Mix Examples - Coordinate Geometry Questions in English

(D) 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 $A(2, 3)$ and $B(-3, -9)$.
Here,$x_1 = 2, y_1 = 3, x_2 = -3, y_2 = -9$.
Substituting these values in the formula:
$AB = \sqrt{(-3 - 2)^2 + (-9 - 3)^2}$
$AB = \sqrt{(-5)^2 + (-12)^2}$
$AB = \sqrt{25 + 144}$
$AB = \sqrt{169}$
$AB = 13$.
Thus,the distance between the points is $13$ units.

(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 $P(3, 2)$,$Q(7, k)$,and $d(P, Q) = 5$.
Squaring both sides,we get $PQ^2 = 5^2 = 25$.
Using the distance formula: $(7 - 3)^2 + (k - 2)^2 = 25$.
$(4)^2 + (k - 2)^2 = 25$.
$16 + (k - 2)^2 = 25$.
$(k - 2)^2 = 25 - 16 = 9$.
Taking the square root on both sides: $k - 2 = \pm 3$.
Case $1$: $k - 2 = 3 \implies k = 5$.
Case $2$: $k - 2 = -3 \implies k = -1$.
Since $5$ is one of the options provided,the correct value is $k = 5$.

(A) Let the point on the $X-$ axis be $P(x, 0)$.
Since $P$ is equidistant from $A(-1, 2)$ and $B(5, 4)$,we have $PA = PB$,which implies $PA^2 = PB^2$.
Using the distance formula,we get:
$[x - (-1)]^2 + (0 - 2)^2 = (x - 5)^2 + (0 - 4)^2$
$(x + 1)^2 + 4 = (x - 5)^2 + 16$
$x^2 + 2x + 1 + 4 = x^2 - 10x + 25 + 16$
$2x + 5 = -10x + 41$
$12x = 36$
$x = 3$
Therefore,the required point on the $X-$ axis is $(3, 0)$.

(B) Suppose the required point on the $Y-$ axis is $R(0, y)$.
Since the point is equidistant from $P(3, 2)$ and $Q(-1, -5)$,we have $PR = QR$.
Squaring both sides,we get $PR^2 = QR^2$.
Using the distance formula,$(3 - 0)^2 + (2 - y)^2 = (-1 - 0)^2 + (-5 - y)^2$.
Expanding the squares,$9 + (4 - 4y + y^2) = 1 + (25 + 10y + y^2)$.
Simplifying the equation,$13 - 4y + y^2 = 26 + 10y + y^2$.
Subtracting $y^2$ from both sides,$13 - 4y = 26 + 10y$.
Rearranging the terms,$-13 = 14y$.
Therefore,$y = -\frac{13}{14}$.
The required point on the $Y-$ axis is $\left(0, -\frac{13}{14}\right)$.

(A) To determine the type of 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 $PQ$: $PQ^2 = (2-2)^2 + (5-3)^2 = 0^2 + 2^2 = 4$. So,$PQ = 2$.
$2$. Length of $QR$: $QR^2 = (2+\sqrt{3}-2)^2 + (4-5)^2 = (\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$. So,$QR = 2$.
$3$. Length of $PR$: $PR^2 = (2+\sqrt{3}-2)^2 + (4-3)^2 = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4$. So,$PR = 2$.
Since $PQ = QR = PR = 2$,all three sides are equal.
Therefore,$\Delta PQR$ is an equilateral triangle.

(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 $A(4, 7)$ and $B(7, 3)$.
Substituting the values into the formula:
$AB = \sqrt{(7 - 4)^2 + (3 - 7)^2}$
$AB = \sqrt{(3)^2 + (-4)^2}$
$AB = \sqrt{9 + 16}$
$AB = \sqrt{25}$
$AB = 5$.

(D) 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 $P(2, -3)$ and $Q(7, 9)$.
Substituting the values into the formula:
$PQ = \sqrt{(7 - 2)^2 + (9 - (-3))^2}$
$PQ = \sqrt{(5)^2 + (12)^2}$
$PQ = \sqrt{25 + 144}$
$PQ = \sqrt{169}$
$PQ = 13$ units.

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

(B) The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
First,find the coordinates of $M$,which is the midpoint of $\overline{AB}$ where $A(0, 0)$ and $B(4, 8)$:
$M = \left(\frac{0+4}{2}, \frac{0+8}{2}\right) = (2, 4)$.
Next,find the coordinates of $N$,which is the midpoint of $\overline{BM}$ where $B(4, 8)$ and $M(2, 4)$:
$N = \left(\frac{4+2}{2}, \frac{8+4}{2}\right) = \left(\frac{6}{2}, \frac{12}{2}\right) = (3, 6)$.

(A) Let the coordinates of $D$ be $(x, y)$.
In a parallelogram,the diagonals bisect each other,meaning the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
The midpoint of $AC = \left(\frac{1+3}{2}, \frac{2+(-4)}{2}\right) = \left(\frac{4}{2}, \frac{-2}{2}\right) = (2, -1)$.
The midpoint of $BD = \left(\frac{2+x}{2}, \frac{1+y}{2}\right)$.
Equating the midpoints: $(2, -1) = \left(\frac{2+x}{2}, \frac{1+y}{2}\right)$.
For the $x$-coordinate: $\frac{2+x}{2} = 2 \implies 2+x = 4 \implies x = 2$.
For the $y$-coordinate: $\frac{1+y}{2} = -1 \implies 1+y = -2 \implies y = -3$.
Therefore,the coordinates of $D$ are $(2, -3)$.

(B) Given that points $A, B$ and $C$ are collinear and $AB = BC$.
This implies that point $B$ is the midpoint of the line segment $\overline{AC}$.
Using the midpoint formula,the coordinates of $B$ are given by $\left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right)$.
Substituting the given values: $(3, 4) = \left(\frac{5 + x}{2}, \frac{2 + y}{2}\right)$.
Equating the $x$-coordinates: $\frac{5 + x}{2} = 3 \implies 5 + x = 6 \implies x = 1$.
Equating the $y$-coordinates: $\frac{2 + y}{2} = 4 \implies 2 + y = 8 \implies y = 6$.
Therefore,the coordinates of point $C$ are $(1, 6)$.

(A) Let the coordinates of point $P$ on the $X-$axis be $(x, 0)$.
Given that the distance $PA = 13$ units.
Using the distance formula,$PA^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
$13^2 = (x - 11)^2 + (0 - 12)^2$
$169 = (x - 11)^2 + (-12)^2$
$169 = (x - 11)^2 + 144$
$(x - 11)^2 = 169 - 144$
$(x - 11)^2 = 25$
Taking the square root on both sides,$x - 11 = \pm 5$.
Case $1$: $x - 11 = 5 \implies x = 16$.
Case $2$: $x - 11 = -5 \implies x = 6$.
Therefore,the coordinates of $P$ are $(16, 0)$ or $(6, 0)$.

(C) The coordinates of point $A$ are $(a \cos \theta, a \sin \theta)$ and the origin $O$ is $(0, 0)$.
Using the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the values,we get $d = \sqrt{(a \cos \theta - 0)^2 + (a \sin \theta - 0)^2}$.
$d = \sqrt{a^2 \cos^2 \theta + a^2 \sin^2 \theta}$.
$d = \sqrt{a^2(\cos^2 \theta + \sin^2 \theta)}$.
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have $d = \sqrt{a^2(1)} = \sqrt{a^2}$.
Since $a \in R^{+}$,the distance is $a$.

(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 $A(\cos \theta, 0)$ and $B(0, \sin \theta)$.
Substituting the coordinates into the formula:
$d = \sqrt{(0 - \cos \theta)^2 + (\sin \theta - 0)^2}$
$d = \sqrt{(-\cos \theta)^2 + (\sin \theta)^2}$
$d = \sqrt{\cos^2 \theta + \sin^2 \theta}$
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$d = \sqrt{1} = 1$.

(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 $A(a+b, b-a)$ and $B(a-b, a+b)$.
$AB = \sqrt{((a-b) - (a+b))^2 + ((a+b) - (b-a))^2}$
$AB = \sqrt{(a - b - a - b)^2 + (a + b - b + a)^2}$
$AB = \sqrt{(-2b)^2 + (2a)^2}$
$AB = \sqrt{4b^2 + 4a^2}$
$AB = \sqrt{4(a^2 + b^2)}$
$AB = 2\sqrt{a^2 + b^2}$.

(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.

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

(B) For the median $\overline{AD}$,$D$ is the midpoint of $\overline{BC}$,where $B(-3, 2)$ and $C(5, 0)$.
The coordinates of $D$ are calculated using the midpoint formula: $D = \left(\frac{-3+5}{2}, \frac{2+0}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)$.
Now,we find the length of the median $\overline{AD}$ using the distance formula between $A(1, 4)$ and $D(1, 1)$:
$AD = \sqrt{(1-1)^2 + (4-1)^2}$
$AD = \sqrt{0^2 + 3^2}$
$AD = \sqrt{9} = 3$.
Thus,the length of the median $\overline{AD}$ is $3$.

(A) The coordinates of the vertices are $A(0, 0)$,$B(4, 0)$,and $C(0, 3)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$ units.
$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ units.
$AC = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{3^2} = 3$ units.
Since $AB^2 + AC^2 = 4^2 + 3^2 = 16 + 9 = 25$ and $BC^2 = 5^2 = 25$,we have $AB^2 + AC^2 = BC^2$.
This satisfies the Pythagorean theorem,indicating that $\Delta ABC$ is a right-angled triangle at vertex $A$. Since $AB \neq AC$,it is not isosceles.

(D) The perpendicular distance of any point $(x, y)$ from the $X-$ axis is given by the absolute value of its $y-$ coordinate.
For the point $(5, 3)$,the $x-$ coordinate is $5$ and the $y-$ coordinate is $3$.
Therefore,the perpendicular distance from the $X-$ axis $= |y| = |3| = 3$ units.

(C) The perpendicular distance of a point $(x, y)$ from the $Y$-axis is given by the absolute value of its $x$-coordinate.
For the point $(-2, 5)$,the $x$-coordinate is $-2$.
Therefore,the perpendicular distance from the $Y$-axis is $|-2| = 2$ units.

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(1, 3)$ and $B(a, 0)$ with distance $d = 3$.
Substituting the values: $3 = \sqrt{(a - 1)^2 + (0 - 3)^2}$.
Squaring both sides: $9 = (a - 1)^2 + (-3)^2$.
$9 = (a - 1)^2 + 9$.
$(a - 1)^2 = 0$.
Therefore,$a - 1 = 0$,which gives $a = 1$.

(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 $A(-4, -3)$ and $B(6, a)$ with distance $d = 10$.
$10 = \sqrt{(6 - (-4))^2 + (a - (-3))^2}$
$10 = \sqrt{(10)^2 + (a + 3)^2}$
Squaring both sides:
$100 = 100 + (a + 3)^2$
$(a + 3)^2 = 0$
$a + 3 = 0$
$a = -3$

(B) To determine the type of triangle, we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(0 - 0)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.
$BC = \sqrt{(\sqrt{3} - 0)^2 + (1 - 2)^2} = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$AC = \sqrt{(\sqrt{3} - 0)^2 + (1 - 0)^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Since $AB = BC = AC = 2$, all three sides are equal.
Therefore, $\Delta ABC$ is an equilateral triangle.

(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 $(2, 3)$.

(B) Since $\overline{AB}$ is the diameter of the circle,the centre $P$ is the midpoint of the diameter $\overline{AB}$.
Using the midpoint formula,the coordinates of $P$ are given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Substituting the values $A(5, -6)$ and $B(3, -2)$:
$P = \left( \frac{5 + 3}{2}, \frac{-6 + (-2)}{2} \right)$
$P = \left( \frac{8}{2}, \frac{-8}{2} \right)$
$P = (4, -4)$.

(D) The given points are $A(8, 10)$ and $B(4, 12)$.
Since the point divides the line segment $\overline{AB}$ in the ratio $1:1$,it is the midpoint of the segment.
The midpoint formula is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Substituting the values,we get $\left(\frac{8 + 4}{2}, \frac{10 + 12}{2}\right) = \left(\frac{12}{2}, \frac{22}{2}\right) = (6, 11)$.

Solution:
Let the ratio in which the point $P(4,4)$ divides the line segment joining $A(2,6)$ and $B(6,2)$ be $k:1$.
Using the section formula, the coordinates of point $P$ are:
$P = \left( \frac{k(6) + 1(2)}{k+1}, \frac{k(2) + 1(6)}{k+1} \right) = (4,4)$
Equating the $x$-coordinates:
$\frac{6k + 2}{k+1} = 4$
$6k + 2 = 4k + 4$
$2k = 2$
$k = 1$
Thus, the ratio $k:1$ is $1:1$.
Alternatively:
The midpoint of $AB$ is:
$\left( \frac{2+6}{2}, \frac{6+2}{2} \right) = (4,4)$
Hence, the point $(4,4)$ is the midpoint, dividing the segment in the ratio $1:1$.

(D) The midpoint formula for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Given the points $(3, 4)$ and $(a, 8)$,the midpoint is $\left(\frac{3 + a}{2}, \frac{4 + 8}{2}\right)$.
We are given that the midpoint is $(4, 6)$.
Equating the coordinates,we get $\left(\frac{3 + a}{2}, \frac{12}{2}\right) = (4, 6)$.
This simplifies to $\left(\frac{3 + a}{2}, 6\right) = (4, 6)$.
Comparing the $x$-coordinates: $\frac{3 + a}{2} = 4$.
Multiplying by $2$,we get $3 + a = 8$.
Therefore,$a = 8 - 3 = 5$.

(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 $y = 0$.

(D) The coordinates of the points are $A(-3, 5)$ and $B(2, 5)$.
Since the $y-$coordinate of both points is $5$,the line segment joining these points is a horizontal line represented by the equation $y = 5$.
$A$ horizontal line $y = 5$ is parallel to the $X-$axis and intersects the $Y-$axis at the point where $x = 0$.
Substituting $x = 0$ into the equation $y = 5$,we get the point $(0, 5)$.
Therefore,the line segment intersects the $Y-$axis at $(0, 5)$.

(D) Let the required point be $P(x, y)$ and the ratio be $m:n = 3:2$.
Given $A(x_1, y_1) = (3, -6)$ and $B(x_2, y_2) = (-2, -1)$.
Using the section formula,the coordinates of $P$ are given by:
$x = \frac{mx_2 + nx_1}{m+n} = \frac{3(-2) + 2(3)}{3+2} = \frac{-6 + 6}{5} = \frac{0}{5} = 0$
$y = \frac{my_2 + ny_1}{m+n} = \frac{3(-1) + 2(-6)}{3+2} = \frac{-3 - 12}{5} = \frac{-15}{5} = -3$
Therefore,the coordinates of $P$ are $(0, -3)$.

(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 $A(0,0), B(4,0)$ and $C(0,6)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |0(0 - 6) + 4(6 - 0) + 0(0 - 0)|$
Area $= \frac{1}{2} |0 + 24 + 0|$
Area $= \frac{1}{2} |24| = 12$ 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 $A(3, 0)$,$B(0, 3)$,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 median $\overline{AD}$ connects vertex $A$ to the midpoint $D$ of the opposite side $\overline{BC}$.
Since $B = (0, 0)$ and $C = (6, 0)$,the coordinates of the midpoint $D$ are calculated using the midpoint formula:
$D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + 6}{2}, \frac{0 + 0}{2} \right) = (3, 0)$.
Now,we find the length of the segment $\overline{AD}$ between $A(3, 4)$ and $D(3, 0)$ using the distance formula:
$AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 3)^2 + (0 - 4)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$.
Therefore,the length of the median $\overline{AD}$ is $4$.

(D) Since $\overline{BE}$ is the median of $\Delta ABC$ to the side $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.
The coordinates of $A$ are $(9,5)$ and $C$ are $(3,3)$.
Using the midpoint formula,the coordinates of $E = \left(\frac{9+3}{2}, \frac{5+3}{2}\right) = \left(\frac{12}{2}, \frac{8}{2}\right) = (6,4)$.
The length of median $\overline{BE}$ is the distance between $B(6,7)$ and $E(6,4)$.
Using the distance formula,$BE = \sqrt{(6-6)^2 + (7-4)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.

(C) To determine the type of 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 $PQ$: $PQ^2 = (3 - 6)^2 + (10 - 5)^2 = (-3)^2 + (5)^2 = 9 + 25 = 34$. So,$PQ = \sqrt{34}$.
$2$. Length of $QR$: $QR^2 = (6 - 1)^2 + (5 - 2)^2 = (5)^2 + (3)^2 = 25 + 9 = 34$. So,$QR = \sqrt{34}$.
$3$. Length of $PR$: $PR^2 = (3 - 1)^2 + (10 - 2)^2 = (2)^2 + (8)^2 = 4 + 64 = 68$. So,$PR = \sqrt{68}$.
Since $PQ = QR = \sqrt{34}$,two sides of the triangle are equal in length.
Therefore,$\Delta PQR$ is an isosceles triangle.

(B) To determine the type of $\Delta LMN$,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 $LM = \sqrt{(4 - 1)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.
$2$. Length of $MN = \sqrt{(4 - 4)^2 + (4 - 1)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
$3$. Length of $NL = \sqrt{(4 - 1)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$.
Since $MN = NL = 3$,the triangle is isosceles.
Also,check for a right angle using the Pythagorean theorem: $MN^2 + NL^2 = 3^2 + 3^2 = 9 + 9 = 18$,and $LM^2 = (3\sqrt{2})^2 = 18$.
Since $MN^2 + NL^2 = LM^2$,the triangle is a right-angled triangle.
Therefore,$\Delta LMN$ is an isosceles right-angled triangle.

(B) If a line segment $\overline{PQ}$ is parallel to the $X$-axis,all points on the line segment must have the same $Y$-coordinate.
Since $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ lie on this line,their $Y$-coordinates must be equal.
Therefore,$y_{1} = y_{2}$.

(A) line parallel to the $Y$-axis is a vertical line.
For any vertical line,the $x$-coordinate remains constant for all points on that line.
Since points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ lie on a line parallel to the $Y$-axis,their $x$-coordinates must be equal.
Therefore,$x_{1} = x_{2}$.

(C) The distance formula between two points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ is given by $PQ = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$.
If the line segment $\overline{PQ}$ is parallel to the $X$-axis,the $y$-coordinates of both points must be equal,i.e.,$y_{1} = y_{2}$.
Substituting $y_{1} = y_{2}$ into the distance formula:
$PQ = \sqrt{(x_{2} - x_{1})^2 + (y_{1} - y_{1})^2}$
$PQ = \sqrt{(x_{2} - x_{1})^2 + 0^2}$
$PQ = \sqrt{(x_{2} - x_{1})^2} = |x_{2} - x_{1}| = |x_{1} - x_{2}|$.
Thus,the distance is $|x_{1} - x_{2}|$.

(C) If a line segment $\overline{PQ}$ is parallel to the $Y$-axis,the $x$-coordinates of all points on the line segment must be equal.
Therefore,$x_{1} = x_{2}$.
The distance formula between two points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$.
Substituting $x_{1} = x_{2}$ into the formula:
$PQ = \sqrt{(x_{1}-x_{1})^2 + (y_{2}-y_{1})^2}$
$PQ = \sqrt{0^2 + (y_{2}-y_{1})^2}$
$PQ = \sqrt{(y_{2}-y_{1})^2} = |y_{2}-y_{1}| = |y_{1}-y_{2}|$.
Thus,the distance is $|y_{1}-y_{2}|$.

(A) The distance between two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ in a Cartesian plane is given by the distance formula.
According to the distance formula,the distance $d$ between points $A$ and $B$ is calculated as $d(A, B) = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$.
Since $(x_{2}-x_{1})^{2} = (x_{1}-x_{2})^{2}$ and $(y_{2}-y_{1})^{2} = (y_{1}-y_{2})^{2}$,the formula can also be written as $d(A, B) = \sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2}}$.
Therefore,option $A$ is the correct answer.

(A) point $P$ lying on the line segment $\overline{AB}$ divides the segment into two parts: $AP$ and $PB$.
By definition,the division ratio of a line segment $\overline{AB}$ by a point $P$ (where $P$ is between $A$ and $B$) is the ratio of the lengths of the two segments formed,which is $\frac{AP}{PB}$.
Thus,$P$ divides $\overline{AB}$ from $A$ in the ratio $\frac{AP}{PB}$.

(B) The line segment is $\overline{ AB }$. Point $P$ lies between $A$ and $B$ such that $A-P-B$.
When a point $P$ divides a line segment $\overline{ AB }$ in a ratio,it is typically defined as the ratio of the lengths of the two segments formed.
If the division is specified 'from $B$',it implies the ratio of the segment starting from $B$ to the segment starting from $A$.
Thus,the ratio is $\frac{BP}{PA}$.

(B) According to the section formula,if a point $P(x, y)$ divides the line segment joining points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ internally in the ratio $m : n$,the coordinates of $P$ are given by:
$x = \frac{m x_{2} + n x_{1}}{m + n}$
$y = \frac{m y_{2} + n y_{1}}{m + n}$
Thus,the coordinates of $P$ are $\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$.

(C) Given that $A - M - B$,the point $M$ lies on the line segment $AB$.
We are given the ratio $\frac{AM}{AB} = \frac{3}{5}$.
Since $AB = AM + MB$,we can substitute this into the equation:
$\frac{AM}{AM + MB} = \frac{3}{5}$.
By cross-multiplying or comparing the parts,we get $5(AM) = 3(AM + MB)$.
$5(AM) = 3(AM) + 3(MB)$.
$2(AM) = 3(MB)$.
Therefore,$\frac{AM}{MB} = \frac{3}{2}$.
Thus,the point $M$ divides the segment $AB$ in the ratio $3:2$ internally.