Mix Examples - Coordinate Geometry Questions in English

(B) Let the points be $A(1,5), B(2,3),$ and $C(-2,-11)$.
To check if the points are collinear,we calculate the area of the triangle formed by these points using 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 values:
$\text{Area} = \frac{1}{2} |1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)|$
$\text{Area} = \frac{1}{2} |1(14) + 2(-16) + (-2)(2)|$
$\text{Area} = \frac{1}{2} |14 - 32 - 4|$
$\text{Area} = \frac{1}{2} |-22| = 11 \text{ square units}$.
Since the area of the triangle is not $0$,the points are not collinear.

(C) 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 $A(1, -1), B(-4, 6)$ and $C(-3, -5)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$
Area $= \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$
Area $= \frac{1}{2} |11 + 16 + 21|$
Area $= \frac{1}{2} |48| = 24$
Thus,the area of the triangle is $24$ square units.

(D) 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 $A(5, 2)$,$B(4, 7)$,and $C(7, -4)$.
Substituting the values into the formula:
Area $= \frac{1}{2} |5(7 - (-4)) + 4(-4 - 2) + 7(2 - 7)|$
Area $= \frac{1}{2} |5(11) + 4(-6) + 7(-5)|$
Area $= \frac{1}{2} |55 - 24 - 35|$
Area $= \frac{1}{2} |55 - 59|$
Area $= \frac{1}{2} |-4|$
Area $= \frac{1}{2} \times 4 = 2$ square units.

(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 $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given coordinates $P(-1.5, 3)$,$Q(6, -2)$,and $R(-3, 4)$:
Area $= \frac{1}{2} |-1.5(-2 - 4) + 6(4 - 3) + (-3)(3 - (-2))|$
Area $= \frac{1}{2} |-1.5(-6) + 6(1) - 3(5)|$
Area $= \frac{1}{2} |9 + 6 - 15|$
Area $= \frac{1}{2} |0| = 0$ square units.
Since the area is $0$,the points $P$,$Q$,and $R$ are collinear,meaning they lie on the same straight line and do not form a triangle.

(C) To find the area of the quadrilateral $ABCD$,we can divide it into two triangles by drawing a diagonal $BD$.
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: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
$1$. Area of $\Delta ABD$ with vertices $A(-5, 7)$,$B(-4, -5)$,and $D(4, 5)$:
$\text{Area} = \frac{1}{2} |-5(-5 - 5) + (-4)(5 - 7) + 4(7 - (-5))|$
$= \frac{1}{2} |-5(-10) - 4(-2) + 4(12)|$
$= \frac{1}{2} |50 + 8 + 48| = \frac{106}{2} = 53$ square units.
$2$. Area of $\Delta BCD$ with vertices $B(-4, -5)$,$C(-1, -6)$,and $D(4, 5)$:
$\text{Area} = \frac{1}{2} |-4(-6 - 5) + (-1)(5 - (-5)) + 4(-5 - (-6))|$
$= \frac{1}{2} |-4(-11) - 1(10) + 4(1)|$
$= \frac{1}{2} |44 - 10 + 4| = \frac{38}{2} = 19$ square units.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABD) + \text{Area}(\Delta BCD) = 53 + 19 = 72$ square units.

(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 $(x_1, y_1) = (2, 3), (x_2, y_2) = (-1, 0),$ and $(x_3, y_3) = (2, -4)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$
Area $= \frac{1}{2} |2(4) + (-1)(-7) + 2(3)|$
Area $= \frac{1}{2} |8 + 7 + 6|$
Area $= \frac{1}{2} |21| = 10.5$ square units.

(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 $8$.

(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 $AP = \frac{1}{2} AB$.

(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 $(h, k) = (x, y)$.

(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 $0$.

(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) 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.

(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.

(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.

(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.

(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 $= \frac{1}{2} |0 + 3(6) - 1(-6)| = \frac{1}{2} |18 + 6| = \frac{24}{2} = 12$ $sq. units$.

(B) 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 $(x_1, y_1) = (-5, -1)$,$(x_2, y_2) = (3, -5)$,and $(x_3, y_3) = (5, 2)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |(-5)(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))|$
Area $= \frac{1}{2} |(-5)(-7) + 3(3) + 5(4)|$
Area $= \frac{1}{2} |35 + 9 + 20|$
Area $= \frac{1}{2} |64| = 32$ square units.

(B) For points to be collinear,the area of the triangle formed by them must be equal to $0$.
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 points $(7, -2), (5, 1),$ and $(3, k)$ into the formula:
$\frac{1}{2} |7(1 - k) + 5(k - (-2)) + 3(-2 - 1)| = 0$
$\frac{1}{2} |7 - 7k + 5(k + 2) + 3(-3)| = 0$
$\frac{1}{2} |7 - 7k + 5k + 10 - 9| = 0$
$|8 - 2k| = 0$
$8 - 2k = 0$
$2k = 8$
$k = 4$

(C) For points to be collinear,the area of the triangle formed by them must be $0$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points $(8,1), (k,-4),$ and $(2,-5)$ into the formula:
$\frac{1}{2} |8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))| = 0$
$\frac{1}{2} |8(-4 + 5) + k(-6) + 2(1 + 4)| = 0$
$\frac{1}{2} |8(1) - 6k + 2(5)| = 0$
$|8 - 6k + 10| = 0$
$18 - 6k = 0$
$6k = 18$
$k = 3$

(1:4) Let the vertices of the triangle be $A(0,-1), B(2,1), C(0,3)$.
Let $D, E, F$ be the mid-points of the sides $AB, AC, BC$ respectively. The coordinates of $D, E,$ and $F$ are calculated using the mid-point formula:
$D = \left(\frac{0+2}{2}, \frac{-1+1}{2}\right) = (1,0)$
$E = \left(\frac{0+0}{2}, \frac{-1+3}{2}\right) = (0,1)$
$F = \left(\frac{2+0}{2}, \frac{1+3}{2}\right) = (1,2)$
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area of $\Delta DEF = \frac{1}{2} |1(1-2) + 0(2-0) + 1(0-1)| = \frac{1}{2} |-1 + 0 - 1| = \frac{1}{2} |-2| = 1$ square unit.
Area of $\Delta ABC = \frac{1}{2} |0(1-3) + 2(3-(-1)) + 0(-1-1)| = \frac{1}{2} |0 + 2(4) + 0| = \frac{1}{2} |8| = 4$ square units.
Therefore,the required ratio is $1:4$.

(B) Let the vertices of the quadrilateral be $A(-4,-2), B(-3,-5), C(3,-2),$ and $D(2,3)$. Join $AC$ to form two triangles $\triangle ABC$ and $\triangle ACD$.
Area of a triangle $= \frac{1}{2} |x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$
Area of $\triangle ABC = \frac{1}{2} |(-4){(-5)-(-2)} + (-3){(-2)-(-2)} + 3{(-2)-(-5)}|$
$= \frac{1}{2} |(-4)(-3) + (-3)(0) + 3(3)| = \frac{1}{2} |12 + 0 + 9| = \frac{21}{2} = 10.5$ square units.
Area of $\triangle ACD = \frac{1}{2} |(-4){(-2)-(3)} + 3{(3)-(-2)} + 2{(-2)-(-2)}|$
$= \frac{1}{2} |(-4)(-5) + 3(5) + 2(0)| = \frac{1}{2} |20 + 15 + 0| = \frac{35}{2} = 17.5$ square units.
Area of quadrilateral $ABCD = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD$
$= 10.5 + 17.5 = 28$ square units.

(C) Let the vertices of the triangle be $A (4, -6), B (3, -2),$ and $C (5, 2)$.
Let $D$ be the mid-point of side $BC$ of $\triangle ABC$. Therefore,$AD$ is the median of $\triangle ABC$.
Coordinates of point $D = \left(\frac{3+5}{2}, \frac{-2+2}{2}\right) = (4, 0)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area of $\triangle ABD = \frac{1}{2} |4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))|$
$= \frac{1}{2} |4(-2) + 3(6) + 4(-4)|$
$= \frac{1}{2} |-8 + 18 - 16| = \frac{1}{2} |-6| = 3$ square units.
Area of $\triangle ADC = \frac{1}{2} |4(0 - 2) + 4(2 - (-6)) + 5(-6 - 0)|$
$= \frac{1}{2} |4(-2) + 4(8) + 5(-6)|$
$= \frac{1}{2} |-8 + 32 - 30| = \frac{1}{2} |-6| = 3$ square units.
Since the area of $\triangle ABD = \text{Area of } \triangle ADC = 3$ square units,the median $AD$ divides $\triangle ABC$ into two triangles of equal areas.

(N/A) If the given points are collinear,then the area of the triangle formed by these points must be $0$.
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 $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given points $(x, y), (1, 2)$ and $(7, 0)$ into the formula:
$0 = \frac{1}{2} |x(2 - 0) + 1(0 - y) + 7(y - 2)|$
$0 = \frac{1}{2} |2x - y + 7y - 14|$
$0 = \frac{1}{2} |2x + 6y - 14|$
Multiplying by $2$ on both sides:
$2x + 6y - 14 = 0$
Dividing the entire equation by $2$:
$x + 3y - 7 = 0$
Thus,the required relation between $x$ and $y$ is $x + 3y - 7 = 0$.

(N/A) Taking $C$ as the origin $(0,0)$,$CB$ as the $x$-axis,and $CD$ as the $y$-axis:
$1$. The coordinates of the vertices are determined by counting the units from the origin $C$ along the axes.
$2$. The coordinates of the vertices $P, Q,$ and $R$ are $P(12, 2), Q(13, 6),$ and $R(10, 3)$.
$3$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$4$. Substituting the values:
$\text{Area} = \frac{1}{2} |12(6 - 3) + 13(3 - 2) + 10(2 - 6)|$
$\text{Area} = \frac{1}{2} |12(3) + 13(1) + 10(-4)|$
$\text{Area} = \frac{1}{2} |36 + 13 - 40|$
$\text{Area} = \frac{1}{2} |9| = 4.5 \text{ square units}$.

(C) $P$ is the mid-point of side $AB$. Therefore,the coordinates of $P$ are $\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right) = \left(-1, \frac{3}{2}\right)$.
Similarly,the coordinates of $Q, R$,and $S$ are $(2, 4)$,$\left(5, \frac{3}{2}\right)$,and $(2, -1)$ respectively.
Length of $PQ = \sqrt{(-1-2)^2 + \left(\frac{3}{2}-4\right)^2} = \sqrt{(-3)^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $QR = \sqrt{(2-5)^2 + \left(4-\frac{3}{2}\right)^2} = \sqrt{(-3)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $RS = \sqrt{(5-2)^2 + \left(\frac{3}{2}-(-1)\right)^2} = \sqrt{3^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $SP = \sqrt{(2-(-1))^2 + \left(-1-\frac{3}{2}\right)^2} = \sqrt{3^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of diagonal $PR = \sqrt{(-1-5)^2 + \left(\frac{3}{2}-\frac{3}{2}\right)^2} = \sqrt{(-6)^2 + 0^2} = 6$.
Length of diagonal $QS = \sqrt{(2-2)^2 + (4-(-1))^2} = \sqrt{0^2 + 5^2} = 5$.
Since all sides are equal $(PQ = QR = RS = SP = \sqrt{\frac{61}{4}})$ and the diagonals are not equal $(PR \neq QS)$,the quadrilateral $PQRS$ is a rhombus.

(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 $(x_{1}+x_{3}-x_{2}, y_{1}+y_{3}-y_{2})$.

(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.

(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:
$\frac{1}{2} |5(-3 - 2m) + (-2)(2m - 1) + 8(1 - (-3))| = 0$
$5(-3 - 2m) - 2(2m - 1) + 8(4) = 0$
$-15 - 10m - 4m + 2 + 32 = 0$
$-14m + 19 = 0$
$14m = 19$
$m = \frac{19}{14}$

(C) Given the vertices of the triangle are $(x_1, y_1) = (-8, 4)$,$(x_2, y_2) = (-6, 6)$,and $(x_3, y_3) = (-3, 9)$.
The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given values into the formula:
$\text{Area} = \frac{1}{2} |-8(6 - 9) + (-6)(9 - 4) + (-3)(4 - 6)|$
$= \frac{1}{2} |-8(-3) - 6(5) - 3(-2)|$
$= \frac{1}{2} |24 - 30 + 6|$
$= \frac{1}{2} |30 - 30| = \frac{1}{2}(0) = 0$
Since the area is $0$,the points are collinear,and the area of the triangle formed by them is $0$.

(D) Since $(a, b)$ is the mid-point of the line segment $AB$,we use the mid-point formula:
$(a, b) = \left(\frac{10 + k}{2}, \frac{-6 + 4}{2}\right)$
$(a, b) = \left(\frac{10 + k}{2}, -1\right)$
Equating the coordinates,we get:
$a = \frac{10 + k}{2}$ ... $(i)$
$b = -1$ ... $(ii)$
Given the equation $a - 2b = 18$,substitute $b = -1$:
$a - 2(-1) = 18$
$a + 2 = 18 \Rightarrow a = 16$
Now,substitute $a = 16$ into equation $(i)$:
$16 = \frac{10 + k}{2}$
$32 = 10 + k \Rightarrow k = 22$
Thus,the coordinates of $B$ are $(22, 4)$.
Now,calculate the distance $AB$ using the distance formula:
$AB = \sqrt{(22 - 10)^2 + (4 - (-6))^2}$
$AB = \sqrt{(12)^2 + (10)^2}$
$AB = \sqrt{144 + 100} = \sqrt{244}$
$AB = 2\sqrt{61}$ units.

(A-D) Given that the centre of the circle is $C(2a, a-7)$ and it passes through the point $P(11, -9)$.
The distance between the centre $C$ and the point $P$ on the circle is equal to the radius $r$.
Using the distance formula,$r = \sqrt{(11 - 2a)^2 + (-9 - (a - 7))^2} = \sqrt{(11 - 2a)^2 + (-2 - a)^2}$.
The diameter of the circle is $10\sqrt{2}$ units,so the radius $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ units.
Equating the two expressions for the radius:
$5\sqrt{2} = \sqrt{(11 - 2a)^2 + (-2 - a)^2}$.
Squaring both sides:
$(5\sqrt{2})^2 = (11 - 2a)^2 + (-2 - a)^2$
$50 = (121 - 44a + 4a^2) + (4 + 4a + a^2)$
$50 = 5a^2 - 40a + 125$
$5a^2 - 40a + 75 = 0$.
Dividing by $5$:
$a^2 - 8a + 15 = 0$
$a^2 - 5a - 3a + 15 = 0$
$a(a - 5) - 3(a - 5) = 0$
$(a - 5)(a - 3) = 0$.
Therefore,the values of $a$ are $a = 5$ or $a = 3$.

(B) Let $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ be the vertices of $\triangle ABC$.
Given that $D\left(-\frac{1}{2}, \frac{5}{2}\right)$,$E(7, 3)$,and $F\left(\frac{7}{2}, \frac{7}{2}\right)$ are the midpoints of sides $BC$,$CA$,and $AB$ respectively.
Using the midpoint formula,we have:
For $BC$: $\frac{x_2+x_3}{2} = -\frac{1}{2} \Rightarrow x_2+x_3 = -1$ $(i)$ and $\frac{y_2+y_3}{2} = \frac{5}{2} \Rightarrow y_2+y_3 = 5$ $(ii)$.
For $CA$: $\frac{x_3+x_1}{2} = 7 \Rightarrow x_3+x_1 = 14$ $(iii)$ and $\frac{y_3+y_1}{2} = 3 \Rightarrow y_3+y_1 = 6$ $(iv)$.
For $AB$: $\frac{x_1+x_2}{2} = \frac{7}{2} \Rightarrow x_1+x_2 = 7$ $(v)$ and $\frac{y_1+y_2}{2} = \frac{7}{2} \Rightarrow y_1+y_2 = 7$ $(vi)$.
Adding $(i), (iii), (v)$: $2(x_1+x_2+x_3) = 20 \Rightarrow x_1+x_2+x_3 = 10$ $(vii)$.
Subtracting $(i), (iii), (v)$ from $(vii)$ gives $x_1=11, x_2=-4, x_3=3$.
Adding $(ii), (iv), (vi)$: $2(y_1+y_2+y_3) = 18 \Rightarrow y_1+y_2+y_3 = 9$ $(viii)$.
Subtracting $(ii), (iv), (vi)$ from $(viii)$ gives $y_1=4, y_2=3, y_3=2$.
Vertices are $A(11, 4), B(-4, 3), C(3, 2)$.
Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$= \frac{1}{2} |11(3-2) + (-4)(2-4) + 3(4-3)| = \frac{1}{2} |11(1) + (-4)(-2) + 3(1)| = \frac{1}{2} |11+8+3| = \frac{22}{2} = 11$ sq units.

(A) Given that the points $A(2, 9)$,$B(a, 5)$,and $C(5, 5)$ are the vertices of a $\triangle ABC$ right-angled at $B$.
By the Pythagoras theorem,$AC^2 = AB^2 + BC^2$ $(i)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(a - 2)^2 + (5 - 9)^2} = \sqrt{a^2 - 4a + 4 + 16} = \sqrt{a^2 - 4a + 20}$.
$BC = \sqrt{(5 - a)^2 + (5 - 5)^2} = \sqrt{(5 - a)^2} = |5 - a|$.
$AC = \sqrt{(5 - 2)^2 + (5 - 9)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Substituting these into $(i)$:
$5^2 = (\sqrt{a^2 - 4a + 20})^2 + (5 - a)^2$.
$25 = a^2 - 4a + 20 + 25 - 10a + a^2$.
$2a^2 - 14a + 20 = 0$.
$a^2 - 7a + 10 = 0$.
$(a - 2)(a - 5) = 0$.
So,$a = 2$ or $a = 5$.
If $a = 5$,then $B$ coincides with $C$,which is not possible for a triangle. Thus,$a = 2$.
With $a = 2$,the vertices are $A(2, 9)$,$B(2, 5)$,and $C(5, 5)$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB$.
$BC = |5 - 2| = 3$ units.
$AB = |9 - 5| = 4$ units.
Area $= \frac{1}{2} \times 3 \times 4 = 6$ sq units.

(A) We know that if three points are collinear,the area of the triangle formed by these points is zero.
Since the points $A(k+1, 2k)$,$B(3k, 2k+3)$,and $C(5k-1, 5k)$ are collinear,the area of $\triangle ABC = 0$.
The formula for the area of a triangle is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Here,$x_1 = k+1, x_2 = 3k, x_3 = 5k-1$ and $y_1 = 2k, y_2 = 2k+3, y_3 = 5k$.
Substituting these values:
$\frac{1}{2} [(k+1)(2k+3 - 5k) + 3k(5k - 2k) + (5k-1)(2k - (2k+3))] = 0$
$\frac{1}{2} [(k+1)(-3k+3) + 3k(3k) + (5k-1)(-3)] = 0$
$(-3k^2 + 3k - 3k + 3) + 9k^2 - 15k + 3 = 0$
$6k^2 - 15k + 6 = 0$
Dividing by $3$,we get $2k^2 - 5k + 2 = 0$.
Factoring the quadratic equation: $2k^2 - 4k - k + 2 = 0 \Rightarrow 2k(k-2) - 1(k-2) = 0$.
$(k-2)(2k-1) = 0$.
Thus,$k = 2$ or $k = \frac{1}{2}$.

(N/A) Let the vertices of the triangle be $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3).$ The mid-points are given as $D(3, 4)$ on $AB,$ $E(8, 9)$ on $BC,$ and $F(6, 7)$ on $AC.$
Using the mid-point formula:
$1) \frac{x_1 + x_2}{2} = 3, \frac{y_1 + y_2}{2} = 4 \implies x_1 + x_2 = 6, y_1 + y_2 = 8$
$2) \frac{x_2 + x_3}{2} = 8, \frac{y_2 + y_3}{2} = 9 \implies x_2 + x_3 = 16, y_2 + y_3 = 18$
$3) \frac{x_1 + x_3}{2} = 6, \frac{y_1 + y_3}{2} = 7 \implies x_1 + x_3 = 12, y_1 + y_3 = 14$
Adding these equations:
$(x_1 + x_2) + (x_2 + x_3) + (x_1 + x_3) = 6 + 16 + 12 = 34 \implies 2(x_1 + x_2 + x_3) = 34 \implies x_1 + x_2 + x_3 = 17$
$(y_1 + y_2) + (y_2 + y_3) + (y_1 + y_3) = 8 + 18 + 14 = 40 \implies 2(y_1 + y_2 + y_3) = 40 \implies y_1 + y_2 + y_3 = 20$
Solving for individual coordinates:
$x_1 = (x_1 + x_2 + x_3) - (x_2 + x_3) = 17 - 16 = 1$
$x_2 = (x_1 + x_2 + x_3) - (x_1 + x_3) = 17 - 12 = 5$
$x_3 = (x_1 + x_2 + x_3) - (x_1 + x_2) = 17 - 6 = 11$
$y_1 = (y_1 + y_2 + y_3) - (y_2 + y_3) = 20 - 18 = 2$
$y_2 = (y_1 + y_2 + y_3) - (y_1 + y_3) = 20 - 14 = 6$
$y_3 = (y_1 + y_2 + y_3) - (y_1 + y_2) = 20 - 8 = 12$
Thus,the vertices are $A(1, 2), B(5, 6),$ and $C(11, 12).$

(D) Let the vertices of the equilateral triangle be $A(-4,3)$,$B(4,3)$,and $C(x,y)$.
Since the triangle is equilateral,$AB = BC = CA$,which implies $AB^2 = BC^2 = CA^2$.
First,calculate $AB^2 = (4 - (-4))^2 + (3 - 3)^2 = 8^2 + 0^2 = 64$.
For $BC^2 = CA^2$,we have $(x-4)^2 + (y-3)^2 = (x+4)^2 + (y-3)^2$.
This simplifies to $(x-4)^2 = (x+4)^2$,which gives $x^2 - 8x + 16 = x^2 + 8x + 16$,so $16x = 0$,hence $x = 0$.
Now,use $AB^2 = BC^2$ with $x=0$: $64 = (0-4)^2 + (y-3)^2$.
$64 = 16 + (y-3)^2$,so $(y-3)^2 = 48$.
$y-3 = \pm \sqrt{48} = \pm 4\sqrt{3}$.
Thus,$y = 3 \pm 4\sqrt{3}$.
The two possible vertices are $C_1(0, 3+4\sqrt{3})$ and $C_2(0, 3-4\sqrt{3})$.
The origin $(0,0)$ lies in the interior of the triangle. For the triangle with vertices $(-4,3)$,$(4,3)$,and $(0, y)$,the $y$-coordinate of the third vertex must be less than $3$ for the origin to be inside. Since $3+4\sqrt{3} > 3$ and $3-4\sqrt{3} < 3$,the correct vertex is $(0, 3-4\sqrt{3})$.

(A) Given that $A(6,1), B(8,2)$ and $C(9,4)$ are three vertices of a parallelogram $ABCD$.
Let the fourth vertex $D$ be $(x, y)$.
Since the diagonals of a parallelogram bisect each other,the midpoint of $BD$ is the same as the midpoint of $AC$.
$\text{Midpoint of } AC = \left(\frac{6+9}{2}, \frac{1+4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)$.
$\text{Midpoint of } BD = \left(\frac{8+x}{2}, \frac{2+y}{2}\right)$.
Equating the coordinates: $\frac{8+x}{2} = \frac{15}{2} \Rightarrow x = 7$ and $\frac{2+y}{2} = \frac{5}{2} \Rightarrow y = 3$.
So,the fourth vertex is $D(7,3)$.
$E$ is the midpoint of $DC$,so $E = \left(\frac{7+9}{2}, \frac{3+4}{2}\right) = \left(8, \frac{7}{2}\right)$.
The area of $\triangle ADE$ with vertices $A(6,1), D(7,3)$ and $E(8, 3.5)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |6(3 - 3.5) + 7(3.5 - 1) + 8(1 - 3)|$
$\text{Area} = \frac{1}{2} |6(-0.5) + 7(2.5) + 8(-2)|$
$\text{Area} = \frac{1}{2} |-3 + 17.5 - 16|$
$\text{Area} = \frac{1}{2} |-1.5| = 0.75 = \frac{3}{4} \text{ sq. units.}$

(N/A) Given that the points $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ are the vertices of $\triangle ABC$.
$(i)$ We know that the median bisects the line segment into two equal parts,i.e.,$D$ is the mid-point of $BC$.
$\therefore$ Coordinate of mid-point of $BC = \left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$.
$\Rightarrow D = \left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$.
$(ii)$ Let the coordinates of point $P$ be $(x, y)$.
Given that the point $P(x, y)$ divides the line segment joining $A(x_{1}, y_{1})$ and $D\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$ in the ratio $2 : 1$.
Using the internal section formula $\left(\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right)$:
$P = \left[\frac{2\left(\frac{x_{2}+x_{3}}{2}\right) + 1(x_{1})}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right) + 1(y_{1})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
$(iii)$ Similarly,for median $BE$,$E$ is the mid-point of $AC$,so $E = \left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)$.
Point $Q$ divides $BE$ in ratio $2 : 1$. Using the section formula:
$Q = \left[\frac{2\left(\frac{x_{1}+x_{3}}{2}\right) + 1(x_{2})}{2+1}, \frac{2\left(\frac{y_{1}+y_{3}}{2}\right) + 1(y_{2})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
For median $CF$,$F$ is the mid-point of $AB$,so $F = \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$.
Point $R$ divides $CF$ in ratio $2 : 1$. Using the section formula:
$R = \left[\frac{2\left(\frac{x_{1}+x_{2}}{2}\right) + 1(x_{3})}{2+1}, \frac{2\left(\frac{y_{1}+y_{2}}{2}\right) + 1(y_{3})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
$(iv)$ The centroid of a triangle is the point where all medians intersect. Since $P, Q,$ and $R$ are all the same point,the coordinates of the centroid are $\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.

(A) In a parallelogram,the diagonals bisect each other,which means the mid-point of $AC$ is equal to the mid-point of $BD$.
$\Rightarrow \left(\frac{1+a}{2}, \frac{-2+2}{2}\right) = \left(\frac{2-4}{2}, \frac{3-3}{2}\right)$
$\Rightarrow \frac{1+a}{2} = \frac{-2}{2} = -1$
$1+a = -2 \Rightarrow a = -3$
So,the value of $a$ is $-3$.
Given $AB$ as the base,let $DP$ be the height where $P$ is the foot of the perpendicular from $D$ to $AB$.
The equation of line $AB$ passing through $(1, -2)$ and $(2, 3)$ is:
$(y - (-2)) = \frac{3 - (-2)}{2 - 1}(x - 1)$
$y + 2 = 5(x - 1) \Rightarrow 5x - y = 7$ ... $(i)$
The slope of $AB$ is $m_1 = 5$. Since $DP \perp AB$,the slope of $DP$ is $m_2 = -1/5$.
The equation of line $DP$ passing through $D(-4, -3)$ with slope $-1/5$ is:
$(y + 3) = -\frac{1}{5}(x + 4)$
$5y + 15 = -x - 4 \Rightarrow x + 5y = -19$ ... $(ii)$
Solving $(i)$ and $(ii)$ for intersection point $P$:
$x + 5(5x - 7) = -19 \Rightarrow 26x = 16 \Rightarrow x = 8/13$
$y = 5(8/13) - 7 = -51/13$
$P = (8/13, -51/13)$. The height $DP$ is the distance between $D(-4, -3)$ and $P(8/13, -51/13)$:
$DP = \sqrt{(\frac{8}{13} + 4)^2 + (-\frac{51}{13} + 3)^2} = \sqrt{(\frac{60}{13})^2 + (-\frac{12}{13})^2} = \frac{1}{13}\sqrt{3600 + 144} = \frac{\sqrt{3744}}{13} = \frac{12\sqrt{26}}{13}$ units.

(D) Yes,from the figure,we observe that the positions of four students $A, B, C$ and $D$ are $(3, 5), (7, 9), (11, 5)$ and $(7, 1)$ respectively. These are the four vertices of a quadrilateral.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(7 - 3)^2 + (9 - 5)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$BC = \sqrt{(11 - 7)^2 + (5 - 9)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$CD = \sqrt{(7 - 11)^2 + (1 - 5)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$DA = \sqrt{(3 - 7)^2 + (5 - 1)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Since all sides are equal,it is a rhombus.
Next,we calculate the lengths of the diagonals:
$AC = \sqrt{(11 - 3)^2 + (5 - 5)^2} = \sqrt{8^2 + 0^2} = 8$
$BD = \sqrt{(7 - 7)^2 + (1 - 9)^2} = \sqrt{0^2 + (-8)^2} = 8$
Since the diagonals are also equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.
In a square,the point equidistant from all four vertices is the intersection point of the diagonals,which is the midpoint of either diagonal.
Using the midpoint formula for $AC$:
$P = \left(\frac{3 + 11}{2}, \frac{5 + 5}{2}\right) = \left(\frac{14}{2}, \frac{10}{2}\right) = (7, 5)$
Thus,Jaspal should be placed at position $(7, 5)$.

(A) Let the vertices be $A(-2,-3), B(6,3), C(3,7),$ and $D(-5,1).$
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (-2-6)^2 + (-3-3)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100 \implies AB = 10$
$BC^2 = (6-3)^2 + (3-7)^2 = (3)^2 + (-4)^2 = 9 + 16 = 25 \implies BC = 5$
$CD^2 = (3 - (-5))^2 + (7-1)^2 = (8)^2 + (6)^2 = 64 + 36 = 100 \implies CD = 10$
$DA^2 = (-5 - (-2))^2 + (1 - (-3))^2 = (-3)^2 + (4)^2 = 9 + 16 = 25 \implies DA = 5$
Since opposite sides are equal ($AB=CD=10$ and $BC=DA=5$),the quadrilateral is a parallelogram.
Next,we check the diagonal $AC$:
$AC^2 = (-2-3)^2 + (-3-7)^2 = (-5)^2 + (-10)^2 = 25 + 100 = 125$
In $\triangle ABC$,$AB^2 + BC^2 = 100 + 25 = 125 = AC^2$.
By the converse of the Pythagorean theorem,$\angle B = 90^\circ$.
Since the parallelogram has one right angle,it is a rectangle.

(D) Three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ are collinear if the slope of the line segment joining the first two points is equal to the slope of the line segment joining the last two points.
Slope $m_1$ of the line segment joining $(-8, -4)$ and $(-2, 4)$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-4)}{-2 - (-8)} = \frac{8}{6} = \frac{4}{3}$.
Slope $m_2$ of the line segment joining $(-2, 4)$ and $(5, a)$ is given by $m_2 = \frac{a - 4}{5 - (-2)} = \frac{a - 4}{7}$.
Since the points are collinear,$m_1 = m_2$.
Therefore,$\frac{4}{3} = \frac{a - 4}{7}$.
Multiplying both sides by $21$,we get $4 \times 7 = 3(a - 4)$.
$28 = 3a - 12$.
$3a = 28 + 12 = 40$.
$a = \frac{40}{3}$.

(D) Given that $\angle A = 90^{\circ}$,the sides $AB$ and $AC$ are perpendicular to each other.
Therefore,the product of their slopes must be $-1$,i.e.,$m_{AB} \times m_{AC} = -1$.
The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $AB$ $(m_{AB})$ = $\frac{4 - 7}{2 - 1} = \frac{-3}{1} = -3$.
Slope of $AC$ $(m_{AC})$ = $\frac{5 - 7}{k - 1} = \frac{-2}{k - 1}$.
Since $m_{AB} \times m_{AC} = -1$,we have $(-3) \times \left( \frac{-2}{k - 1} \right) = -1$.
$\frac{6}{k - 1} = -1$.
$6 = -(k - 1)$.
$6 = -k + 1$.
$k = 1 - 6 = -5$.

(A) Let $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ be the vertices of $\Delta ABC$. Given that $P(0, 1/2)$,$Q(1/2, 1/2)$,and $R(1/2, 0)$ are the midpoints of sides $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively.
For $x$-coordinates:
$(x_1 + x_2)/2 = 0 \implies x_1 + x_2 = 0$ $(i)$
$(x_2 + x_3)/2 = 1/2 \implies x_2 + x_3 = 1$ (ii)
$(x_3 + x_1)/2 = 1/2 \implies x_3 + x_1 = 1$ (iii)
Adding $(i)$,(ii),and (iii): $2(x_1 + x_2 + x_3) = 2 \implies x_1 + x_2 + x_3 = 1$.
Subtracting (ii) from the sum: $x_1 = 1 - 1 = 0$.
Subtracting (iii) from the sum: $x_2 = 1 - 1 = 0$.
Subtracting $(i)$ from the sum: $x_3 = 1 - 0 = 1$.
For $y$-coordinates:
$(y_1 + y_2)/2 = 1/2 \implies y_1 + y_2 = 1$ (iv)
$(y_2 + y_3)/2 = 1/2 \implies y_2 + y_3 = 1$ $(v)$
$(y_3 + y_1)/2 = 0 \implies y_3 + y_1 = 0$ (vi)
Adding (iv),$(v)$,and (vi): $2(y_1 + y_2 + y_3) = 2 \implies y_1 + y_2 + y_3 = 1$.
Subtracting $(v)$ from the sum: $y_1 = 1 - 1 = 0$.
Subtracting (vi) from the sum: $y_2 = 1 - 0 = 1$.
Subtracting (iv) from the sum: $y_3 = 1 - 1 = 0$.
Thus,the vertices are $A(0, 0)$,$B(0, 1)$,and $C(1, 0)$.