Mix Examples - Coordinate Geometry Questions in English

(N/A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Let the midpoints be $D(3,1)$,$E(5,6)$,and $F(-3,2)$.
Using the midpoint formula,we have:
$(x_1+x_2)/2 = 3, (y_1+y_2)/2 = 1 \implies x_1+x_2 = 6, y_1+y_2 = 2$ (Eq. $1$)
$(x_2+x_3)/2 = 5, (y_2+y_3)/2 = 6 \implies x_2+x_3 = 10, y_2+y_3 = 12$ (Eq. $2$)
$(x_3+x_1)/2 = -3, (y_3+y_1)/2 = 2 \implies x_3+x_1 = -6, y_3+y_1 = 4$ (Eq. $3$)
Adding all equations: $2(x_1+x_2+x_3) = 6+10-6 = 10 \implies x_1+x_2+x_3 = 5$.
Similarly,$y_1+y_2+y_3 = (2+12+4)/2 = 9$.
Subtracting Eq. $2$ from the sum: $x_1 = 5-10 = -5$ and $y_1 = 9-12 = -3$.
Subtracting Eq. $3$ from the sum: $x_2 = 5-(-6) = 11$ and $y_2 = 9-4 = 5$.
Subtracting Eq. $1$ from the sum: $x_3 = 5-6 = -1$ and $y_3 = 9-2 = 7$.
The vertices are $(-5,-3), (11,5), (-1,7)$.

(A) In a parallelogram,the diagonals bisect each other. This means the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = ((-2+x)/2, (1+3)/2) = ((-2+x)/2, 2)$.
Midpoint of $BD = ((1+1)/2, (0+y)/2) = (1, y/2)$.
Equating the coordinates of the midpoints:
$(-2+x)/2 = 1 \implies -2+x = 2 \implies x = 4$.
$y/2 = 2 \implies y = 4$.
Therefore,the values are $x=4$ and $y=4$.

(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 $A(8, 2)$,$B(2, 4)$,and $C(6, 7)$.
Here,$x_1 = 8, y_1 = 2, x_2 = 2, y_2 = 4, x_3 = 6, y_3 = 7$.
Substituting these values into the formula:
Area $= \frac{1}{2} |8(4 - 7) + 2(7 - 2) + 6(2 - 4)|$
Area $= \frac{1}{2} |8(-3) + 2(5) + 6(-2)|$
Area $= \frac{1}{2} |-24 + 10 - 12|$
Area $= \frac{1}{2} |-26|$
Area $= \frac{26}{2} = 13$.
Therefore,the area of $\Delta ABC$ is $13$ square units.

(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 $A(a, 5), B(6, 7),$ and $C(2, 3)$ with Area $= 10$.
Substituting the values into the formula:
$10 = \frac{1}{2} |a(7 - 3) + 6(3 - 5) + 2(5 - 7)|$
$20 = |a(4) + 6(-2) + 2(-2)|$
$20 = |4a - 12 - 4|$
$20 = |4a - 16|$
This implies two cases:
Case $1$: $4a - 16 = 20 \implies 4a = 36 \implies a = 9$
Case $2$: $4a - 16 = -20 \implies 4a = -4 \implies a = -1$
Thus,the possible values for $a$ are $9$ or $-1$.

(B) Let the vertices of the pentagon be $A(4,3), B(-5,6), C(-7,-2), D(0,-7),$ and $E(3,-6)$.
We can divide the pentagon into three triangles: $\Delta ABE, \Delta BCE,$ and $\Delta CDE$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
$1$. Area of $\Delta ABE$ with vertices $A(4,3), B(-5,6), E(3,-6)$:
$= \frac{1}{2} |4(6 - (-6)) + (-5)(-6 - 3) + 3(3 - 6)|$
$= \frac{1}{2} |4(12) + (-5)(-9) + 3(-3)|$
$= \frac{1}{2} |48 + 45 - 9| = \frac{1}{2} |84| = 42$ sq units.
$2$. Area of $\Delta BCE$ with vertices $B(-5,6), C(-7,-2), E(3,-6)$:
$= \frac{1}{2} |-5(-2 - (-6)) + (-7)(-6 - 6) + 3(6 - (-2))|$
$= \frac{1}{2} |-5(4) + (-7)(-12) + 3(8)|$
$= \frac{1}{2} |-20 + 84 + 24| = \frac{1}{2} |88| = 44$ sq units.
$3$. Area of $\Delta CDE$ with vertices $C(-7,-2), D(0,-7), E(3,-6)$:
$= \frac{1}{2} |-7(-7 - (-6)) + 0(-6 - (-2)) + 3(-2 - (-7))|$
$= \frac{1}{2} |-7(-1) + 0 + 3(5)|$
$= \frac{1}{2} |7 + 15| = \frac{1}{2} |22| = 11$ sq units.
Total area of pentagon $ABCDE = \text{Area}(\Delta ABE) + \text{Area}(\Delta BCE) + \text{Area}(\Delta CDE)$
$= 42 + 44 + 11 = 97$ sq units.

(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 $A(3, 2), B(11, 8)$ and $C(8, 12)$.
Here,$x_1 = 3, y_1 = 2, x_2 = 11, y_2 = 8, x_3 = 8, y_3 = 12$.
Substituting these values into the formula:
Area $= \frac{1}{2} |3(8 - 12) + 11(12 - 2) + 8(2 - 8)|$
Area $= \frac{1}{2} |3(-4) + 11(10) + 8(-6)|$
Area $= \frac{1}{2} |-12 + 110 - 48|$
Area $= \frac{1}{2} |50|$
Area $= 25$ square units.

(N/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 $A(5, a)$,$B(2, 5)$,and $C(2, 3)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |5(5 - 3) + 2(3 - a) + 2(a - 5)|$
Area $= \frac{1}{2} |5(2) + 6 - 2a + 2a - 10|$
Area $= \frac{1}{2} |10 + 6 - 10|$
Area $= \frac{1}{2} |6|$
Area $= 3$ square units.
Since the result is independent of $a$,the area of the triangle is $3$ square units for every $a \in R$.

(A) Let the vertices of the quadrilateral be $A(-1, -1), B(-4, -2), C(-5, -4),$ and $D(-2, -3)$.
To find the area,we divide the quadrilateral into two triangles,$\triangle ABC$ and $\triangle ADC$.
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 $\triangle ABC = \frac{1}{2} |(-1)(-2 - (-4)) + (-4)(-4 - (-1)) + (-5)(-1 - (-2))|$
$= \frac{1}{2} |(-1)(2) + (-4)(-3) + (-5)(1)| = \frac{1}{2} |-2 + 12 - 5| = \frac{1}{2} |5| = 2.5$.
Area of $\triangle ADC = \frac{1}{2} |(-1)(-4 - (-3)) + (-5)(-3 - (-1)) + (-2)(-1 - (-4))|$
$= \frac{1}{2} |(-1)(-1) + (-5)(-2) + (-2)(3)| = \frac{1}{2} |1 + 10 - 6| = \frac{1}{2} |5| = 2.5$.
Total Area = Area of $\triangle ABC +$ Area of $\triangle ADC = 2.5 + 2.5 = 5$ square 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 $A(5, 3)$,$B(4, 5)$,and $C(3, 1)$.
Here,$x_1 = 5, y_1 = 3, x_2 = 4, y_2 = 5, x_3 = 3, y_3 = 1$.
Substituting these values into the formula:
Area $= \frac{1}{2} |5(5 - 1) + 4(1 - 3) + 3(3 - 5)|$
Area $= \frac{1}{2} |5(4) + 4(-2) + 3(-2)|$
Area $= \frac{1}{2} |20 - 8 - 6|$
Area $= \frac{1}{2} |6| = 3 \text{ square units}$.
Thus,the correct option is $B$.

(N/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 $(x_1, y_1) = (t, t-2)$,$(x_2, y_2) = (t+2, t+2)$,and $(x_3, y_3) = (t+3, t)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |t((t+2) - t) + (t+2)(t - (t-2)) + (t+3)((t-2) - (t+2))|$
Area $= \frac{1}{2} |t(2) + (t+2)(2) + (t+3)(-4)|$
Area $= \frac{1}{2} |2t + 2t + 4 - 4t - 12|$
Area $= \frac{1}{2} |(2t + 2t - 4t) + (4 - 12)|$
Area $= \frac{1}{2} |0 - 8| = \frac{1}{2} |-8| = 4$ square units.
Since the result is $4$,which is a constant,the area of the triangle does not depend on the value of $t$.

(D) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base is the length of side $\overline{BC}$ and the height is the length of the altitude from $A$ to $\overline{BC}$.
First,calculate the length of $\overline{BC}$ using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
For $B(-2, 1)$ and $C(6, 3)$,$BC = \sqrt{(6 - (-2))^2 + (3 - 1)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$.
Given the area of $\Delta ABC = 28$,we have: $28 = \frac{1}{2} \times (2\sqrt{17}) \times h$.
$28 = \sqrt{17} \times h$.
Therefore,the length of the altitude $h = \frac{28}{\sqrt{17}}$.

(A) To find the area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, we use the shoelace formula:
Area $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1)|$
Given vertices: $(1, 5), (-2, 4), (-3, -1), (2, -3), (5, 1)$.
Sum $1 = (1 \times 4) + (-2 \times -1) + (-3 \times -3) + (2 \times 1) + (5 \times 5) = 4 + 2 + 9 + 2 + 25 = 42$
Sum $2 = (5 \times -2) + (4 \times -3) + (-1 \times 2) + (-3 \times 5) + (1 \times 1) = -10 - 12 - 2 - 15 + 1 = -38$
Area $= \frac{1}{2} |42 - (-38)| = \frac{1}{2} |42 + 38| = \frac{1}{2} |80| = 40$ square units.

(B) The area of the triangle formed by joining the midpoints of the sides of a triangle is one-fourth the area of the original triangle.
First,calculate the area of $\Delta DEF$ using the coordinates $D(2, 1), E(-2, 3),$ and $F(4, -3)$.
Area of $\Delta DEF = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area of $\Delta DEF = \frac{1}{2} |2(3 - (-3)) + (-2)(-3 - 1) + 4(1 - 3)|$
Area of $\Delta DEF = \frac{1}{2} |2(6) + (-2)(-4) + 4(-2)|$
Area of $\Delta DEF = \frac{1}{2} |12 + 8 - 8| = \frac{1}{2} |12| = 6 \text{ sq units}$.
Since $\text{Area}(\Delta ABC) = 4 \times \text{Area}(\Delta DEF)$,
$\text{Area}(\Delta ABC) = 4 \times 6 = 24 \text{ sq units}$.

(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 $A(h, k), B(1, 1),$ and $C(2, 1)$.
Area $= \frac{1}{2} |h(1 - 1) + 1(1 - k) + 2(k - 1)| = 1$
$\frac{1}{2} |0 + 1 - k + 2k - 2| = 1$
$\frac{1}{2} |k - 1| = 1$
$|k - 1| = 2$
This implies $k - 1 = 2$ or $k - 1 = -2$.
If $k - 1 = 2$,then $k = 3$.
If $k - 1 = -2$,then $k = -1$.
Therefore,the possible values of $k$ are $3$ or $-1$.

(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: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $A(2, 1)$,$B(3, -2)$,and $C(x, x+3)$.
Substituting the values into the formula: $5 = \frac{1}{2} |2(-2 - (x+3)) + 3((x+3) - 1) + x(1 - (-2))|$.
$10 = |2(-x - 5) + 3(x + 2) + x(3)|$.
$10 = |-2x - 10 + 3x + 6 + 3x|$.
$10 = |4x - 4|$.
This gives two cases:
Case $1$: $4x - 4 = 10 \Rightarrow 4x = 14 \Rightarrow x = \frac{7}{2}$. Then $y = \frac{7}{2} + 3 = \frac{13}{2}$.
Case $2$: $4x - 4 = -10 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$. Then $y = -\frac{3}{2} + 3 = \frac{3}{2}$.
Thus,the coordinates of the third vertex are $(\frac{7}{2}, \frac{13}{2})$ and $(-\frac{3}{2}, \frac{3}{2})$.

(A) The coordinates of the vertices of the triangle are $(x_1, y_1) = (1, 2)$,$(x_2, y_2) = (3, 3)$,and $(x_3, y_3) = (5, 1)$.
The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2)$,and $(x_3, y_3)$ is given by:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Substituting the given values:
$G = \left( \frac{1 + 3 + 5}{3}, \frac{2 + 3 + 1}{3} \right)$
$G = \left( \frac{9}{3}, \frac{6}{3} \right)$
$G = (3, 2)$
Thus,the centroid of the triangle is $(3, 2)$.

(A) Let the vertices be $A(3,0), B(-1,-6),$ and $C(4,-1)$. Let the circumcentre be $O(x,y)$.
Since $O$ is the circumcentre,$OA^2 = OB^2 = OC^2$.
$OA^2 = (x-3)^2 + (y-0)^2 = x^2 - 6x + 9 + y^2$
$OB^2 = (x+1)^2 + (y+6)^2 = x^2 + 2x + 1 + y^2 + 12y + 36 = x^2 + y^2 + 2x + 12y + 37$
$OC^2 = (x-4)^2 + (y+1)^2 = x^2 - 8x + 16 + y^2 + 2y + 1 = x^2 + y^2 - 8x + 2y + 17$
Equating $OA^2 = OB^2$: $x^2 - 6x + 9 + y^2 = x^2 + y^2 + 2x + 12y + 37 \implies -8x - 12y = 28 \implies 2x + 3y = -7$ (Equation $1$).
Equating $OB^2 = OC^2$: $x^2 + y^2 + 2x + 12y + 37 = x^2 + y^2 - 8x + 2y + 17 \implies 10x + 10y = -20 \implies x + y = -2$ (Equation $2$).
Solving Equations $1$ and $2$: From $2$,$x = -2 - y$. Substituting in $1$: $2(-2-y) + 3y = -7 \implies -4 - 2y + 3y = -7 \implies y = -3$.
Then $x = -2 - (-3) = 1$. So,the circumcentre is $(1, -3)$.
The circumradius $R = OA = \sqrt{(1-3)^2 + (-3-0)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$.

(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$OA^2 = OB^2 = OC^2$.
Given $A(1, 2)$,$B(-2, 2)$,and $C(1, 5)$.
$OA^2 = (x - 1)^2 + (y - 2)^2$
$OB^2 = (x + 2)^2 + (y - 2)^2$
$OC^2 = (x - 1)^2 + (y - 5)^2$
Equating $OA^2 = OB^2$:
$(x - 1)^2 + (y - 2)^2 = (x + 2)^2 + (y - 2)^2$
$(x - 1)^2 = (x + 2)^2$
$x^2 - 2x + 1 = x^2 + 4x + 4$
$-6x = 3 \implies x = -0.5$
Equating $OA^2 = OC^2$:
$(x - 1)^2 + (y - 2)^2 = (x - 1)^2 + (y - 5)^2$
$(y - 2)^2 = (y - 5)^2$
$y^2 - 4y + 4 = y^2 - 10y + 25$
$6y = 21 \implies y = 3.5$
Thus,the circumcentre is $(-0.5, 3.5)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$.

(D) Let the vertices of the triangle be $A(3, -5)$,$B(-7, 4)$,and $C(x, y)$.
The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
$x_g = \frac{x_1 + x_2 + x_3}{3}$ and $y_g = \frac{y_1 + y_2 + y_3}{3}$.
Given $G = (2, -1)$,we have:
$2 = \frac{3 + (-7) + x}{3} \implies 6 = -4 + x \implies x = 10$.
$-1 = \frac{-5 + 4 + y}{3} \implies -3 = -1 + y \implies y = -2$.
Thus,the coordinates of the third vertex are $(10, -2)$.

(D) The given vertices are $A(8,6)$,$B(8,-2)$,and $C(2,-2)$.
First,observe the nature of the triangle. The side $AB$ is vertical (constant $x=8$) and the side $BC$ is horizontal (constant $y=-2$).
Since $AB$ is perpendicular to $BC$,the triangle $ABC$ is a right-angled triangle with the right angle at vertex $B(8,-2)$.
In a right-angled triangle,the circumcentre is the midpoint of the hypotenuse.
The hypotenuse is the side $AC$ connecting vertices $A(8,6)$ and $C(2,-2)$.
The midpoint of $AC$ is calculated as $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{8+2}{2}, \frac{6-2}{2}) = (\frac{10}{2}, \frac{4}{2}) = (5,2)$.
Thus,the circumcentre is $(5,2)$.

(A) The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula: $G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$.
Given vertices are $(-5, 3), (a, -1),$ and $(6, b)$,and the centroid is $(1, -1)$.
Equating the $x$-coordinates: $\frac{-5 + a + 6}{3} = 1 \implies a + 1 = 3 \implies a = 2$.
Equating the $y$-coordinates: $\frac{3 - 1 + b}{3} = -1 \implies 2 + b = -3 \implies b = -5$.
Therefore,the values are $a = 2$ and $b = -5$.

(N/A) The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Given the vertices $(1, a), (2, b), (c^2, -3)$,the $x$-coordinate of the centroid is $x_G = \frac{1 + 2 + c^2}{3} = \frac{3 + c^2}{3} = 1 + \frac{c^2}{3}$.
For the centroid to lie on the $Y$-axis,its $x$-coordinate must be $0$.
Setting $x_G = 0$,we get $1 + \frac{c^2}{3} = 0$,which implies $\frac{c^2}{3} = -1$,or $c^2 = -3$.
Since the square of any real number $c$ cannot be negative $(c^2 \geq 0)$,there is no real value of $c$ for which $x_G = 0$.
Therefore,the centroid cannot lie on the $Y$-axis.

(N/A) Let the vertices of $\Delta ABC$ be $A(3h, 3k)$,$B(-3a, 0)$,and $C(3a, 0)$.
The coordinates of the centroid $G$ are given by:
$G = \left( \frac{3h - 3a + 3a}{3}, \frac{3k + 0 + 0}{3} \right) = (h, k)$.
Now,calculate the sum of the squares of the sides:
$AB^2 = (3h - (-3a))^2 + (3k - 0)^2 = (3h + 3a)^2 + 9k^2 = 9(h+a)^2 + 9k^2 = 9(h^2 + 2ha + a^2 + k^2)$
$BC^2 = (3a - (-3a))^2 + (0 - 0)^2 = (6a)^2 = 36a^2$
$AC^2 = (3h - 3a)^2 + (3k - 0)^2 = 9(h-a)^2 + 9k^2 = 9(h^2 - 2ha + a^2 + k^2)$
Summing these:
$AB^2 + BC^2 + AC^2 = 9(h^2 + 2ha + a^2 + k^2) + 36a^2 + 9(h^2 - 2ha + a^2 + k^2)$
$= 9(2h^2 + 2a^2 + 2k^2) + 36a^2 = 18h^2 + 18k^2 + 18a^2 + 36a^2 = 18(h^2 + k^2 + 3a^2)$ ... $(1)$
Now,calculate $3(GA^2 + GB^2 + GC^2)$:
$GA^2 = (h - 3h)^2 + (k - 3k)^2 = (-2h)^2 + (-2k)^2 = 4h^2 + 4k^2$
$GB^2 = (h - (-3a))^2 + (k - 0)^2 = (h + 3a)^2 + k^2 = h^2 + 6ha + 9a^2 + k^2$
$GC^2 = (h - 3a)^2 + (k - 0)^2 = (h - 3a)^2 + k^2 = h^2 - 6ha + 9a^2 + k^2$
Summing these:
$GA^2 + GB^2 + GC^2 = (4h^2 + 4k^2) + (h^2 + 6ha + 9a^2 + k^2) + (h^2 - 6ha + 9a^2 + k^2) = 6h^2 + 6k^2 + 18a^2$
$3(GA^2 + GB^2 + GC^2) = 3(6h^2 + 6k^2 + 18a^2) = 18(h^2 + k^2 + 3a^2)$ ... $(2)$
From $(1)$ and $(2)$,we have $AB^2 + BC^2 + AC^2 = 3(GA^2 + GB^2 + GC^2)$.

(A) Let the coordinates of point $P$ be $(x, y)$.
Given that $PA = PB$,we have $PA^2 = PB^2$.
Using the distance formula: $(x - 2)^2 + (y - 2)^2 = (x - 6)^2 + (y - 6)^2$.
Expanding both sides: $x^2 - 4x + 4 + y^2 - 4y + 4 = x^2 - 12x + 36 + y^2 - 12y + 36$.
Simplifying: $8x + 8y = 64$,which gives $x + y = 8$ (Equation $1$).
The area of $\Delta PAB$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$.
Substituting the coordinates $P(x, y)$,$A(2, 2)$,and $B(6, 6)$: $\frac{1}{2} |x(2 - 6) + 2(6 - y) + 6(y - 2)| = 4$.
$\frac{1}{2} |-4x + 12 - 2y + 6y - 12| = 4$.
$\frac{1}{2} |-4x + 4y| = 4$,which simplifies to $|-x + y| = 2$.
This implies $-x + y = 2$ or $-x + y = -2$ (Equation $2$).
Solving $x + y = 8$ and $-x + y = 2$ gives $2y = 10 \Rightarrow y = 5$ and $x = 3$.
Solving $x + y = 8$ and $-x + y = -2$ gives $2y = 6 \Rightarrow y = 3$ and $x = 5$.
Thus,the coordinates of $P$ are $(3, 5)$ or $(5, 3)$.

(N/A) Let the vertices of the quadrilateral $\square ABCD$ be $A(x_1, y_1)$,$B(x_2, y_2)$,$C(x_3, y_3)$,and $D(x_4, y_4)$.
Let $P, Q, R$,and $S$ be the midpoints of sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$,and $\overline{DA}$ respectively.
Using the midpoint formula,the coordinates are:
$P = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$,$Q = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$,$R = \left(\frac{x_3+x_4}{2}, \frac{y_3+y_4}{2}\right)$,$S = \left(\frac{x_4+x_1}{2}, \frac{y_4+y_1}{2}\right)$.
$A$ quadrilateral is a parallelogram if its diagonals bisect each other,meaning they share the same midpoint.
Midpoint of diagonal $\overline{PR}$:
$= \left( \frac{\frac{x_1+x_2}{2} + \frac{x_3+x_4}{2}}{2}, \frac{\frac{y_1+y_2}{2} + \frac{y_3+y_4}{2}}{2} \right) = \left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4} \right)$.
Midpoint of diagonal $\overline{QS}$:
$= \left( \frac{\frac{x_2+x_3}{2} + \frac{x_4+x_1}{2}}{2}, \frac{\frac{y_2+y_3}{2} + \frac{y_4+y_1}{2}}{2} \right) = \left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4} \right)$.
Since the midpoints of diagonals $\overline{PR}$ and $\overline{QS}$ are identical,the diagonals bisect each other.
Therefore,$\square PQRS$ is a parallelogram.

(A) Let the coordinates of $C$ be $(x, y)$.
The coordinates of the centroid $G$ of $\Delta ABC$ are given by $\left(\frac{2+3+x}{3}, \frac{-3-2+y}{3}\right) = \left(\frac{x+5}{3}, \frac{y-5}{3}\right)$.
Given that the $Y$-coordinate of the centroid is $8$ less than three times its $X$-coordinate:
$\frac{y-5}{3} = 3\left(\frac{x+5}{3}\right) - 8$
$\frac{y-5}{3} = x + 5 - 8$
$\frac{y-5}{3} = x - 3$
$y - 5 = 3x - 9 \implies 3x - y = 4$ ... $(1)$
The area of $\Delta ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{3}{2}$.
$\frac{1}{2} |2(-2 - y) + 3(y - (-3)) + x(-3 - (-2))| = \frac{3}{2}$
$|-4 - 2y + 3y + 9 - x| = 3$
$|y - x + 5| = 3$
This gives two cases: $y - x + 5 = 3$ or $y - x + 5 = -3$.
Case $1$: $y - x = -2 \implies x - y = 2$ ... $(2)$
Case $2$: $y - x = -8 \implies x - y = 8$ ... $(3)$
Solving $(1)$ and $(2)$: $3x - y = 4$ and $x - y = 2$. Subtracting gives $2x = 2 \implies x = 1$. Then $y = -1$.
Solving $(1)$ and $(3)$: $3x - y = 4$ and $x - y = 8$. Subtracting gives $2x = -4 \implies x = -2$. Then $y = -10$.
Thus,the coordinates of $C$ are $(1, -1)$ or $(-2, -10)$.

(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 area of $\triangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Let $D, E,$ and $F$ be the midpoints of sides $BC, AC,$ and $AB$ respectively.
The coordinates of the midpoints are $D = (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$,$E = (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$,and $F = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
Using the area formula for $\triangle DEF$,we calculate the area as $\frac{1}{2} |x_D(y_E - y_F) + x_E(y_F - y_D) + x_F(y_D - y_E)|$.
Substituting the coordinates and simplifying,we get the area of $\triangle DEF = \frac{1}{4} \times \text{Area of } \triangle ABC$.
Alternatively,by the Midpoint Theorem,$\triangle AFE \sim \triangle ABC$ with ratio $1:2$,so $\text{Area}(\triangle AFE) = \frac{1}{4} \text{Area}(\triangle ABC)$. Similarly for other small triangles,the central triangle $\triangle DEF$ also has an area equal to $\frac{1}{4} \text{Area}(\triangle ABC)$.

(B) Let the vertices of the square be $A(-4, 3)$,$B(10, 5)$,$C(12, -9)$,and $D(x, y)$.
In a square,the diagonals bisect each other at the same midpoint.
Let the diagonals be $AC$ and $BD$.
The midpoint of $AC = ((-4 + 12)/2, (3 - 9)/2) = (8/2, -6/2) = (4, -3)$.
The midpoint of $BD = ((10 + x)/2, (5 + y)/2)$.
Equating the midpoints: $(10 + x)/2 = 4 \implies 10 + x = 8 \implies x = -2$.
$(5 + y)/2 = -3 \implies 5 + y = -6 \implies y = -11$.
Thus,the fourth vertex is $(-2, -11)$.

(C) $1$. Since $PA = PB$,point $P(x, y)$ lies on the perpendicular bisector of segment $AB$.
$2$. The midpoint of $AB$ is $M = (\frac{3+5}{2}, \frac{4-2}{2}) = (4, 1)$.
$3$. The slope of $AB$ is $m_{AB} = \frac{-2-4}{5-3} = \frac{-6}{2} = -3$.
$4$. The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{-3} = \frac{1}{3}$.
$5$. The equation of the perpendicular bisector is $y - 1 = \frac{1}{3}(x - 4) \implies 3y - 3 = x - 4 \implies x = 3y + 1$.
$6$. The area of $\Delta PAB = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 10$.
$7$. Substituting $P(3y+1, y)$,$A(3, 4)$,and $B(5, -2)$: $10 = \frac{1}{2} |(3y+1)(4 - (-2)) + 3(-2 - y) + 5(y - 4)|$.
$8$. $20 = |(3y+1)(6) - 6 - 3y + 5y - 20| = |18y + 6 - 26 + 2y| = |20y - 20|$.
$9$. $|20y - 20| = 20 \implies |y - 1| = 1$.
$10$. Case $1: y - 1 = 1 \implies y = 2$. Then $x = 3(2) + 1 = 7$. Point $P = (7, 2)$.
$11$. Case $2: y - 1 = -1 \implies y = 0$. Then $x = 3(0) + 1 = 1$. Point $P = (1, 0)$.

(N/A) Let the vertices be $A(7,9), B(10,8)$,and $C(12,10)$.
$1$. Area of the triangle: Using the formula $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$,we get $\text{Area} = \frac{1}{2} |7(8 - 10) + 10(10 - 9) + 12(9 - 8)| = \frac{1}{2} |-14 + 10 + 12| = \frac{1}{2} |8| = 4 \text{ sq units}$.
$2$. Circumcentre $(h, k)$: The circumcentre is equidistant from the vertices. Thus,$(h-7)^2 + (k-9)^2 = (h-10)^2 + (k-8)^2 = (h-12)^2 + (k-10)^2$.
Solving $(h-7)^2 + (k-9)^2 = (h-10)^2 + (k-8)^2$ gives $6h - 2k = 32$ or $3h - k = 16$.
Solving $(h-10)^2 + (k-8)^2 = (h-12)^2 + (k-10)^2$ gives $4h + 4k = 84$ or $h + k = 21$.
Adding the two equations: $4h = 37 \implies h = \frac{37}{4}$.
Substituting $h$: $\frac{37}{4} + k = 21 \implies k = 21 - \frac{37}{4} = \frac{47}{4}$.
Thus,the circumcentre is $(\frac{37}{4}, \frac{47}{4})$ and the area is $4$.

(A) Let the coordinates of vertex $A$ be $(x, y)$.
The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
Given $G(3, 3)$,$B(-1, 2)$,and $C(2, -3)$,we have:
$3 = \frac{x - 1 + 2}{3} \implies 9 = x + 1 \implies x = 8$.
$3 = \frac{y + 2 - 3}{3} \implies 9 = y - 1 \implies y = 10$.
So,vertex $A$ is $(8, 10)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |8(2 - (-3)) + (-1)(-3 - 10) + 2(10 - 2)|$.
Area $= \frac{1}{2} |8(5) + (-1)(-13) + 2(8)|$.
Area $= \frac{1}{2} |40 + 13 + 16| = \frac{1}{2} |69| = \frac{69}{2}$ square units.

(C) Let the vertices of $\Delta ABC$ be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Given midpoints are $D(1, 2)$,$E(2, 1)$,and $F(3, 3)$.
The centroid of $\Delta ABC$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
$A$ key property of triangles is that the centroid of the triangle formed by the midpoints of the sides is the same as the centroid of the original triangle.
Thus,the centroid $G$ is the average of the coordinates of the midpoints $D, E,$ and $F$.
$G = (\frac{1+2+3}{3}, \frac{2+1+3}{3}) = (\frac{6}{3}, \frac{6}{3}) = (2, 2)$.

(D) Given that $\angle A = 90^\circ$,the line segments $AB$ 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 $AB$ is $m_{AB} = \frac{4 - 7}{2 - 1} = \frac{-3}{1} = -3$.
The slope of $AC$ is $m_{AC} = \frac{5 - 7}{k - 1} = \frac{-2}{k - 1}$.
Substituting these into the condition: $(-3) \times \left( \frac{-2}{k - 1} \right) = -1$.
$\frac{6}{k - 1} = -1$.
$6 = -(k - 1)$.
$6 = -k + 1$.
$k = 1 - 6 = -5$.

(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 $(a, 5), (6, 7)$ and $(2, 3)$,and the area is $10$.
Substituting the values:
$10 = \frac{1}{2} |a(7 - 3) + 6(3 - 5) + 2(5 - 7)|$
$20 = |a(4) + 6(-2) + 2(-2)|$
$20 = |4a - 12 - 4|$
$20 = |4a - 16|$
This gives two cases:
Case $1$: $4a - 16 = 20 \implies 4a = 36 \implies a = 9$
Case $2$: $4a - 16 = -20 \implies 4a = -4 \implies a = -1$
Since it is given that $a \in N$ (natural number),we discard $a = -1$.
Therefore,$a = 9$.

(A) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Given vertices are $(x_1, y_1) = (2, 3)$,$(x_2, y_2) = (4, 7)$,and $(x_3, y_3) = (-3, 2)$.
Substituting these values into the formula:
$x = \frac{2 + 4 + (-3)}{3} = \frac{3}{3} = 1$
$y = \frac{3 + 7 + 2}{3} = \frac{12}{3} = 4$
Therefore,the coordinates of the centroid are $(1, 4)$.

(B) Let the vertices of the triangle be $A(3, 1)$,$B(1, 5)$,and $C(x, y)$.
The centroid $G$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Given that the centroid $G$ is the origin $(0, 0)$,we have:
$\frac{3 + 1 + x}{3} = 0 \implies 4 + x = 0 \implies x = -4$.
$\frac{1 + 5 + y}{3} = 0 \implies 6 + y = 0 \implies y = -6$.
Therefore,the third vertex is $(-4, -6)$.

(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 $A(-4, -1)$,$B(1, 2)$,and $C(4, -3)$.
Here,$x_1 = -4, y_1 = -1, x_2 = 1, y_2 = 2, x_3 = 4, y_3 = -3$.
Substituting these values into the formula:
Area $= \frac{1}{2} |-4(2 - (-3)) + 1(-3 - (-1)) + 4(-1 - 2)|$
Area $= \frac{1}{2} |-4(5) + 1(-2) + 4(-3)|$
Area $= \frac{1}{2} |-20 - 2 - 12|$
Area $= \frac{1}{2} |-34|$
Area $= \frac{34}{2} = 17$ square units.

(B) line segment is parallel to the $Y$-axis if and only if all points on the line have the same $x$-coordinate.
Given that $\overline{CD}$ is parallel to the $Y$-axis and the point $C$ has coordinates $(4, -5)$,the $x$-coordinate of any point $D$ on this line must be $4$.
Therefore,the coordinates of $D$ must be of the form $(4, y)$,where $y$ is any real number.
Comparing this with the given options,option $(B)$ $(4, 6)$ is the only point that satisfies the condition of having an $x$-coordinate of $4$.

(C) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Given vertices are $(a, 2)$,$(2, -a)$,and $(0, 2)$.
Substituting these values into the formula:
$G = \left( \frac{a + 2 + 0}{3}, \frac{2 - a + 2}{3} \right) = \left( \frac{a + 2}{3}, \frac{4 - a}{3} \right)$
According to the problem,both coordinates are equal:
$\frac{a + 2}{3} = \frac{4 - a}{3}$
$a + 2 = 4 - a$
$2a = 2$
$a = 1$

(C) The $x$-coordinates of both given points are the same $(x = 2)$.
Therefore,the line passing through these two points is a vertical line parallel to the $Y$-axis,defined by the equation $x = 2$.
For a third point to be collinear with these two points,it must also lie on the line $x = 2$. This means the $x$-coordinate of the third point must be $2$.
Among the given options,only the point $(2,-4)$ has an $x$-coordinate of $2$.
Thus,the points $(2,3)$,$(2,5)$,and $(2,-4)$ are collinear.

(A) The $y$-coordinates of both given points $(3,4)$ and $(-1,4)$ are equal to $4$.
This implies that the line passing through these two points is a horizontal line defined by the equation $y = 4$.
For a third point to be collinear with these two points,it must also lie on the same line $y = 4$.
Among the given options,only the point $(-5,4)$ has a $y$-coordinate of $4$.
Therefore,the point $(-5,4)$ is collinear with the given points.

(C) First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sqrt{9} = 3$.
$BC = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.
$CD = \sqrt{(4-1)^2 + (3-3)^2} = \sqrt{3^2 + 0^2} = 3$.
$AD = \sqrt{(4-3)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.
Since opposite sides are equal ($AB = CD = 3$ and $BC = AD = \sqrt{10}$),the quadrilateral is a parallelogram.
Next,we check the diagonals:
$AC = \sqrt{(1-3)^2 + (3-0)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$.
$BD = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$.
Since the diagonals are not equal $(AC \neq BD)$,it is not a rectangle.
Therefore,$\square ABCD$ is a parallelogram.

(D) Let the coordinates of point $P$ be $(x, y)$.
Since $P$ is equidistant from $A(0, 0)$,$B(4, 0)$,and $C(0, 6)$,we have $PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.
From $PA^2 = PB^2$,we get $x^2 + y^2 = (x - 4)^2 + (y - 0)^2$.
$x^2 + y^2 = x^2 - 8x + 16 + y^2$.
$8x = 16 \implies x = 2$.
From $PA^2 = PC^2$,we get $x^2 + y^2 = (x - 0)^2 + (y - 6)^2$.
$x^2 + y^2 = x^2 + y^2 - 12y + 36$.
$12y = 36 \implies y = 3$.
Thus,the coordinates of point $P$ are $(2, 3)$.

(A) The coordinates of the points are $A(2, 2)$ and $B(2, -2)$.
Since the $x-$coordinate of both points is $2$,the line segment joining these points lies on the vertical line $x = 2$.
$A$ line segment is parallel to the $Y-$ axis if its $x-$coordinates are constant.
To find where it intersects the $X-$ axis,we set $y = 0$ on the line $x = 2$.
The point of intersection with the $X-$ axis is $(2, 0)$ because the line $x = 2$ crosses the $X-$ axis at $y = 0$.

(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 $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |3(3 - 3) + 0(3 - 0) + 3(0 - 3)|$
Area $= \frac{1}{2} |3(0) + 0(3) + 3(-3)|$
Area $= \frac{1}{2} |0 + 0 - 9|$
Area $= \frac{1}{2} |-9| = \frac{9}{2} = 4.5$ 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 $A(2, 4)$,$B(5, 0)$,and $C(5, 4)$.
Substituting the values into the formula:
Area $= \frac{1}{2} |2(0 - 4) + 5(4 - 4) + 5(4 - 0)|$
Area $= \frac{1}{2} |2(-4) + 5(0) + 5(4)|$
Area $= \frac{1}{2} |-8 + 0 + 20|$
Area $= \frac{1}{2} |12| = 6$ square units.
Alternatively,since $AC$ is a horizontal line (both have $y=4$) and $BC$ is a vertical line (both have $x=5$),the triangle is a right-angled triangle at $C$.
Base $AC = |5 - 2| = 3$ units.
Height $BC = |4 - 0| = 4$ units.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6$ square units.

(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 $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
This formula is derived from the shoelace formula or by calculating the area of trapezoids formed by dropping perpendiculars from the vertices to the $x$-axis.
Comparing this with the given options,option $C$ is the correct representation.

(A) The formula for the coordinates of the centroid of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by:
Centroid $= \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
Given vertices are $A(3, 2)$,$B(7, 5)$,and $C(2, 2)$.
Substituting the values into the formula:
Centroid $= \left(\frac{3 + 7 + 2}{3}, \frac{2 + 5 + 2}{3}\right)$
Centroid $= \left(\frac{12}{3}, \frac{9}{3}\right)$
Centroid $= (4, 3)$

(C) $1$. Given coordinates: $A(3, 0)$,$B(0, 0)$,$C(0, -4)$.
$2$. Calculate lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sqrt{9} = 3$.
$BC = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{16} = 4$.
$AC = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$3$. Check for right-angled triangle: $AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 = AC^2$. Since $AB^2 + BC^2 = AC^2$,$\Delta ABC$ is a right-angled triangle at $\angle B$. (Option $A$ is true).
$4$. Midpoint of $\overline{AC} = (\frac{3+0}{2}, \frac{0-4}{2}) = (\frac{3}{2}, -2)$. (Option $B$ is true).
$5$. Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6$. (Option $C$ states area is $12$,which is false).
$6$. $AB=3, BC=4, AC=5$ is true. (Option $D$ is true).
Therefore,the false statement is $C$.