Mix Examples - Coordinate Geometry Questions in English

(C) Let the point on the $X$-axis be $P(x, 0)$.
Since $P$ is equidistant from $A(5, 4)$ and $B(-2, 3)$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(x - 5)^2 + (0 - 4)^2 = (x - (-2))^2 + (0 - 3)^2$
$(x - 5)^2 + 16 = (x + 2)^2 + 9$
$x^2 - 10x + 25 + 16 = x^2 + 4x + 4 + 9$
$-10x + 41 = 4x + 13$
$41 - 13 = 4x + 10x$
$28 = 14x$
$x = 2$.
Therefore,the point is $(2, 0)$.

(NONE) Let the point on the $Y$-axis be $P(0, y)$.
Since $P$ is equidistant from $A(3, 1)$ and $B(-2, 5)$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(3 - 0)^2 + (1 - y)^2 = (-2 - 0)^2 + (5 - y)^2$.
$9 + (1 - 2y + y^2) = 4 + (25 - 10y + y^2)$.
$10 - 2y + y^2 = 29 - 10y + y^2$.
Subtracting $y^2$ from both sides and rearranging terms:
$10y - 2y = 29 - 10$.
$8y = 19$.
$y = \frac{19}{8}$.
Thus,the point is $(0, \frac{19}{8})$.

(N/A) Let the vertices of the triangle be $A(10, -18), B(3, 6)$ and $C(-5, 2)$.
To determine if the triangle is isosceles,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 side $AB$:
$AB = \sqrt{(3 - 10)^2 + (6 - (-18))^2} = \sqrt{(-7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ units.
$2$. Length of side $BC$:
$BC = \sqrt{(-5 - 3)^2 + (2 - 6)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$ units.
$3$. Length of side $AC$:
$AC = \sqrt{(-5 - 10)^2 + (2 - (-18))^2} = \sqrt{(-15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25$ units.
Since the lengths of sides $AB$ and $AC$ are equal ($AB = AC = 25$ units),the triangle is an isosceles triangle.

(B) Let the point $P (-4, 3)$ divide the line segment joining $A (1, -2)$ and $B (-6, 5)$ in the ratio $k: 1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1} \right)$
Substituting the values $x_1 = 1, y_1 = -2, x_2 = -6, y_2 = 5$:
$-4 = \frac{k(-6) + 1(1)}{k + 1}$
$-4(k + 1) = -6k + 1$
$-4k - 4 = -6k + 1$
$2k = 5$
$k = \frac{5}{2}$
Thus,the ratio is $5: 2$.

(C) Let the vertices of $\Delta ABC$ be $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3).$
The midpoints of the sides are given as $D(4,6), E(2,2),$ and $F(3,1).$
The centroid of a triangle formed by the midpoints of the sides of a triangle is the same as the centroid of the original triangle.
The centroid $(G)$ of $\Delta DEF$ is given by the formula: $G = \left( \frac{x_D + x_E + x_F}{3}, \frac{y_D + y_E + y_F}{3} \right).$
Substituting the given values: $G = \left( \frac{4 + 2 + 3}{3}, \frac{6 + 2 + 1}{3} \right).$
$G = \left( \frac{9}{3}, \frac{9}{3} \right) = (3, 3).$
Thus,the coordinates of the centroid of $\Delta ABC$ are $(3, 3).$

(D) Let the vertices of the triangle be $A(4,6), B(5,1),$ and $C(4,0)$. Let the circumcentre be $P(x,y)$.
Since the circumcentre is equidistant from all vertices,$PA^2 = PB^2 = PC^2$.
First,equate $PA^2 = PC^2$:
$(x-4)^2 + (y-6)^2 = (x-4)^2 + (y-0)^2$
$(y-6)^2 = y^2$
$y^2 - 12y + 36 = y^2$
$12y = 36 \implies y = 3$.
Now,equate $PA^2 = PB^2$ with $y=3$:
$(x-4)^2 + (3-6)^2 = (x-5)^2 + (3-1)^2$
$(x-4)^2 + (-3)^2 = (x-5)^2 + (2)^2$
$x^2 - 8x + 16 + 9 = x^2 - 10x + 25 + 4$
$-8x + 25 = -10x + 29$
$2x = 4 \implies x = 2$.
Thus,the circumcentre is $(2,3)$.

(A) Let the points of trisection be $P$ and $Q$. Point $P$ divides $AB$ in the ratio $1:2$ and point $Q$ divides $AB$ in the ratio $2:1$.
Using the section formula,$P = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$.
For point $P$ $(m_1=1, m_2=2)$: $P = \left( \frac{1(7) + 2(-2)}{1+2}, \frac{1(8) + 2(-1)}{1+2} \right) = \left( \frac{7-4}{3}, \frac{8-2}{3} \right) = \left( \frac{3}{3}, \frac{6}{3} \right) = (1, 2)$.
For point $Q$ $(m_1=2, m_2=1)$: $Q = \left( \frac{2(7) + 1(-2)}{2+1}, \frac{2(8) + 1(-1)}{2+1} \right) = \left( \frac{14-2}{3}, \frac{16-1}{3} \right) = \left( \frac{12}{3}, \frac{15}{3} \right) = (4, 5)$.
Thus,the coordinates of the trisection points are $(1, 2)$ and $(4, 5)$.

(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 $(-2, -11)$.

(N/A) Let the vertices be $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$.
To show that $ABCD$ is a rhombus,we must prove that all four sides are equal in length $(AB = BC = CD = DA)$ and the diagonals are not equal $(AC \neq BD)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(3 - 7)^2 + (0 - 3)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$BC = \sqrt{(0 - 3)^2 + (-4 - 0)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$CD = \sqrt{(4 - 0)^2 + (-1 - (-4))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$DA = \sqrt{(7 - 4)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Since $AB = BC = CD = DA = 5$,all sides are equal.
Now,calculate the diagonals:
$AC = \sqrt{(0 - 7)^2 + (-4 - 3)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}$.
$BD = \sqrt{(4 - 3)^2 + (-1 - 0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,$ABCD$ is a rhombus.

(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 $4$.

(N/A) Given coordinates are $A(b, c)$,$B(-a, 0)$,and $C(a, 0)$.
Since $D$ is the midpoint of $\overline{BC}$,the coordinates of $D$ are $(\frac{-a+a}{2}, \frac{0+0}{2}) = (0, 0)$.
Now,calculate the squares of the lengths:
$AB^2 = (b - (-a))^2 + (c - 0)^2 = (b+a)^2 + c^2 = b^2 + 2ab + a^2 + c^2$.
$AC^2 = (b - a)^2 + (c - 0)^2 = b^2 - 2ab + a^2 + c^2$.
Adding them: $AB^2 + AC^2 = (b^2 + 2ab + a^2 + c^2) + (b^2 - 2ab + a^2 + c^2) = 2(a^2 + b^2 + c^2)$.
Now calculate the right side:
$AD^2 = (b - 0)^2 + (c - 0)^2 = b^2 + c^2$.
$BD^2 = (-a - 0)^2 + (0 - 0)^2 = a^2$.
Thus,$2(AD^2 + BD^2) = 2(b^2 + c^2 + a^2)$.
Since both sides are equal to $2(a^2 + b^2 + c^2)$,the identity is proven.

(C) Let the third vertex be $C(x, y)$. Since $ABC$ is an equilateral triangle,$AB = BC = AC$.
The distance $AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$.
Since $AB = BC = AC = 2$,we have:
$x^2 + y^2 = 2^2 = 4$ ---$(1)$
$(x-1)^2 + (y-\sqrt{3})^2 = 4$ ---$(2)$
Expanding $(2)$: $x^2 - 2x + 1 + y^2 - 2\sqrt{3}y + 3 = 4$.
Substituting $x^2 + y^2 = 4$ into the equation: $4 - 2x + 1 - 2\sqrt{3}y + 3 = 4$,which simplifies to $2x + 2\sqrt{3}y = 4$ or $x + \sqrt{3}y = 2$.
Thus,$x = 2 - \sqrt{3}y$.
Substituting this into $(1)$: $(2 - \sqrt{3}y)^2 + y^2 = 4$.
$4 - 4\sqrt{3}y + 3y^2 + y^2 = 4$,so $4y^2 - 4\sqrt{3}y = 0$.
$4y(y - \sqrt{3}) = 0$,giving $y = 0$ or $y = \sqrt{3}$.
If $y = 0$,$x = 2 - 0 = 2$. If $y = \sqrt{3}$,$x = 2 - \sqrt{3}(\sqrt{3}) = 2 - 3 = -1$.
Therefore,the third vertex $C$ can be $(2, 0)$ or $(-1, \sqrt{3})$.

(A) Given points are $A(3, 3)$ and $B(6, 1)$.
Let the coordinates of point $C$ be $(x, y)$.
The condition $A-B-C$ implies that $B$ lies between $A$ and $C$.
The condition $3 AB = BC$ implies that the ratio $AB : BC = 1 : 3$.
Since $B$ divides $AC$ externally in the ratio $1 : 3$ (or internally in the ratio $1 : 3$ if we consider $B$ as a point on the segment $AC$),we use the section formula.
Alternatively,using vectors: $\vec{B} - \vec{A} = (6-3, 1-3) = (3, -2)$.
Since $BC = 3 AB$,the vector $\vec{BC} = 3 \vec{AB} = 3(3, -2) = (9, -6)$.
Thus,$\vec{C} = \vec{B} + \vec{BC} = (6, 1) + (9, -6) = (15, -5)$.

(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.

(B) According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $\overline{BC}$ in the ratio of the lengths of the other two sides,$AB$ and $AC$.
Step $1$: Calculate the lengths of sides $AB$ and $AC$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(-2 - 6)^2 + (3 - 7)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.
$AC = \sqrt{(9 - 6)^2 + (1 - 7)^2} = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.
Step $2$: The ratio $m:n$ in which the bisector divides $\overline{BC}$ is $AB:AC = 4\sqrt{5}:3\sqrt{5} = 4:3$.
Step $3$: Use the section formula to find the coordinates of the point $D(x, y)$ on $\overline{BC}$ that divides it in the ratio $4:3$ internally.
$x = \frac{mx_2 + nx_1}{m + n} = \frac{4(9) + 3(-2)}{4 + 3} = \frac{36 - 6}{7} = \frac{30}{7}$.
$y = \frac{my_2 + ny_1}{m + n} = \frac{4(1) + 3(3)}{4 + 3} = \frac{4 + 9}{7} = \frac{13}{7}$.
Thus,the coordinates are $\left(\frac{30}{7}, \frac{13}{7}\right)$.

(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) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given points are $(6, 8)$ and $(3, 4)$.
Here,$x_1 = 6, y_1 = 8$ and $x_2 = 3, y_2 = 4$.
Substituting these values into the formula:
$d = \sqrt{(3 - 6)^2 + (4 - 8)^2}$
$d = \sqrt{(-3)^2 + (-4)^2}$
$d = \sqrt{9 + 16}$
$d = \sqrt{25}$
$d = 5$
Therefore,the distance between the points is $5$.

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:
$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given $P(4, -7)$ and $Q(-1, 5)$,we have $x_1 = 4, y_1 = -7, x_2 = -1, y_2 = 5$.
Substituting these values into the formula:
$PQ = \sqrt{(-1 - 4)^2 + (5 - (-7))^2}$
$PQ = \sqrt{(-5)^2 + (5 + 7)^2}$
$PQ = \sqrt{(-5)^2 + (12)^2}$
$PQ = \sqrt{25 + 144}$
$PQ = \sqrt{169}$
$PQ = 13$
Therefore,the distance $PQ$ is $13$.

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

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(a, 1)$ and $(8, a)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(8 - a)^2 + (a - 1)^2}$
Squaring both sides:
$25 = (8 - a)^2 + (a - 1)^2$
$25 = (64 - 16a + a^2) + (a^2 - 2a + 1)$
$25 = 2a^2 - 18a + 65$
$2a^2 - 18a + 40 = 0$
Dividing by $2$:
$a^2 - 9a + 20 = 0$
Factoring the quadratic equation:
$(a - 4)(a - 5) = 0$
Thus,$a = 4$ or $a = 5$.
Since $5$ is one of the given options,the correct answer is $5$.

(C) To determine the type of triangle $\Delta PQR$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $PQ$: $PQ = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{0^2 + 5^2} = 5 \text{ units}$.
$2$. Length of $QR$: $QR = \sqrt{(6 - 0)^2 + (0 - 5)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ units}$.
$3$. Length of $PR$: $PR = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{6^2 + 0^2} = 6 \text{ units}$.
Since the vertices $P(0,0), Q(0,5),$ and $R(6,0)$ lie on the axes,the angle at the origin $P(0,0)$ is $90^\circ$ because the $y$-axis $(PQ)$ is perpendicular to the $x$-axis $(PR)$.
Since one angle is $90^\circ$,$\Delta PQR$ is a right-angled triangle.

(A) To check if the points $A(-1, 4), B(2, 3),$ and $C(8, 1)$ are collinear,we calculate the slopes of the segments $AB$ and $BC$.
Slope of $AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{2 - (-1)} = \frac{-1}{3}$.
Slope of $BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{8 - 2} = \frac{-2}{6} = \frac{-1}{3}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.
Now,we check the distances between the points:
$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$BC = \sqrt{(8 - 2)^2 + (1 - 3)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
$AC = \sqrt{(8 - (-1))^2 + (1 - 4)^2} = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$.
Since $AB + BC = \sqrt{10} + 2\sqrt{10} = 3\sqrt{10} = AC$,the point $B$ lies between $A$ and $C$.
Therefore,the order of the points is $A-B-C$.

(B) The diameter of the circle is the distance between the points $(2, 4)$ and $(2, 6)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we get:
Diameter $= \sqrt{(2 - 2)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.
The radius $r$ is half of the diameter.
$r = \frac{\text{Diameter}}{2} = \frac{2}{2} = 1$.
Therefore, the radius of the circle is $1$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(5, 0)$ and $(x, 4)$ with distance $d = \sqrt{17}$.
Substituting the values: $\sqrt{17} = \sqrt{(x - 5)^2 + (4 - 0)^2}$.
Squaring both sides: $17 = (x - 5)^2 + 4^2$.
$17 = (x - 5)^2 + 16$.
$(x - 5)^2 = 17 - 16 = 1$.
Taking the square root: $x - 5 = \pm 1$.
Case $1$: $x - 5 = 1 \implies x = 6$.
Case $2$: $x - 5 = -1 \implies x = 4$.
Thus,$x = 4, 6$.

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

(A) The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Given points are $A(5, 3)$ and $B(3, 3)$.
Here,$x_1 = 5, y_1 = 3$ and $x_2 = 3, y_2 = 3$.
Substituting these values into the formula:
Midpoint $= \left( \frac{5 + 3}{2}, \frac{3 + 3}{2} \right)$
Midpoint $= \left( \frac{8}{2}, \frac{6}{2} \right)$
Midpoint $= (4, 3)$.

(B) The midpoint formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Given points are $(1, 1)$ and $(3, 3)$.
Substituting the values into the formula:
Midpoint $= \left(\frac{1 + 3}{2}, \frac{1 + 3}{2}\right)$
Midpoint $= \left(\frac{4}{2}, \frac{4}{2}\right)$
Midpoint $= (2, 2)$.

(C) The point dividing the line segment in the ratio $1:1$ is the midpoint of the segment.
Using the midpoint formula,the coordinates $(x, y)$ are given by:
$x = \frac{x_1 + x_2}{2} = \frac{2 + (-2)}{2} = \frac{0}{2} = 0$
$y = \frac{y_1 + y_2}{2} = \frac{1 + (-1)}{2} = \frac{0}{2} = 0$
Therefore,the coordinates of the point are $(0, 0)$.

(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 $(-4, -6)$.

(C) The midpoint formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Given points are $(20, 10)$ and $(6, 8)$.
Here,$x_1 = 20, y_1 = 10$ and $x_2 = 6, y_2 = 8$.
Substituting these values into the formula:
Midpoint $= \left( \frac{20 + 6}{2}, \frac{10 + 8}{2} \right)$
Midpoint $= \left( \frac{26}{2}, \frac{18}{2} \right)$
Midpoint $= (13, 9)$.

(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.

(A) The center $P$ of a circle is the midpoint of its diameter $\overline{AB}$.
Given $A(x_1, y_1) = (4, 0)$ and $B(x_2, y_2) = (-2, 2)$.
The midpoint formula is $P(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Substituting the values: $P(x, y) = \left( \frac{4 + (-2)}{2}, \frac{0 + 2}{2} \right)$.
$P(x, y) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$.
Therefore,the coordinates of $P$ are $(1, 1)$.

(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 a point are given as $(x, y)$.
Here,the point is $(6, 2)$,where $x = 6$ and $y = 2$.
The perpendicular distance of any point $(x, y)$ from the $X-$ axis is given by the absolute value of its $y-$ coordinate,which is $|y|$.
Therefore,the perpendicular distance of the point $(6, 2)$ from the $X-$ axis is $|2| = 2$ units.

(D) The coordinates of the given point are $(x, y) = (-3, 8)$.
In a Cartesian coordinate system,the perpendicular distance of any point $(x, y)$ from the $Y$-axis is given by the absolute value of its $x$-coordinate,i.e.,$|x|$.
Here,the $x$-coordinate is $-3$.
Therefore,the perpendicular distance from the $Y$-axis is $|-3| = 3$.
Thus,the correct option is $D$.

(A) The coordinates of point $B$ are $(0, 12)$ and the coordinates of point $C$ are $(5, 0)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,we calculate the distance $BC$:
$BC = \sqrt{(5 - 0)^2 + (0 - 12)^2}$
$BC = \sqrt{(5)^2 + (-12)^2}$
$BC = \sqrt{25 + 144}$
$BC = \sqrt{169}$
$BC = 13$
Thus,the distance $BC$ is $13$ units.

(B) Let the endpoints of the line segment be $A(0,0)$ and $B(6,6)$.
To divide the segment into four congruent parts,we need three points $P, Q,$ and $R$ that divide the segment in the ratio $1:3, 1:1,$ and $3:1$ respectively.
The midpoint $Q$ is calculated as $(\frac{0+6}{2}, \frac{0+6}{2}) = (3,3)$.
The point $P$ is the midpoint of $AQ$,which is $(\frac{0+3}{2}, \frac{0+3}{2}) = (1.5, 1.5)$.
The point $R$ is the midpoint of $QB$,which is $(\frac{3+6}{2}, \frac{3+6}{2}) = (4.5, 4.5)$.
Comparing these points with the given options,$(4.5, 4.5)$ is one of the dividing points.

(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:
$\frac{a + 2}{3} = \frac{4 - a}{3}$
$a + 2 = 4 - a$
$2a = 2$
$a = 1$

(D) Let the vertices be $A(1, 1)$, $B(-1, -1)$, and $C(-\sqrt{3}, \sqrt{3})$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
$BC = \sqrt{(-\sqrt{3} - (-1))^2 + (\sqrt{3} - (-1))^2} = \sqrt{(1 - \sqrt{3})^2 + (\sqrt{3} + 1)^2} = \sqrt{(1 - 2\sqrt{3} + 3) + (3 + 2\sqrt{3} + 1)} = \sqrt{4 - 2\sqrt{3} + 4 + 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
$CA = \sqrt{(1 - (-\sqrt{3}))^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + 2\sqrt{3} + 3) + (1 - 2\sqrt{3} + 3)} = \sqrt{4 + 2\sqrt{3} + 4 - 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
Since $AB = BC = CA = 2\sqrt{2}$, all three sides are equal.
Therefore, the triangle is an equilateral triangle.

(A) Let the coordinates of the other end point of the diameter be $(x,y).$
Since the centre of a circle is the midpoint of its diameter,we use the midpoint formula:
Centre $= (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$
Given the centre is $(-2,4)$ and one end point is $(0,0)$,we have:
$-2 = \frac{0 + x}{2} \implies x = -4$
$4 = \frac{0 + y}{2} \implies y = 8$
Therefore,the other end point is $(-4,8).$

(B) The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Given points are $A(6, 5)$ and $B(4, 3)$.
Here,$x_1 = 6, y_1 = 5$ and $x_2 = 4, y_2 = 3$.
Midpoint $= \left( \frac{6 + 4}{2}, \frac{5 + 3}{2} \right)$.
Midpoint $= \left( \frac{10}{2}, \frac{8}{2} \right)$.
Midpoint $= (5, 4)$.

(C) Let the vertices of the triangle be $A(6,0)$,$B(0,0)$,and $C(0,8)$.
Since the vertices $A(6,0)$ and $C(0,8)$ lie on the $x$-axis and $y$-axis respectively,and the vertex $B$ is at the origin $(0,0)$,the triangle is a right-angled triangle with the right angle at $B(0,0)$.
For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse.
The hypotenuse is the line segment joining $A(6,0)$ and $C(0,8)$.
The midpoint of the hypotenuse is given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Substituting the coordinates of $A$ and $C$: $\left( \frac{6 + 0}{2}, \frac{0 + 8}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3,4)$.
Thus,the circumcentre is $(3,4)$.

(D) Let the vertices of the triangle be $A(0,0)$,$B(4.2,0)$,and $C(0,9.1)$.
Since vertex $A$ lies at the origin $(0,0)$,side $AB$ lies along the $x$-axis and side $AC$ lies along the $y$-axis.
The angle between the $x$-axis and the $y$-axis is $90^\circ$.
Therefore,the angle $\angle BAC = 90^\circ$.
Since one angle of the triangle is $90^\circ$,it is a right-angled triangle.

(A) Let the coordinates of $B$ be $(x, y)$.
Given that $P(1, 12)$ divides the line segment $AB$ in the ratio $m:n = 2:1$,where $A(x_1, y_1) = (3, 8)$ and $B(x_2, y_2) = (x, y)$.
Using the section formula: $P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$.
Substituting the values: $1 = \frac{2(x) + 1(3)}{2+1} \implies 1 = \frac{2x + 3}{3} \implies 3 = 2x + 3 \implies 2x = 0 \implies x = 0$.
Similarly,$12 = \frac{2(y) + 1(8)}{2+1} \implies 12 = \frac{2y + 8}{3} \implies 36 = 2y + 8 \implies 2y = 28 \implies y = 14$.
Thus,the coordinates of $B$ are $(0, 14)$.

(B) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$.
Given points are $P(x_{1}, 0)$ and $Q(x_{2}, 0)$.
Substituting these values into the distance formula:
$PQ = \sqrt{(x_{2}-x_{1})^2 + (0-0)^2}$
$PQ = \sqrt{(x_{2}-x_{1})^2}$
Since the distance must always be non-negative,we take the absolute value:
$PQ = |x_{2}-x_{1}| = |x_{1}-x_{2}|$.
Therefore,the correct option is $B$.

(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 the points $A(0, y_{1})$ and $B(0, y_{2})$,we substitute these coordinates into the formula:
$AB = \sqrt{(0 - 0)^{2} + (y_{2} - y_{1})^{2}}$
$AB = \sqrt{0^{2} + (y_{2} - y_{1})^{2}}$
$AB = \sqrt{(y_{2} - y_{1})^{2}}$
Since the distance must be non-negative,we take the absolute value:
$AB = |y_{2} - y_{1}|$ or $|y_{1} - y_{2}|$.
Therefore,the correct option is $C$.

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