Mix Examples - Coordinate Geometry Questions in English

(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 $0$.

(6:7) Let the required ratio be $\lambda : 1$. The coordinates of the point $M$ dividing the line segment joining $A(-4, -6)$ and $B(-1, 7)$ are given by the section formula:
$\left( \frac{\lambda x_2 + 1 \cdot x_1}{\lambda + 1}, \frac{\lambda y_2 + 1 \cdot y_1}{\lambda + 1} \right)$
Substituting the values $x_1 = -4, x_2 = -1, y_1 = -6, y_2 = 7$:
$\left( \frac{\lambda(-1) + 1(-4)}{\lambda + 1}, \frac{\lambda(7) + 1(-6)}{\lambda + 1} \right) = \left( \frac{-\lambda - 4}{\lambda + 1}, \frac{7\lambda - 6}{\lambda + 1} \right)$
Since the line segment is divided by the $x$-axis,the $y$-coordinate of the point of division must be $0$:
$\frac{7\lambda - 6}{\lambda + 1} = 0 \implies 7\lambda - 6 = 0 \implies \lambda = \frac{6}{7}$
Thus,the required ratio is $6:7$.
Now,substituting $\lambda = \frac{6}{7}$ into the $x$-coordinate expression:
$x = \frac{-\frac{6}{7} - 4}{\frac{6}{7} + 1} = \frac{-\frac{34}{7}}{\frac{13}{7}} = -\frac{34}{13}$
Therefore,the point of division is $\left( -\frac{34}{13}, 0 \right)$.

(A) Let the point $P \left(\frac{3}{4}, \frac{5}{12}\right)$ divide the line segment joining $A \left(\frac{1}{2}, \frac{3}{2}\right)$ and $B(2, -5)$ in the ratio $k: 1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left(\frac{k(2) + 1(\frac{1}{2})}{k+1}, \frac{k(-5) + 1(\frac{3}{2})}{k+1}\right)$
Equating the $x$-coordinate:
$\frac{3}{4} = \frac{2k + \frac{1}{2}}{k+1}$
$3(k+1) = 4(2k + \frac{1}{2})$
$3k + 3 = 8k + 2$
$3 - 2 = 8k - 3k$
$1 = 5k$
$k = \frac{1}{5}$
Thus,the ratio $k: 1$ is $\frac{1}{5}: 1$,which is $1: 5$.
Checking with the $y$-coordinate:
$y = \frac{-5(\frac{1}{5}) + \frac{3}{2}}{\frac{1}{5} + 1} = \frac{-1 + \frac{3}{2}}{\frac{6}{5}} = \frac{\frac{1}{2}}{\frac{6}{5}} = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$.
Since this matches the given $y$-coordinate,the ratio is $1: 5$.

(B) Given that point $P(9a-2, -b)$ divides the line segment joining $A(3a+1, -3)$ and $B(8a, 5)$ in the ratio $m_1:m_2 = 3:1$.
Using the section formula,the coordinates of point $P$ are given by:
$P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2} \right)$
Equating the $x$-coordinates:
$9a - 2 = \frac{3(8a) + 1(3a+1)}{3+1}$
$9a - 2 = \frac{24a + 3a + 1}{4}$
$4(9a - 2) = 27a + 1$
$36a - 8 = 27a + 1$
$36a - 27a = 1 + 8$
$9a = 9 \implies a = 1$
Equating the $y$-coordinates:
$-b = \frac{3(5) + 1(-3)}{3+1}$
$-b = \frac{15 - 3}{4}$
$-b = \frac{12}{4}$
$-b = 3 \implies b = -3$
Thus,the values are $a = 1$ and $b = -3$.

(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 $a = 3$.

(A) Given that the line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1: 2$.
Using the section formula for internal division,the coordinates of point $P$ are given by:
$P = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$
Substituting the values $m_1 = 1, m_2 = 2, x_1 = 3, y_1 = 2, x_2 = 5, y_2 = 1$:
$P = \left( \frac{1(5) + 2(3)}{1 + 2}, \frac{1(1) + 2(2)}{1 + 2} \right) = \left( \frac{5 + 6}{3}, \frac{1 + 4}{3} \right) = \left( \frac{11}{3}, \frac{5}{3} \right)$
Since the point $P\left( \frac{11}{3}, \frac{5}{3} \right)$ lies on the line $3x - 18y + k = 0$,it must satisfy the equation:
$3\left( \frac{11}{3} \right) - 18\left( \frac{5}{3} \right) + k = 0$
$11 - 6(5) + k = 0$
$11 - 30 + k = 0$
$-19 + k = 0$
$k = 19$
Thus,the required value of $k$ is $19$.

(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.

(D) Given that $PR = \frac{3}{5} PQ$.
This implies that the point $R$ divides the line segment $PQ$ in the ratio $PR : RQ = 3 : (5 - 3) = 3 : 2$.
Let the coordinates of point $R$ be $(x, y)$.
Using the section formula,the coordinates of a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ are given by:
$x = \frac{mx_2 + nx_1}{m + n}$,$y = \frac{my_2 + ny_1}{m + n}$
Here,$(x_1, y_1) = (-1, 3)$,$(x_2, y_2) = (2, 5)$,$m = 3$,and $n = 2$.
$x = \frac{3(2) + 2(-1)}{3 + 2} = \frac{6 - 2}{5} = \frac{4}{5}$
$y = \frac{3(5) + 2(3)}{3 + 2} = \frac{15 + 6}{5} = \frac{21}{5}$
Thus,the coordinates of point $R$ are $(\frac{4}{5}, \frac{21}{5})$.

(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 $k = \frac{1}{2}$.

(A) Let the line $2x + 3y - 5 = 0$ divide the line segment joining the points $A(8, -9)$ and $B(2, 1)$ in the ratio $\lambda : 1$ at point $P$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{2\lambda + 8}{\lambda + 1}, \frac{\lambda - 9}{\lambda + 1} \right)$
Since point $P$ lies on the line $2x + 3y - 5 = 0$,we substitute these coordinates into the equation:
$2\left( \frac{2\lambda + 8}{\lambda + 1} \right) + 3\left( \frac{\lambda - 9}{\lambda + 1} \right) - 5 = 0$
Multiplying by $(\lambda + 1)$:
$2(2\lambda + 8) + 3(\lambda - 9) - 5(\lambda + 1) = 0$
$4\lambda + 16 + 3\lambda - 27 - 5\lambda - 5 = 0$
$2\lambda - 16 = 0$
$2\lambda = 16 \Rightarrow \lambda = 8$
Thus,the ratio is $8 : 1$.
Now,find the coordinates of $P$ by substituting $\lambda = 8$:
$x = \frac{2(8) + 8}{8 + 1} = \frac{16 + 8}{9} = \frac{24}{9} = \frac{8}{3}$
$y = \frac{8 - 9}{8 + 1} = \frac{-1}{9}$
The coordinates of the point of division are $(\frac{8}{3}, -\frac{1}{9})$.

(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) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Distance from House $(2,4)$ to Bank $(5,8)$:
$d_1 = \sqrt{(5-2)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \, km$.
$2$. Distance from Bank $(5,8)$ to School $(13,14)$:
$d_2 = \sqrt{(13-5)^2 + (14-8)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10 \, km$.
$3$. Distance from School $(13,14)$ to Office $(13,26)$:
$d_3 = \sqrt{(13-13)^2 + (26-14)^2} = \sqrt{0^2 + 12^2} = 12 \, km$.
Total distance travelled by Ayush $= 5 + 10 + 12 = 27 \, km$.
$4$. Direct distance from House $(2,4)$ to Office $(13,26)$:
$d_{direct} = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{11^2 + 22^2} = \sqrt{121 + 484} = \sqrt{605} \approx 24.6 \, km$.
Extra distance travelled $= 27 - 24.6 = 2.4 \, km$.

(B) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(5, 8)$ and $B(1, 2)$.
Here,$x_1 = 5, y_1 = 8, x_2 = 1, y_2 = 2$.
Substituting these values into the formula:
$AB = \sqrt{(1 - 5)^2 + (2 - 8)^2}$
$AB = \sqrt{(-4)^2 + (-6)^2}$
$AB = \sqrt{16 + 36}$
$AB = \sqrt{52}$
$AB = \sqrt{4 \times 13}$
$AB = 2 \sqrt{13}$.

(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (5-2)^2 + (2-2)^2 = 3^2 + 0^2 = 9 \implies AB = 3$.
$BC^2 = (5-5)^2 + (5-2)^2 = 0^2 + 3^2 = 9 \implies BC = 3$.
$CD^2 = (2-5)^2 + (5-5)^2 = (-3)^2 + 0^2 = 9 \implies CD = 3$.
$DA^2 = (2-2)^2 + (2-5)^2 = 0^2 + (-3)^2 = 9 \implies DA = 3$.
Since $AB = BC = CD = DA = 3$,the quadrilateral is a rhombus.
Now,check the diagonals:
$AC^2 = (5-2)^2 + (5-2)^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies AC = \sqrt{18} = 3\sqrt{2}$.
$BD^2 = (2-5)^2 + (5-2)^2 = (-3)^2 + 3^2 = 9 + 9 = 18 \implies BD = \sqrt{18} = 3\sqrt{2}$.
Since all sides are equal and the diagonals are equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.

(N/A) Let $A(0, -3), B(1, -1),$ and $C(2, 1)$ be the given points.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$BC = \sqrt{(2 - 1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$AC = \sqrt{(2 - 0)^2 + (1 - (-3))^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
Since $AB + BC = \sqrt{5} + \sqrt{5} = 2\sqrt{5} = AC$,the sum of the lengths of two segments is equal to the length of the third segment.
Therefore,the points $A, B,$ and $C$ are collinear.

(A) Given the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Here,$A = (2, 4)$,$B = (-3, b)$,and $AB = \sqrt{26}$.
Squaring both sides,we get $AB^2 = 26$.
Substituting the coordinates: $(2 - (-3))^2 + (4 - b)^2 = 26$.
$(2 + 3)^2 + (4 - b)^2 = 26$.
$5^2 + (16 - 8b + b^2) = 26$.
$25 + 16 - 8b + b^2 = 26$.
$b^2 - 8b + 41 = 26$.
$b^2 - 8b + 15 = 0$.
Factoring the quadratic equation: $(b - 5)(b - 3) = 0$.
Therefore,$b = 5$ or $b = 3$.

(A-D) Let the vertices be $A(1, 7), B(2, 4)$,and $C(k, 5)$.
Calculate the squared distances between the points:
$AB^2 = (1-2)^2 + (7-4)^2 = (-1)^2 + 3^2 = 1 + 9 = 10$
$BC^2 = (2-k)^2 + (4-5)^2 = 4 - 4k + k^2 + 1 = k^2 - 4k + 5$
$AC^2 = (1-k)^2 + (7-5)^2 = 1 - 2k + k^2 + 4 = k^2 - 2k + 5$
For a right triangle,one of the angles must be $90^{\circ}$:
Case $1$: $\angle A = 90^{\circ}$. Then $BC^2 = AB^2 + AC^2$.
$k^2 - 4k + 5 = 10 + k^2 - 2k + 5$
$-2k = 10 \implies k = -5$
Case $2$: $\angle B = 90^{\circ}$. Then $AC^2 = AB^2 + BC^2$.
$k^2 - 2k + 5 = 10 + k^2 - 4k + 5$
$2k = 10 \implies k = 5$
Case $3$: $\angle C = 90^{\circ}$. Then $AB^2 = BC^2 + AC^2$.
$10 = (k^2 - 4k + 5) + (k^2 - 2k + 5)$
$10 = 2k^2 - 6k + 10$
$2k^2 - 6k = 0 \implies 2k(k - 3) = 0$
$k = 0$ or $k = 3$
Thus,the possible values of $k$ are $-5, 5, 0, 3$.

(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.

(N/A) Let the vertices be $A(0,6), B(-5,3)$,and $C(3,1)$.
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:
$AB^2 = (-5 - 0)^2 + (3 - 6)^2 = (-5)^2 + (-3)^2 = 25 + 9 = 34$
$BC^2 = (3 - (-5))^2 + (1 - 3)^2 = (8)^2 + (-2)^2 = 64 + 4 = 68$
$AC^2 = (3 - 0)^2 + (1 - 6)^2 = (3)^2 + (-5)^2 = 9 + 25 = 34$
Since $AB^2 + AC^2 = 34 + 34 = 68 = BC^2$,the triangle satisfies the Pythagorean theorem,so it is a right-angled triangle with $\angle BAC = 90^\circ$.
Also,since $AB^2 = AC^2 = 34$,we have $AB = AC$.
Therefore,the triangle is an isosceles right triangle.

(A) Let the vertices of the equilateral triangle be $A(0,0)$,$B(3, \sqrt{3})$,and $C(x, y)$.
Since it is an equilateral triangle,$AB = BC = AC$,which implies $AB^2 = BC^2 = AC^2$.
First,calculate $AB^2 = (3-0)^2 + (\sqrt{3}-0)^2 = 9 + 3 = 12$.
Since $AC^2 = AB^2$,we have $x^2 + y^2 = 12$ (Equation $1$).
Since $BC^2 = AB^2$,we have $(x-3)^2 + (y-\sqrt{3})^2 = 12$.
Expanding this: $x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12$.
Substituting $x^2 + y^2 = 12$ into this equation: $12 - 6x - 2\sqrt{3}y + 12 = 12$,which simplifies to $6x + 2\sqrt{3}y = 12$,or $y = \sqrt{3}(2-x)$ (Equation $2$).
Substituting Equation $2$ into Equation $1$: $x^2 + [\sqrt{3}(2-x)]^2 = 12$.
$x^2 + 3(4 - 4x + x^2) = 12$.
$x^2 + 12 - 12x + 3x^2 = 12$.
$4x^2 - 12x = 0$,so $4x(x-3) = 0$.
This gives $x = 0$ or $x = 3$.
If $x = 0$,$y = \sqrt{3}(2-0) = 2\sqrt{3}$.
If $x = 3$,$y = \sqrt{3}(2-3) = -\sqrt{3}$.
Thus,the third vertex is $(0, 2\sqrt{3})$ or $(3, -\sqrt{3})$.

(N/A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
First,calculate $SP$:
$SP = \sqrt{(a - at^2)^2 + (0 - 2at)^2} = \sqrt{a^2(1-t^2)^2 + 4a^2t^2} = |a|\sqrt{1 - 2t^2 + t^4 + 4t^2} = |a|\sqrt{(1+t^2)^2} = |a|(1+t^2)$.
Next,calculate $SQ$:
$SQ = \sqrt{(a - \frac{a}{t^2})^2 + (0 - (-\frac{2a}{t}))^2} = \sqrt{a^2(1 - \frac{1}{t^2})^2 + \frac{4a^2}{t^2}} = |a|\sqrt{1 - \frac{2}{t^2} + \frac{1}{t^4} + \frac{4}{t^2}} = |a|\sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} = |a|\sqrt{(1 + \frac{1}{t^2})^2} = |a|(1 + \frac{1}{t^2}) = |a|(\frac{t^2+1}{t^2})$.
Now,evaluate $\frac{1}{SP} + \frac{1}{SQ}$:
$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{|a|(1+t^2)} + \frac{1}{|a|(\frac{t^2+1}{t^2})} = \frac{1}{|a|(1+t^2)} + \frac{t^2}{|a|(1+t^2)} = \frac{1+t^2}{|a|(1+t^2)} = \frac{1}{|a|}$.
Since the result $\frac{1}{|a|}$ does not contain $t$,the expression $\frac{1}{SP} + \frac{1}{SQ}$ is independent of $t$.

(N/A) To show that points $A(-1, 4)$,$B(2, 3)$,and $C(5, 2)$ are collinear,we must show that the sum of the lengths of two segments equals the length of the third segment.
$1$. Calculate the distance $AB$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$2$. Calculate the distance $BC$:
$BC = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Calculate the distance $AC$:
$AC = \sqrt{(5 - (-1))^2 + (2 - 4)^2} = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
$4$. Since $AB + BC = \sqrt{10} + \sqrt{10} = 2\sqrt{10} = AC$,the points $A$,$B$,and $C$ lie on the same line and are therefore collinear.

(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 $4 \times 7 = 3(a - 4)$.
$28 = 3a - 12$.
$3a = 28 + 12 = 40$.
$a = \frac{40}{3}$.

(N/A) To prove that point $P(7, 5)$ is equidistant from $A(2, 4)$ and $B(6, 10)$,we need to show that the distance $PA$ is equal to the distance $PB$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Distance $PA = \sqrt{(7 - 2)^2 + (5 - 4)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
$2$. Distance $PB = \sqrt{(7 - 6)^2 + (5 - 10)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$.
Since $PA = PB = \sqrt{26}$,the point $P(7, 5)$ is equidistant from points $A(2, 4)$ and $B(6, 10)$.

(A) Let the point on the $X$-axis be $P(x, 0)$.
The distance between $P(x, 0)$ and $A(11, 12)$ is given as $13$.
Using the distance formula: $\sqrt{(x - 11)^2 + (0 - 12)^2} = 13$.
Squaring both sides: $(x - 11)^2 + (-12)^2 = 13^2$.
$(x - 11)^2 + 144 = 169$.
$(x - 11)^2 = 169 - 144 = 25$.
Taking the square root: $x - 11 = \pm 5$.
Case $1$: $x - 11 = 5 \implies x = 16$.
Case $2$: $x - 11 = -5 \implies x = 6$.
Therefore,the points are $(16, 0)$ and $(6, 0)$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(a, 2)$ and $B(3, -5)$ with distance $d = \sqrt{53}$.
Substituting the values into the formula:
$\sqrt{(3 - a)^2 + (-5 - 2)^2} = \sqrt{53}$
Squaring both sides:
$(3 - a)^2 + (-7)^2 = 53$
$(3 - a)^2 + 49 = 53$
$(3 - a)^2 = 53 - 49$
$(3 - a)^2 = 4$
Taking the square root on both sides:
$3 - a = 2$ or $3 - a = -2$
If $3 - a = 2$,then $a = 1$.
If $3 - a = -2$,then $a = 5$.
Therefore,the possible values of $a$ are $1$ and $5$.

(N/A) Let the given points be $A(1, 1)$,$B(4, 4)$,and $C(6, 2)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Length of $AB = \sqrt{(4 - 1)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$.
$2$. Length of $BC = \sqrt{(6 - 4)^2 + (2 - 4)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$.
$3$. Length of $AC = \sqrt{(6 - 1)^2 + (2 - 1)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
Now,check for the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 = 18$,$BC^2 = 8$,and $AC^2 = 26$.
Since $18 + 8 = 26$,we have $AB^2 + BC^2 = AC^2$.
Therefore,the triangle is a right-angled triangle with the right angle at vertex $B$.

(N/A) To prove that the points $A(1, 7)$,$B(2, 4)$,and $C(5, 5)$ form an isosceles right triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $AB$: $AB = \sqrt{(2 - 1)^2 + (4 - 7)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.
$2$. Length of $BC$: $BC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Length of $AC$: $AC = \sqrt{(5 - 1)^2 + (5 - 7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Since $AB = BC = \sqrt{10}$,the triangle is isosceles.
Now,check for the right-angled condition using the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 + BC^2 = 10 + 10 = 20$.
$AC^2 = 20$.
Since $AB^2 + BC^2 = AC^2$,the triangle is a right-angled triangle.
Thus,$A, B,$ and $C$ are the vertices of an isosceles right triangle.

(N/A) To prove that the points $A(0,0), B(7,0), C(7,5)$,and $D(0,5)$ form a rectangle,we must show that opposite sides are equal and the diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(7-0)^2 + (0-0)^2} = \sqrt{49} = 7$
$BC = \sqrt{(7-7)^2 + (5-0)^2} = \sqrt{25} = 5$
$CD = \sqrt{(0-7)^2 + (5-5)^2} = \sqrt{49} = 7$
$DA = \sqrt{(0-0)^2 + (0-5)^2} = \sqrt{25} = 5$
Since $AB = CD = 7$ and $BC = DA = 5$,opposite sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(7-0)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
$BD = \sqrt{(0-7)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
Since the opposite sides are equal and the diagonals are equal,the quadrilateral $ABCD$ is a rectangle.

(N/A) Let the vertices be $A(1, -2)$,$B(2, 3)$,$C(-3, 2)$,and $D(-4, -3)$.
To prove that $ABCD$ is a rhombus,we must show that all four sides are equal in length,but the diagonals are not equal.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - 1)^2 + (3 - (-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$
$BC = \sqrt{(-3 - 2)^2 + (2 - 3)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$
$CD = \sqrt{(-4 - (-3))^2 + (-3 - 2)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$
$DA = \sqrt{(1 - (-4))^2 + (-2 - (-3))^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$
Since $AB = BC = CD = DA = \sqrt{26}$,all sides are equal.
Now,calculate the lengths of the diagonals $AC$ and $BD$:
$AC = \sqrt{(-3 - 1)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$BD = \sqrt{(-4 - 2)^2 + (-3 - 3)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$
Since all sides are equal $(AB=BC=CD=DA)$ and the diagonals are not equal $(AC \neq BD)$,the quadrilateral $ABCD$ is a rhombus.

(A) Let the given points be $A(2, 2)$,$B(-2, 2)$,$C(-2, -2)$,and $D(2, -2)$.
To prove that these points form a square,we must show that all four sides are equal and both diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-2 - 2)^2 + (2 - 2)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.
$BC = \sqrt{(-2 - (-2))^2 + (-2 - 2)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$ units.
$CD = \sqrt{(2 - (-2))^2 + (-2 - (-2))^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$ units.
$DA = \sqrt{(2 - 2)^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.
Since $AB = BC = CD = DA = 4$,all sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(-2 - 2)^2 + (-2 - 2)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
$BD = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
Since the sides are equal and the diagonals are equal,the points $A, B, C, D$ form a square.

(A) Let the vertices be $A(1, -3/2)$,$B(-3, -7/2)$,and $C(-4, -3/2)$.
To check if it is a right-angled triangle,we calculate the squares of the lengths of the sides using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
$AB^2 = (-3 - 1)^2 + (-7/2 - (-3/2))^2 = (-4)^2 + (-4/2)^2 = 16 + (-2)^2 = 16 + 4 = 20$.
$BC^2 = (-4 - (-3))^2 + (-3/2 - (-7/2))^2 = (-1)^2 + (4/2)^2 = 1 + 2^2 = 1 + 4 = 5$.
$AC^2 = (-4 - 1)^2 + (-3/2 - (-3/2))^2 = (-5)^2 + (0)^2 = 25 + 0 = 25$.
Since $AB^2 + BC^2 = 20 + 5 = 25$,which is equal to $AC^2$,the triangle satisfies the Pythagorean theorem.
Therefore,the given points form a right-angled triangle.

(N/A) Given that point $P(x, y)$ is equidistant from points $A(a+b, b-a)$ and $B(a-b, a+b).$
By the distance formula,$PA = PB.$
Squaring both sides,$PA^2 = PB^2.$
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2,$
$(x - (a+b))^2 + (y - (b-a))^2 = (x - (a-b))^2 + (y - (a+b))^2.$
Expanding both sides:
$x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(b-a) + (b-a)^2 = x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2.$
Canceling $x^2, y^2, (a+b)^2$ from both sides:
$-2x(a+b) - 2y(b-a) + (b-a)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2.$
Since $(b-a)^2 = (a-b)^2,$ these terms also cancel out:
$-2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2ay - 2by.$
Canceling $-2ax$ and $-2by$ from both sides:
$-2bx + 2ay = 2bx - 2ay.$
Rearranging the terms:
$2ay + 2ay = 2bx + 2bx.$
$4ay = 4bx.$
Dividing by $4,$ we get $bx = ay.$

(C) Let the point on the $Y$-axis be $P(0, y)$.
Given points are $A(-5, -2)$ and $B(3, 2)$.
Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(0 - (-5))^2 + (y - (-2))^2 = (0 - 3)^2 + (y - 2)^2$
$5^2 + (y + 2)^2 = (-3)^2 + (y - 2)^2$
$25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4$
$29 + 4y = 13 - 4y$
$8y = 13 - 29$
$8y = -16$
$y = -2$
Therefore,the required point is $(0, -2)$.

(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 $(-3) \times \left( \frac{-2}{k - 1} \right) = -1$.
$\frac{6}{k - 1} = -1$.
$6 = -(k - 1)$.
$6 = -k + 1$.
$k = 1 - 6 = -5$.

(N/A) Let the vertices be $A(2a, 4a)$, $B(2a, 6a)$, and $C(2a + \sqrt{3}a, 5a)$.
To prove that the triangle is equilateral, we must show that the lengths of all three sides are equal.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = \sqrt{0^2 + (2a)^2} = \sqrt{4a^2} = 2a$.
$2$. Length of side $BC$:
$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{(\sqrt{3}a)^2 + (-a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
$3$. Length of side $AC$:
$AC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{(\sqrt{3}a)^2 + (a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
Since $AB = BC = AC = 2a$, the triangle $ABC$ is an equilateral triangle.

(B) Let the given points be $A (x_1, y_1) = (4, 3)$ and $B (x_2, y_2) = (6, 5)$.
The midpoint formula for a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
Midpoint $= \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Substituting the values of $A$ and $B$ into the formula:
Midpoint $= \left( \frac{4 + 6}{2}, \frac{3 + 5}{2} \right)$
$= \left( \frac{10}{2}, \frac{8}{2} \right)$
$= (5, 4)$
Thus,the coordinates of the midpoint are $(5, 4)$.

(C) Let the points be $A(x_1, y_1) = (3, 2)$ and $B(x_2, y_2) = (-2, -5)$.
Let the point $P(x, y)$ divide the segment $\overline{AB}$ in the ratio $m:n = 3:2$ starting from $A$.
Using the section formula,the coordinates of $P$ are given by:
$x = \frac{mx_2 + nx_1}{m+n} = \frac{3(-2) + 2(3)}{3+2} = \frac{-6 + 6}{5} = \frac{0}{5} = 0$
$y = \frac{my_2 + ny_1}{m+n} = \frac{3(-5) + 2(2)}{3+2} = \frac{-15 + 4}{5} = \frac{-11}{5} = -2.2$
Thus,the coordinates of the point are $(0, -2.2)$.

(D) Let the point $P(-2, 3)$ divide the line segment $\overline{AB}$ in the ratio $m:n$ starting from $A$.
Using the section formula for the $x$-coordinate:
$x = \frac{mx_2 + nx_1}{m + n}$
Given $A(4, -9)$ as $(x_1, y_1)$ and $B(-3, 5)$ as $(x_2, y_2)$,and $P(-2, 3)$ as $(x, y)$:
$-2 = \frac{m(-3) + n(4)}{m + n}$
$-2(m + n) = -3m + 4n$
$-2m - 2n = -3m + 4n$
$-2m + 3m = 4n + 2n$
$m = 6n$
$\frac{m}{n} = \frac{6}{1}$
Therefore,the required ratio is $6: 1$.

(A) Suppose the $Y-$ axis intersects $\overline{AB}$ at point $P(0, y)$ and $P$ divides $\overline{AB}$ in the ratio $m:n$ from $A$.
Using the section formula for the $x-$coordinate:
$x = \frac{mx_2 + nx_1}{m+n}$
Since $P$ lies on the $Y-$ axis,its $x-$coordinate is $0$.
$0 = \frac{m(3) + n(-2)}{m+n}$
$0 = 3m - 2n$
$3m = 2n \implies \frac{m}{n} = \frac{2}{3}$
So,the ratio is $2:3$.
Now,find the $y-$coordinate of $P$ using the ratio $m=2$ and $n=3$:
$y = \frac{my_2 + ny_1}{m+n}$
$y = \frac{2(0) + 3(3)}{2+3} = \frac{9}{5}$
Thus,the point of division is $(0, 9/5)$ and the ratio is $2:3$.

(N/A) Let $A(-2,-1)$ and $B(7,8)$ be the given points and let $P$ and $Q$ be the points of trisection of $\overline{AB}$.
Here,$x_1 = -2, y_1 = -1, x_2 = 7$ and $y_2 = 8$.
For point $P$ on $\overline{AB}$,the ratio $m:n = 1:2$.
Using the section formula,the coordinates of $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
$P = \left(\frac{1(7) + 2(-2)}{1+2}, \frac{1(8) + 2(-1)}{1+2}\right)$.
$P = \left(\frac{7-4}{3}, \frac{8-2}{3}\right) = \left(\frac{3}{3}, \frac{6}{3}\right) = (1, 2)$.
Now,$Q$ is the midpoint of $\overline{PB}$,or it divides $\overline{AB}$ in the ratio $2:1$.
Using the section formula for $Q$ with ratio $2:1$:
$Q = \left(\frac{2(7) + 1(-2)}{2+1}, \frac{2(8) + 1(-1)}{2+1}\right)$.
$Q = \left(\frac{14-2}{3}, \frac{16-1}{3}\right) = \left(\frac{12}{3}, \frac{15}{3}\right) = (4, 5)$.
Thus,the points of trisection of the line segment joining $(-2,-1)$ and $(7,8)$ are $(1, 2)$ and $(4, 5)$.

(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 $C(1, 0)$.

(D) Given that $C - A - B$ and $2 AC = 3 AB$.
This implies that $A$ divides the segment $\overline{CB}$ externally or $A$ lies on the line segment $\overline{CB}$ such that $C$ is outside. Specifically,$A$ is between $C$ and $B$ is not possible here as $C-A-B$ means $A$ is between $C$ and $B$. Wait,the condition $C-A-B$ implies $A$ is between $C$ and $B$. Let's re-evaluate: $CA + AB = CB$. Given $2 AC = 3 AB$,so $AC = 1.5 AB$.
Since $C-A-B$,the distance $CB = CA + AB = 1.5 AB + AB = 2.5 AB$.
Using the section formula where $A(1, 2)$ divides $\overline{CB}$ in the ratio $CA:AB = 3:2$.
Let $C = (x_1, y_1)$ and $B = (2, 3)$. $A(1, 2)$ divides $CB$ in ratio $m:n = 3:2$.
$1 = \frac{3(2) + 2(x_1)}{3+2} \implies 5 = 6 + 2x_1 \implies 2x_1 = -1 \implies x_1 = -0.5$.
$2 = \frac{3(3) + 2(y_1)}{3+2} \implies 10 = 9 + 2y_1 \implies 2y_1 = 1 \implies y_1 = 0.5$.
Thus,$C = (-0.5, 0.5)$.