Mix Examples - Coordinate Geometry Questions in English

(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 $(2, -2)$ and $(-1, x)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(-1 - 2)^2 + (x - (-2))^2}$
$5 = \sqrt{(-3)^2 + (x + 2)^2}$
$5 = \sqrt{9 + (x^2 + 4x + 4)}$
$5 = \sqrt{x^2 + 4x + 13}$
Squaring both sides:
$25 = x^2 + 4x + 13$
$x^2 + 4x - 12 = 0$
Factoring the quadratic equation:
$(x + 6)(x - 2) = 0$
Thus,$x = -6$ or $x = 2$.
Therefore,one of the possible values of $x$ is $2$.

(B) The mid-point formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Given points are $A(-2, 8)$ and $B(-6, -4)$.
Substituting the values into the formula:
Mid-point $= \left(\frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2}\right)$
Mid-point $= \left(\frac{-8}{2}, \frac{4}{2}\right)$
Mid-point $= (-4, 2)$.

(C) To determine the shape,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
$AB = \sqrt{(9-9)^2 + (6-0)^2} = \sqrt{0^2 + 6^2} = 6$
$BC = \sqrt{(-9-9)^2 + (6-6)^2} = \sqrt{(-18)^2 + 0^2} = 18$
$CD = \sqrt{(-9-(-9))^2 + (0-6)^2} = \sqrt{0^2 + (-6)^2} = 6$
$DA = \sqrt{(9-(-9))^2 + (0-0)^2} = \sqrt{18^2 + 0^2} = 18$
Since opposite sides are equal ($AB = CD = 6$ and $BC = DA = 18$) and the adjacent sides are perpendicular (as the sides are parallel to the coordinate axes),the figure is a rectangle.

(D) We know that for any point $P(x, y)$ in the Cartesian plane:
$1$. The perpendicular distance of the point from the $y$-axis is given by the absolute value of its $x$-coordinate (abscissa),i.e.,$|x|$.
$2$. The perpendicular distance of the point from the $x$-axis is given by the absolute value of its $y$-coordinate (ordinate),i.e.,$|y|$.
For the point $P(2, 3)$,the $x$-coordinate is $2$ and the $y$-coordinate is $3$.
Therefore,the distance of the point $P(2, 3)$ from the $x$-axis is equal to the ordinate of the point,which is $3$ 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 $A(0, 6)$ and $B(0, -2)$.
Here,$x_1 = 0, y_1 = 6$ and $x_2 = 0, y_2 = -2$.
Substituting these values into the distance formula:
$AB = \sqrt{(0 - 0)^2 + (-2 - 6)^2}$
$AB = \sqrt{0^2 + (-8)^2}$
$AB = \sqrt{64}$
$AB = 8$
Therefore,the distance between the points $A$ and $B$ is $8$ units.

(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,the point $P$ is $(-6, 8)$ and the origin $O$ is $(0, 0)$.
So,$x_1 = -6, y_1 = 8$ and $x_2 = 0, y_2 = 0$.
Substituting these values into the distance formula:
$PO = \sqrt{(0 - (-6))^2 + (0 - 8)^2}$
$PO = \sqrt{(6)^2 + (-8)^2}$
$PO = \sqrt{36 + 64}$
$PO = \sqrt{100}$
$PO = 10$
Thus,the distance of the point $P(-6, 8)$ from the origin is $10$ units.

(C) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:
$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
Here,$x_{1} = 0, y_{1} = 5$ and $x_{2} = -5, y_{2} = 0$.
Substituting these values into the formula:
$d = \sqrt{(-5 - 0)^{2} + (0 - 5)^{2}}$
$d = \sqrt{(-5)^{2} + (-5)^{2}}$
$d = \sqrt{25 + 25}$
$d = \sqrt{50}$
$d = 5 \sqrt{2}$

(D) In a rectangle $AOBC$,the diagonal is the line segment joining opposite vertices. Here,$AB$ is the diagonal.
The length of the diagonal $AB$ is the distance between the points $A (0, 3)$ and $B (5, 0)$.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,$x_1 = 0, y_1 = 3$ and $x_2 = 5, y_2 = 0$.
Substituting these values into the distance formula:
$AB = \sqrt{(5 - 0)^2 + (0 - 3)^2}$
$AB = \sqrt{(5)^2 + (-3)^2}$
$AB = \sqrt{25 + 9}$
$AB = \sqrt{34}$
Thus,the length of the diagonal is $\sqrt{34}$.

(A) To find the perimeter of a triangle,we calculate the sum of the lengths of all its sides. Let the vertices of the triangle be $A(0,4)$,$O(0,0)$,and $B(3,0)$.
The perimeter of $\triangle AOB = \text{Sum of the lengths of all its sides} = d(AO) + d(OB) + d(AB)$.
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Distance $d(AO) = \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{0 + 16} = 4$.
$2$. Distance $d(OB) = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9 + 0} = 3$.
$3$. Distance $d(AB) = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Perimeter $= 4 + 3 + 5 = 12$.
Thus,the required perimeter of the triangle is $12$.

(B) The area of a triangle with vertices $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(3, 0), B(7, 0)$ and $C(8, 4)$.
Here,$x_1 = 3, y_1 = 0, x_2 = 7, y_2 = 0, x_3 = 8, y_3 = 4$.
Substituting these values into the formula:
$\text{Area} = \frac{1}{2} |3(0 - 4) + 7(4 - 0) + 8(0 - 0)|$
$\text{Area} = \frac{1}{2} |3(-4) + 7(4) + 8(0)|$
$\text{Area} = \frac{1}{2} |-12 + 28 + 0|$
$\text{Area} = \frac{1}{2} |16|$
$\text{Area} = 8 \text{ square units}$.
Thus,the area of the triangle is $8$.

(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$
$BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
$AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
Since $BC = AC = 5$,two sides of the triangle are equal.
Therefore,$\triangle ABC$ is an isosceles triangle.

(D) If $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m: n$,then the coordinates are given by the section formula:
$x = \frac{mx_2 + nx_1}{m + n}$ and $y = \frac{my_2 + ny_1}{m + n}$
Given that $x_1 = 7, y_1 = -6, x_2 = 3, y_2 = 4, m = 1$ and $n = 2$.
Substituting these values into the section formula:
$x = \frac{1(3) + 2(7)}{1 + 2} = \frac{3 + 14}{3} = \frac{17}{3}$
$y = \frac{1(4) + 2(-6)}{1 + 2} = \frac{4 - 12}{3} = -\frac{8}{3}$
Thus,the point is $(\frac{17}{3}, -\frac{8}{3})$.
Since the $x$-coordinate is positive and the $y$-coordinate is negative,the point lies in the $IV$ quadrant.

(A) We know that the perpendicular bisector of any line segment passes through the midpoint of that line segment.
The midpoint of the line segment joining the points $A(-2, -5)$ and $B(2, 5)$ is calculated using the midpoint formula:
Midpoint $= \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Substituting the given coordinates:
Midpoint $= \left( \frac{-2 + 2}{2}, \frac{-5 + 5}{2} \right) = \left( \frac{0}{2}, \frac{0}{2} \right) = (0, 0)$
Since the perpendicular bisector must pass through the midpoint of the segment,the point $(0, 0)$ lies on the perpendicular bisector.

(B) Let the fourth vertex of the parallelogram be $D(x_4, y_4)$.
In a parallelogram,the diagonals bisect each other,which means the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
The midpoint formula for a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Midpoint of $AC = \left(\frac{-2+8}{2}, \frac{3+3}{2}\right) = \left(\frac{6}{2}, \frac{6}{2}\right) = (3,3)$.
Midpoint of $BD = \left(\frac{6+x_4}{2}, \frac{7+y_4}{2}\right)$.
Since the midpoints are equal:
$\frac{6+x_4}{2} = 3 \Rightarrow 6+x_4 = 6 \Rightarrow x_4 = 0$.
$\frac{7+y_4}{2} = 3 \Rightarrow 7+y_4 = 6 \Rightarrow y_4 = -1$.
Thus,the fourth vertex $D$ is $(0,-1)$.

(C) Given that the point $P(2,1)$ lies on the line segment joining the points $A(4,2)$ and $B(8,4)$.
First,we calculate the distance between $A(4,2)$ and $P(2,1)$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AP = \sqrt{(2 - 4)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.
Next,we calculate the distance between $A(4,2)$ and $B(8,4)$:
$AB = \sqrt{(8 - 4)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.
Now,we compare $AP$ and $AB$:
$AB = 2\sqrt{5} = 2 \times (\sqrt{5}) = 2 AP$.
Therefore,$AP = \frac{1}{2} AB$.
Thus,the correct condition is $AP = \frac{1}{2} AB$.

(D) The mid-point formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.
Given points are $Q(-6, 5)$ and $R(-2, 3)$.
Calculating the mid-point $P$ of $QR$:
$P = (\frac{-6 + (-2)}{2}, \frac{5 + 3}{2}) = (\frac{-8}{2}, \frac{8}{2}) = (-4, 4)$.
We are given that $P = (\frac{a}{3}, 4)$.
Comparing the coordinates of $P$,we have:
$\frac{a}{3} = -4$
$a = -4 \times 3$
$a = -12$.
Thus,the value of $a$ is $-12$.

(A) $1$. Find the midpoint $P$ of the line segment $AB$:
$P = \left( \frac{1+4}{2}, \frac{5+6}{2} \right) = \left( \frac{5}{2}, \frac{11}{2} \right)$
$2$. Find the slope of the line segment $AB$ $(m_{AB})$:
$m_{AB} = \frac{6-5}{4-1} = \frac{1}{3}$
$3$. The slope of the perpendicular bisector $(m_{\perp})$ is the negative reciprocal of $m_{AB}$:
$m_{\perp} = -\frac{1}{m_{AB}} = -3$
$4$. Use the point-slope form to find the equation of the perpendicular bisector:
$y - y_1 = m_{\perp}(x - x_1)$
$y - \frac{11}{2} = -3(x - \frac{5}{2})$
$y - \frac{11}{2} = -3x + \frac{15}{2}$
$y = -3x + \frac{15}{2} + \frac{11}{2}$
$y = -3x + \frac{26}{2}$
$y = -3x + 13$
$5$. To find where it cuts the $y$-axis,set $x = 0$:
$y = -3(0) + 13 = 13$
Thus,the perpendicular bisector cuts the $y$-axis at $(0, 13)$.

(B) Let the coordinates of the point $P(h, k)$ be equidistant from the three vertices $O(0, 0)$,$A(0, 2y)$,and $B(2x, 0)$.
Then,$PO = PA = PB$.
$\Rightarrow (PO)^2 = (PA)^2 = (PB)^2$ ..........$(i)$
By the distance formula,we have:
$[\sqrt{(h-0)^2 + (k-0)^2}]^2 = [\sqrt{(h-0)^2 + (k-2y)^2}]^2 = [\sqrt{(h-2x)^2 + (k-0)^2}]^2$
$\Rightarrow h^2 + k^2 = h^2 + (k-2y)^2 = (h-2x)^2 + k^2$
Taking the first two parts:
$h^2 + k^2 = h^2 + (k-2y)^2$
$k^2 = k^2 + 4y^2 - 4yk$
$4y(y - k) = 0 \Rightarrow k = y$ (since $y \neq 0$).
Taking the first and third parts:
$h^2 + k^2 = (h-2x)^2 + k^2$
$h^2 = h^2 + 4x^2 - 4xh$
$4x(x - h) = 0 \Rightarrow h = x$ (since $x \neq 0$).
Therefore,the required point is $(h, k) = (x, y)$.

(C) The center of the circle is $(0,0)$ and it passes through the point $\left(\frac{13}{2}, 0\right)$.
Therefore,the radius of the circle $r$ is the distance between $(0,0)$ and $\left(\frac{13}{2}, 0\right)$.
$r = \sqrt{\left(\frac{13}{2}-0\right)^{2} + (0-0)^{2}} = \sqrt{\left(\frac{13}{2}\right)^{2}} = \frac{13}{2} = 6.5$.
$A$ point $(x, y)$ lies in the interior of the circle if its distance from the center is less than $r$,on the circle if the distance is equal to $r$,and outside if the distance is greater than $r$.
$(a)$ Distance of $\left(-\frac{3}{4}, 1\right)$ from $(0,0) = \sqrt{\left(-\frac{3}{4}\right)^{2} + 1^{2}} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = 1.25 < 6.5$. (Interior)
$(b)$ Distance of $\left(2, \frac{7}{3}\right)$ from $(0,0) = \sqrt{2^{2} + \left(\frac{7}{3}\right)^{2}} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{85}{9}} \approx 3.07 < 6.5$. (Interior)
$(c)$ Distance of $\left(-6, \frac{5}{2}\right)$ from $(0,0) = \sqrt{(-6)^{2} + \left(\frac{5}{2}\right)^{2}} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144+25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5$. (On the circle)
$(d)$ Distance of $\left(5, -\frac{1}{2}\right)$ from $(0,0) = \sqrt{5^{2} + \left(-\frac{1}{2}\right)^{2}} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} \approx 5.02 < 6.5$. (Interior)
Thus,the point $\left(-6, \frac{5}{2}\right)$ does not lie in the interior of the circle.

(D) Let the coordinates of $P$ be $(0, y)$ and $Q$ be $(x, 0)$,respectively.
The mid-point of $P(0, y)$ and $Q(x, 0)$ is given by the formula $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Thus,the mid-point $M$ is $\left(\frac{0+x}{2}, \frac{y+0}{2}\right) = \left(\frac{x}{2}, \frac{y}{2}\right)$.
It is given that the mid-point of $PQ$ is $(2, -5)$.
Equating the coordinates,we get:
$\frac{x}{2} = 2 \Rightarrow x = 4$
$\frac{y}{2} = -5 \Rightarrow y = -10$
Therefore,the coordinates of $P$ are $(0, -10)$ and $Q$ are $(4, 0)$.

(A) Let the vertices of the triangle be $A(x_1, y_1) = (a, b+c)$,$B(x_2, y_2) = (b, c+a)$,and $C(x_3, y_3) = (c, a+b)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula:
$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given coordinates into the formula:
$\Delta = \frac{1}{2} |a(c+a - (a+b)) + b(a+b - (b+c)) + c(b+c - (c+a))|$
Simplify the terms inside the brackets:
$\Delta = \frac{1}{2} |a(c-b) + b(a-c) + c(b-a)|$
Expand the terms:
$\Delta = \frac{1}{2} |ac - ab + ab - bc + bc - ac|$
All terms cancel out:
$\Delta = \frac{1}{2} |0| = 0$
Thus,the area of the triangle is $0$.

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(4, p)$ and $(1, 0)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(1 - 4)^2 + (0 - p)^2}$
$5 = \sqrt{(-3)^2 + (-p)^2}$
$5 = \sqrt{9 + p^2}$
Squaring both sides:
$25 = 9 + p^2$
$p^2 = 25 - 9$
$p^2 = 16$
$p = \pm \sqrt{16}$
$p = \pm 4$
Thus,the value of $p$ is $\pm 4$.

(C) Let the given points be $A(x_1, y_1) = (1, 2)$,$O(x_2, y_2) = (0, 0)$,and $C(x_3, y_3) = (a, b)$.
Since the points are collinear,the area of the triangle formed by these points must be zero.
The formula for the area of a triangle is $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given coordinates:
$0 = \frac{1}{2} |1(0 - b) + 0(b - 2) + a(2 - 0)|$.
$0 = \frac{1}{2} |-b + 0 + 2a|$.
$0 = \frac{1}{2} |2a - b|$.
This implies $2a - b = 0$,or $2a = b$.
Hence,the required relation is $2a = b$.

(A) The statement is true.
$A$ quadrilateral is a parallelogram if its diagonals bisect each other,which means they share the same midpoint.
$1$. Midpoint of diagonal $AC$:
Midpoint $= (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) = (\frac{-1 + 2}{2}, \frac{0 + 2}{2}) = (\frac{1}{2}, 1)$.
$2$. Midpoint of diagonal $BD$:
Midpoint $= (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) = (\frac{3 - 2}{2}, \frac{1 + 1}{2}) = (\frac{1}{2}, 1)$.
Since the midpoints of both diagonals $AC$ and $BD$ are the same $(\frac{1}{2}, 1)$,the diagonals bisect each other. Therefore,the points $A, B, C,$ and $D$ form a parallelogram.

(B) False.
To check if the points $A(4,5), B(7,6)$ and $C(6,3)$ are collinear,we calculate the area of the triangle formed by these points using the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
Area $= \frac{1}{2} |4(6 - 3) + 7(3 - 5) + 6(5 - 6)|$
Area $= \frac{1}{2} |4(3) + 7(-2) + 6(-1)|$
Area $= \frac{1}{2} |12 - 14 - 6|$
Area $= \frac{1}{2} |-8| = 4 \text{ sq. units}$.
Since the area of the triangle is not $0$,the points are not collinear.

(TRUE) The statement is True.
$1$. $A$ point on the $y$-axis has coordinates $(0, y)$. Since $P (0, -7)$ lies on the $y$-axis,it satisfies the first condition.
$2$. For $P$ to be on the perpendicular bisector of $AB$,it must be equidistant from $A (-1, 0)$ and $B (7, -6)$.
$3$. Calculate the distance $PA = \sqrt{(0 - (-1))^2 + (-7 - 0)^2} = \sqrt{1^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$ units.
$4$. Calculate the distance $PB = \sqrt{(0 - 7)^2 + (-7 - (-6))^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50}$ units.
$5$. Since $PA = PB$,point $P$ lies on the perpendicular bisector of $AB$. Thus,the statement is true.

(TRUE) True.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
For $\triangle ABC$:
$AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2 + 0^2} = 4$
$BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$AC = \sqrt{(0 - (-2))^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
For $\triangle DEF$:
$DE = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$
$EF = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$DF = \sqrt{(0 - (-4))^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Comparing the ratios of corresponding sides:
$\frac{AB}{DE} = \frac{4}{8} = \frac{1}{2}$
$\frac{BC}{EF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
$\frac{AC}{DF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
Since $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{1}{2}$,by the $SSS$ similarity criterion,$\triangle ABC \sim \triangle DEF$.

(TRUE) True.
Analytical Justification:
$1$. The points $A (-4, 6)$ and $B (-4, -6)$ have the same $x$-coordinate,which is $-4$. This means the line segment $AB$ is a vertical line represented by the equation $x = -4$.
$2$. The point $P (-4, 2)$ also has an $x$-coordinate of $-4$,so it lies on the line $x = -4$.
$3$. For $P$ to lie on the line segment $AB$,its $y$-coordinate must be between the $y$-coordinates of $A$ and $B$. The $y$-coordinates of $A$ and $B$ are $6$ and $-6$ respectively.
$4$. Since $-6 < 2 < 6$,the point $P$ lies between $A$ and $B$ on the line segment $AB$.

(B) False.
Let the points be $A(0, 5)$,$B(0, -9)$,and $C(3, 6)$.
For points to be collinear,the area of the triangle formed by them must be zero.
The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$,and $(x_{3}, y_{3})$ is given by:
$\Delta = \frac{1}{2} |x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|$
Substituting the values $x_{1}=0, y_{1}=5, x_{2}=0, y_{2}=-9, x_{3}=3, y_{3}=6$:
$\Delta = \frac{1}{2} |0(-9 - 6) + 0(6 - 5) + 3(5 - (-9))|$
$\Delta = \frac{1}{2} |0 + 0 + 3(5 + 9)|$
$\Delta = \frac{1}{2} |3 \times 14| = \frac{42}{2} = 21$
Since the area of the triangle is $21 \neq 0$,the points are not collinear.

(B) False.
We know that any point lying on the perpendicular bisector of a line segment is equidistant from the endpoints of that segment.
Let us calculate the distance of point $P(0, 2)$ from points $A(-1, 1)$ and $B(3, 3)$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$PA = \sqrt{(0 - (-1))^2 + (2 - 1)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
$PB = \sqrt{(0 - 3)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Since $PA \neq PB$,the point $P(0, 2)$ is not equidistant from $A$ and $B$.
Therefore,the point $P(0, 2)$ does not lie on the perpendicular bisector of the line segment $AB$.

(A) True.
Let the coordinates be $A(x_1, y_1) = (3, 1)$,$B(x_2, y_2) = (12, -2)$,and $C(x_3, y_3) = (0, 2)$.
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:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given values:
$\text{Area} = \frac{1}{2} |3(-2 - 2) + 12(2 - 1) + 0(1 - (-2))|$
$\text{Area} = \frac{1}{2} |3(-4) + 12(1) + 0(3)|$
$\text{Area} = \frac{1}{2} |-12 + 12 + 0|$
$\text{Area} = \frac{1}{2} |0| = 0$
Since the area of the triangle formed by these points is $0$,the points are collinear (they lie on the same straight line).
Therefore,these points cannot form a triangle. Thus,the statement is true.

(B) False.
To determine if the points form a parallelogram,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$
$BC = \sqrt{(5-6)^2 + (-6-4)^2} = \sqrt{(-1)^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101}$
$CD = \sqrt{(-3-5)^2 + (5 - (-6))^2} = \sqrt{(-8)^2 + 11^2} = \sqrt{64 + 121} = \sqrt{185}$
$DA = \sqrt{(4 - (-3))^2 + (3-5)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53}$
In a parallelogram,opposite sides must be equal ($AB = CD$ and $BC = DA$). Since all sides are unequal,the given points do not form a parallelogram.

(TRUE) True.
Given that the circle has its centre at the origin $O(0,0)$ and the point $P(5,0)$ lies on the circle.
The radius $r$ of the circle is the distance between the centre $O(0,0)$ and the point $P(5,0)$ on the circle.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$r = OP = \sqrt{(5 - 0)^2 + (0 - 0)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5$.
Now,we calculate the distance of point $Q(6,8)$ from the centre $O(0,0)$:
$OQ = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
We know that if the distance of a point from the centre is greater than the radius,the point lies outside the circle.
Since $OQ = 10$ and $r = 5$,we have $OQ > r$.
Therefore,the point $Q(6,8)$ lies outside the circle. Hence,the statement is true.

(B) False.
If a point $A$ lies on the perpendicular bisector of a line segment $PQ$,then it must be equidistant from the endpoints $P$ and $Q$,i.e.,$AP = AQ$.
First,calculate the distance $AP$ using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AP = \sqrt{(6-2)^2 + (5-7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Next,calculate the distance $AQ$:
$AQ = \sqrt{(0-2)^2 + (-4-7)^2} = \sqrt{(-2)^2 + (-11)^2} = \sqrt{4 + 121} = \sqrt{125}$.
Since $AP \neq AQ$ (as $\sqrt{20} \neq \sqrt{125}$),the point $A (2,7)$ does not lie on the perpendicular bisector of the line segment $PQ$.

(A) True.
Let point $P(5, -3)$ divide the line segment joining $A(7, -2)$ and $B(1, -5)$ in the ratio $k: 1$ internally.
By the section formula,the coordinates of point $P$ are given by:
$\left( \frac{k(1) + 1(7)}{k + 1}, \frac{k(-5) + 1(-2)}{k + 1} \right) = \left( \frac{k + 7}{k + 1}, \frac{-5k - 2}{k + 1} \right)$.
Equating these to the given coordinates of $P(5, -3)$:
$\frac{k + 7}{k + 1} = 5 \implies k + 7 = 5k + 5 \implies 2 = 4k \implies k = \frac{1}{2}$.
Checking the $y$-coordinate with $k = \frac{1}{2}$:
$\frac{-5(1/2) - 2}{1/2 + 1} = \frac{-2.5 - 2}{1.5} = \frac{-4.5}{1.5} = -3$.
Since the ratio is $1: 2$,point $P$ divides the segment $AB$ such that it is one of the two points of trisection (the other being at ratio $2: 1$). Thus,the statement is true.

(A) True.
First,we check if the points are collinear by calculating the area of the triangle formed by them.
Area of triangle $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Here,$x_1 = -6, x_2 = -4, x_3 = 3$ and $y_1 = 10, y_2 = 6, y_3 = -8$.
Area $= \frac{1}{2} | -6(6 - (-8)) + (-4)(-8 - 10) + 3(10 - 6) |$.
Area $= \frac{1}{2} | -6(14) + (-4)(-18) + 3(4) |$.
Area $= \frac{1}{2} | -84 + 72 + 12 | = \frac{1}{2} | 0 | = 0$.
Since the area is $0$,the points are collinear.
Next,we calculate the distances $AB$ and $AC$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(-4 - (-6))^2 + (6 - 10)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
$AC = \sqrt{(3 - (-6))^2 + (-8 - 10)^2} = \sqrt{9^2 + (-18)^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$.
Now,check the relation $AB = \frac{2}{9} AC$:
$\frac{2}{9} AC = \frac{2}{9} \times 9\sqrt{5} = 2\sqrt{5}$.
Since $AB = 2\sqrt{5}$,the relation holds true.

(B) False.
$A$ point lies on a circle if and only if the distance between the point and the centre of the circle is equal to the radius of the circle.
Let the centre be $C(3, 5)$ and the point be $P(-2, 4)$. The distance $CP$ is calculated using the distance formula:
$CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$CP = \sqrt{(3 - (-2))^2 + (5 - 4)^2}$
$CP = \sqrt{(3 + 2)^2 + (1)^2}$
$CP = \sqrt{5^2 + 1^2}$
$CP = \sqrt{25 + 1} = \sqrt{26}$
Since $\sqrt{26} \neq 6$,the distance between the point $P$ and the centre $C$ is not equal to the radius of the circle.
Therefore,the point $P(-2, 4)$ does not lie on the circle.

(A) True.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Length of sides:
$AB = \sqrt{(4 - (-1))^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$BC = \sqrt{(2 - 4)^2 + (5 - 3)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$CD = \sqrt{(-3 - 2)^2 + (0 - 5)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
Since opposite sides are equal ($AB = CD$ and $BC = DA$),it is a parallelogram.
$2$. Length of diagonals:
$AC = \sqrt{(2 - (-1))^2 + (5 - (-2))^2} = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$
$BD = \sqrt{(-3 - 4)^2 + (0 - 3)^2} = \sqrt{(-7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58}$
Since the diagonals are equal $(AC = BD)$,the parallelogram is a rectangle.

(C) The mid-point formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}).$
Given points are $A (3, 4)$ and $B (k, 6).$
Thus,the mid-point $P (x, y) = (\frac{3+k}{2}, \frac{4+6}{2}) = (\frac{3+k}{2}, 5).$
Comparing the coordinates,we get $x = \frac{3+k}{2}$ and $y = 5.$
We are given the equation $x+y-10=0.$
Substituting the values of $x$ and $y$ into the equation:
$\frac{3+k}{2} + 5 - 10 = 0$
$\frac{3+k}{2} - 5 = 0$
$\frac{3+k}{2} = 5$
$3+k = 10$
$k = 10 - 3$
$k = 7.$

(D) Let the coordinates of vertices $B$ and $C$ be $(a, b)$ and $(x, y)$ respectively.
The mid-point of side $AB$ is given as $(2, -1)$. Thus,$\left(\frac{1+a}{2}, \frac{-4+b}{2}\right) = (2, -1)$.
Equating the coordinates,we get $1+a = 4 \implies a = 3$ and $-4+b = -2 \implies b = 2$. So,$B = (3, 2)$.
The mid-point of side $AC$ is given as $(0, -1)$. Thus,$\left(\frac{1+x}{2}, \frac{-4+y}{2}\right) = (0, -1)$.
Equating the coordinates,we get $1+x = 0 \implies x = -1$ and $-4+y = -2 \implies y = 2$. So,$C = (-1, 2)$.
The vertices of $\triangle ABC$ are $A(1, -4)$,$B(3, 2)$,and $C(-1, 2)$.
The area of $\triangle ABC = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |1(2 - 2) + 3(2 - (-4)) + (-1)(-4 - 2)|$.
Area $= \frac{1}{2} |0 + 3(6) - 1(-6)| = \frac{1}{2} |18 + 6| = \frac{24}{2} = 12$ $sq. units$.

(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$
$PR = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2} = \sqrt{(6 + 2 + 2\sqrt{12}) + (6 + 2 - 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$
$RQ = \sqrt{(-\sqrt{6} - (-\sqrt{2}))^2 + (\sqrt{6} - (-\sqrt{2}))^2} = \sqrt{(-\sqrt{6} + \sqrt{2})^2 + (\sqrt{6} + \sqrt{2})^2} = \sqrt{(6 + 2 - 2\sqrt{12}) + (6 + 2 + 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$
Since $PQ = PR = RQ = 4$,all three sides are equal. Therefore,the triangle $PQR$ is an equilateral triangle.

(N/A) Let the coordinates of vertex $D$ be $(x, y)$.
We know that the diagonals of a parallelogram bisect each other at their midpoint.
Therefore,the midpoint of diagonal $AC$ is equal to the midpoint of diagonal $BD$.
The midpoint of $AC$ is given by $\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)$.
The midpoint of $BD$ is given by $\left(\frac{x_{2}+x}{2}, \frac{y_{2}+y}{2}\right)$.
Equating the coordinates,we get:
$\frac{x_{1}+x_{3}}{2} = \frac{x_{2}+x}{2} \implies x_{1}+x_{3} = x_{2}+x \implies x = x_{1}+x_{3}-x_{2}$.
$\frac{y_{1}+y_{3}}{2} = \frac{y_{2}+y}{2} \implies y_{1}+y_{3} = y_{2}+y \implies y = y_{1}+y_{3}-y_{2}$.
Thus,the coordinates of vertex $D$ are $(x_{1}+x_{3}-x_{2}, y_{1}+y_{3}-y_{2})$.

(C) To determine the type of triangle,we calculate the lengths of all three sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(-4 - (-5))^2 + (-2 - 6)^2} = \sqrt{(1)^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$ units.
$2$. Length of side $BC$:
$BC = \sqrt{(7 - (-4))^2 + (5 - (-2))^2} = \sqrt{(11)^2 + (7)^2} = \sqrt{121 + 49} = \sqrt{170}$ units.
$3$. Length of side $CA$:
$CA = \sqrt{(-5 - 7)^2 + (6 - 5)^2} = \sqrt{(-12)^2 + (1)^2} = \sqrt{144 + 1} = \sqrt{145}$ units.
Since $AB \neq BC \neq CA$,all three sides have different lengths.
Therefore,the triangle formed by these points is a scalene triangle.

(A) We know that every point on the $x$-axis is of the form $(x, 0)$. Let $P(x, 0)$ be a point on the $x$-axis at a distance of $2\sqrt{5}$ from the point $Q(7, -4)$.
Using the distance formula,$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $PQ = 2\sqrt{5}$,so $(PQ)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$.
Substituting the coordinates:
$(x - 7)^2 + (0 - (-4))^2 = 20$
$(x - 7)^2 + (4)^2 = 20$
$x^2 - 14x + 49 + 16 = 20$
$x^2 - 14x + 65 = 20$
$x^2 - 14x + 45 = 0$
Factoring the quadratic equation:
$x^2 - 9x - 5x + 45 = 0$
$x(x - 9) - 5(x - 9) = 0$
$(x - 9)(x - 5) = 0$
Thus,$x = 5$ or $x = 9$.
The points are $(5, 0)$ and $(9, 0)$.
There are $2$ such points.

(B) To determine the type of quadrilateral,we calculate the lengths of all four sides and the two diagonals using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Sides:
$AB = \sqrt{(7 - 2)^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$CD = \sqrt{(6 - 11)^2 + (-6 - (-1))^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$
$DA = \sqrt{(2 - 6)^2 + (-2 - (-6))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Diagonals:
$AC = \sqrt{(11 - 2)^2 + (-1 - (-2))^2} = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82}$
$BD = \sqrt{(6 - 7)^2 + (-6 - 3)^2} = \sqrt{(-1)^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}$
Since opposite sides are equal ($AB = CD$ and $BC = DA$) and the diagonals are equal $(AC = BD)$,the quadrilateral is a rectangle.

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(-3, -14)$ and $B(a, -5)$ with distance $d = 9$.
Substituting the values into the formula:
$9 = \sqrt{(a - (-3))^2 + (-5 - (-14))^2}$
$9 = \sqrt{(a + 3)^2 + (-5 + 14)^2}$
$9 = \sqrt{(a + 3)^2 + (9)^2}$
Squaring both sides:
$81 = (a + 3)^2 + 81$
Subtracting $81$ from both sides:
$(a + 3)^2 = 0$
Taking the square root:
$a + 3 = 0$
$a = -3$.
Thus,the required value of $a$ is $-3$.

(D) Let $P(h, k)$ be a point equidistant from $A(-5, 4)$ and $B(-1, 6)$.
By the distance formula,$PA = PB$,which implies $PA^2 = PB^2$.
$(-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
Expanding both sides:
$25 + h^2 + 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
Simplifying the equation:
$10h - 2h - 8k + 12k + 41 - 37 = 0$
$8h + 4k + 4 = 0$
Dividing by $4$,we get:
$2h + k + 1 = 0$
This is the equation of the perpendicular bisector of the line segment $AB$. Any point lying on this line will be equidistant from $A$ and $B$. Since a line contains an infinite number of points,there are infinitely many such points.
Replacing $h$ and $k$ with $x$ and $y$,the locus of all such points is $2x + y + 1 = 0$.

(N/A) The points are $A(-5, -2)$ and $B(4, -2)$.
Since the $y$-coordinates of both points are the same $(-2)$,the line segment $AB$ is a horizontal line parallel to the $x$-axis.
The perpendicular bisector of a horizontal line is a vertical line passing through the midpoint of the segment.
The midpoint $R$ of $AB$ is given by $\left(\frac{-5+4}{2}, \frac{-2-2}{2}\right) = \left(-\frac{1}{2}, -2\right)$.
The perpendicular bisector is the vertical line $x = -\frac{1}{2}$.
The point $Q$ lies on the $x$-axis,so its $y$-coordinate is $0$. Since it lies on the perpendicular bisector $x = -\frac{1}{2}$,the coordinates of $Q$ are $\left(-\frac{1}{2}, 0\right)$.
To identify the triangle $QAB$,we calculate the lengths of the sides:
$AB = \sqrt{(4 - (-5))^2 + (-2 - (-2))^2} = \sqrt{9^2 + 0^2} = 9$.
$QA = \sqrt{(-5 - (-0.5))^2 + (-2 - 0)^2} = \sqrt{(-4.5)^2 + (-2)^2} = \sqrt{20.25 + 4} = \sqrt{24.25}$.
$QB = \sqrt{(4 - (-0.5))^2 + (-2 - 0)^2} = \sqrt{(4.5)^2 + (-2)^2} = \sqrt{20.25 + 4} = \sqrt{24.25}$.
Since $QA = QB$,the triangle $QAB$ is an isosceles triangle.

(A) Let the points be $A(5, 1)$,$B(-2, -3)$,and $C(8, 2m)$.
Since the points are collinear,the area of the triangle formed by them must be $0$.
The formula for the area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given coordinates:
$\frac{1}{2} |5(-3 - 2m) + (-2)(2m - 1) + 8(1 - (-3))| = 0$
$5(-3 - 2m) - 2(2m - 1) + 8(4) = 0$
$-15 - 10m - 4m + 2 + 32 = 0$
$-14m + 19 = 0$
$14m = 19$
$m = \frac{19}{14}$

(N/A) According to the question,point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$.
This means $PA = QA$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,we have:
$\sqrt{(2 - 3)^2 + (-4 - 8)^2} = \sqrt{(2 - (-10))^2 + (-4 - y)^2}$
$\sqrt{(-1)^2 + (-12)^2} = \sqrt{(12)^2 + (-(4 + y))^2}$
$\sqrt{1 + 144} = \sqrt{144 + (4 + y)^2}$
$\sqrt{145} = \sqrt{144 + 16 + 8y + y^2}$
Squaring both sides:
$145 = 160 + 8y + y^2$
$y^2 + 8y + 15 = 0$
$(y + 5)(y + 3) = 0$
So,$y = -5$ or $y = -3$.
Now,calculating distance $PQ = \sqrt{(-10 - 3)^2 + (y - 8)^2} = \sqrt{(-13)^2 + (y - 8)^2} = \sqrt{169 + (y - 8)^2}$.
For $y = -3$,$PQ = \sqrt{169 + (-3 - 8)^2} = \sqrt{169 + 121} = \sqrt{290}$.
For $y = -5$,$PQ = \sqrt{169 + (-5 - 8)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}$.