 = 10$.
$a^2 - 2a - 8 = 0$.
Factoring the quadratic equation: $(a - 4)(a + 2) = 0$.
Thus,$a = 4$ or $a = -2$.
Since $4$ is provided in option $A$,the correct value is $4$.

(B) The given vertices are $A(0, 0)$,$B(3.1, 0)$,and $C(0, 4.5)$.
Point $A(0, 0)$ is the origin.
Point $B(3.1, 0)$ lies on the $X$-axis.
Point $C(0, 4.5)$ lies on the $Y$-axis.
Since the $X$-axis and $Y$-axis are perpendicular to each other at the origin,the angle formed at vertex $A(0, 0)$ is $90^{\circ}$.
Therefore,the triangle formed by these vertices is a right-angled triangle.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(-4, -6)$ and $B(6, b)$ with distance $d = 10$.
Substituting the values into the formula:
$10 = \sqrt{(6 - (-4))^2 + (b - (-6))^2}$
$10 = \sqrt{(6 + 4)^2 + (b + 6)^2}$
$10 = \sqrt{10^2 + (b + 6)^2}$
Squaring both sides:
$100 = 100 + (b + 6)^2$
$(b + 6)^2 = 100 - 100$
$(b + 6)^2 = 0$
Taking the square root of both sides:
$b + 6 = 0$
$b = -6$.

(A) To determine the type of quadrilateral $ABCD$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $AB = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{2^2 + 0^2} = 2$.
$2$. Length of $BC = \sqrt{(2 - 2)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = 2$.
$3$. Length of $CD = \sqrt{(0 - 2)^2 + (2 - 2)^2} = \sqrt{(-2)^2 + 0^2} = 2$.
$4$. Length of $DA = \sqrt{(0 - 0)^2 + (0 - 2)^2} = \sqrt{0^2 + (-2)^2} = 2$.
Since all sides are equal $(AB = BC = CD = DA = 2)$,the quadrilateral is either a square or a rhombus.
Next,we check the diagonals $AC$ and $BD$:
Diagonal $AC = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Diagonal $BD = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Since all sides are equal and the diagonals are equal,$\square ABCD$ is a square.

(B) The perimeter of a square is given by $4 \times \text{side}$.
Given,$4 \times \text{side} = 20$.
Therefore,$\text{side} = 20 / 4 = 5$.
In a square $ABCD$,the diagonal $AC$ forms a right-angled triangle $ABC$ with sides $AB$ and $BC$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Since $AB = BC = 5$,we have $AC^2 = 5^2 + 5^2 = 25 + 25 = 50$.
Thus,$AC = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.

(D) Let the starting point $A$ be at the origin $(0, 0)$ in a Cartesian coordinate system.
Moving $800 \, m$ north takes the person to point $B(0, 800)$.
Moving $500 \, m$ east from $B$ takes the person to point $C(500, 800)$.
Moving $400 \, m$ north from $C$ takes the person to point $D(500, 800 + 400) = D(500, 1200)$.
The direct distance from $A(0, 0)$ to $D(500, 1200)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$d = \sqrt{(500 - 0)^2 + (1200 - 0)^2} = \sqrt{500^2 + 1200^2}$.
$d = \sqrt{250000 + 1440000} = \sqrt{1690000}$.
$d = 1300 \, m$.

(A) Let the airport be at the origin $(0, 0)$.
After $t = 1.5 \, \text{hours}$,the distance covered by the first plane moving North is $d_1 = 800 \times 1.5 = 1200 \, km$.
The distance covered by the second plane moving East is $d_2 = 840 \times 1.5 = 1260 \, km$.
Since North and East directions are perpendicular,the distance between the two planes forms the hypotenuse of a right-angled triangle.
Using the Pythagorean theorem,the distance $D = \sqrt{d_1^2 + d_2^2}$.
$D = \sqrt{1200^2 + 1260^2} = \sqrt{1440000 + 1587600} = \sqrt{3027600}$.
$D = 1740 \, km$.

(N/A) Let the rectangle be $ABCD$ in the Cartesian plane. Let $A = (0, b)$,$B = (a, b)$,$C = (a, 0)$,and $D = (0, 0)$.
Let the coordinates of point $P$ be $(x, y)$.
Using the distance formula,the squared distances are:
$PA^2 = (x - 0)^2 + (y - b)^2 = x^2 + y^2 - 2by + b^2$
$PC^2 = (x - a)^2 + (y - 0)^2 = x^2 - 2ax + a^2 + y^2$
$PB^2 = (x - a)^2 + (y - b)^2 = x^2 - 2ax + a^2 + y^2 - 2by + b^2$
$PD^2 = (x - 0)^2 + (y - 0)^2 = x^2 + y^2$
Now,calculate the sums:
$PA^2 + PC^2 = (x^2 + y^2 - 2by + b^2) + (x^2 - 2ax + a^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
$PB^2 + PD^2 = (x^2 - 2ax + a^2 + y^2 - 2by + b^2) + (x^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
Since both sums are equal,$PA^2 + PC^2 = PB^2 + PD^2$ is proved.

(B) In a rectangle $ABCD$,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle BCD$ (or $\triangle ABD$).
By the Pythagorean theorem,in $\triangle BCD$,we have $BD^2 = BC^2 + CD^2$.
Since $ABCD$ is a rectangle,$CD = AB = 9$.
Substituting the values,$BD^2 = 12^2 + 9^2$.
$BD^2 = 144 + 81 = 225$.
Taking the square root,$BD = \sqrt{225} = 15$.

(D) Let the side of the square be $a$.
We know that the diagonal $d$ of a square is given by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 5 \sqrt{2}$.
Equating the two expressions: $a \sqrt{2} = 5 \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we get $a = 5$.
Therefore,the length of the side of the square is $5$.

(B) Let the side of the square be $a$.
We know that the length of the diagonal of a square is given by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 8 \sqrt{2}$.
Equating the two,we get $a \sqrt{2} = 8 \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we find the side length $a = 8$.
The perimeter of a square is given by the formula $P = 4 \times a$.
Substituting the value of $a$,we get $P = 4 \times 8 = 32$.