Section Formula Questions in English

(N/A) The coordinates of the vertices of the triangle are $(x_1, y_1) = (a, b-c)$,$(x_2, y_2) = (b, c-a)$,and $(x_3, y_3) = (c, a-b)$.
The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Substituting the given values into the formula:
$x$-coordinate $= \frac{a + b + c}{3}$
$y$-coordinate $= \frac{(b-c) + (c-a) + (a-b)}{3}$
Simplifying the $y$-coordinate:
$y$-coordinate $= \frac{b - c + c - a + a - b}{3} = \frac{0}{3} = 0$.
Since the $y$-coordinate of the centroid is $0$,the centroid lies on the $X$-axis.

(D) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.
Let the coordinates of vertex $D$ be $(x, y)$.
The midpoint of diagonal $AC$ is given by $(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}) = (\frac{3+2}{2}, \frac{2+3}{2}) = (2.5, 2.5)$.
The midpoint of diagonal $BD$ is given by $(\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}) = (\frac{4+x}{2}, \frac{5+y}{2})$.
Since the midpoints are the same,we equate the coordinates:
$\frac{4+x}{2} = 2.5 \implies 4+x = 5 \implies x = 1$.
$\frac{5+y}{2} = 2.5 \implies 5+y = 5 \implies y = 0$.
Therefore,the coordinates of vertex $D$ are $(1, 0)$.

(N/A) To find the midpoint of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the midpoint formula:
Midpoint $= \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Given points are $(x_1, y_1) = (3, 5)$ and $(x_2, y_2) = (7, 5)$.
Substituting these values into the formula:
Midpoint $= \left( \frac{3 + 7}{2}, \frac{5 + 5}{2} \right)$
Midpoint $= \left( \frac{10}{2}, \frac{10}{2} \right)$
Midpoint $= (5, 5)$
Since the calculated midpoint is $(5, 5)$,which matches the given point,it is proven that $(5, 5)$ is the midpoint of the line segment joining $(3, 5)$ and $(7, 5)$.

(A) Let the $Y$-axis divide the line segment $AB$ at point $P(0, y)$ in the ratio $k:1$.
Using the section formula,the coordinates of point $P$ are given by:
$P = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$
Here,$m_1 = k$,$m_2 = 1$,$x_1 = -2$,$y_1 = 4$,$x_2 = 1$,$y_2 = 1$.
Since the point lies on the $Y$-axis,its $x$-coordinate is $0$:
$0 = \frac{k(1) + 1(-2)}{k + 1}$
$0 = k - 2 \implies k = 2$.
So,the ratio is $2:1$.
Now,find the $y$-coordinate:
$y = \frac{2(1) + 1(4)}{2 + 1} = \frac{2 + 4}{3} = \frac{6}{3} = 2$.
Thus,the point of division is $(0, 2)$ and the ratio is $2:1$.

(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 $(3, 3).$

(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 $(4, 5)$.

(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,$\vec{C} = \vec{B} + \vec{BC} = (6, 1) + (9, -6) = (15, -5)$.

(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 $\left(\frac{30}{7}, \frac{13}{7}\right)$.

(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 $(0, 0)$.

(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 $= (13, 9)$.

(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) 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.

(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 $(3,4)$.

(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 $(0, 14)$.

(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 $(1, 6)$.

(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,$a = 8 - 3 = 5$.

(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 $(0, -3)$.

(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.

(D) Given that $P-R-Q$ is a line segment where $R$ lies between $P$ and $Q$.
We are given the ratio $\frac{PR}{PQ} = \frac{2}{7}$.
Since $PQ = PR + RQ$,we have $RQ = PQ - PR$.
Substituting the values,we get $\frac{PR}{RQ} = \frac{PR}{PQ - PR} = \frac{2}{7 - 2} = \frac{2}{5}$.
This means $R$ divides $PQ$ in the ratio $2:5$ from $P$.
However,the question asks for the ratio from $Q$,which is $QR:RP$.
Therefore,$\frac{QR}{RP} = \frac{5}{2}$.
Thus,$R$ divides $\overline{PQ}$ from $Q$ in the ratio $5:2$.

(B) Given that point $P$ lies on the line segment $\overline{AB}$ such that $A-P-B$.
The ratio in which $P$ divides $\overline{AB}$ from $A$ is given as $\frac{AP}{PB} = \frac{2}{3}$.
We need to find the ratio in which $P$ divides $\overline{AB}$ from $B$,which is $\frac{BP}{PA}$.
Since $\frac{AP}{PB} = \frac{2}{3}$,taking the reciprocal gives $\frac{PB}{AP} = \frac{3}{2}$.
Therefore,the required ratio is $3:2$.

(A) The centroid of a triangle is the point where the three medians of the triangle intersect.
For a triangle with vertices $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$,and $C(x_{3}, y_{3})$,the coordinates of the centroid $(G)$ are calculated by taking the arithmetic mean of the $x$-coordinates and the $y$-coordinates of the vertices.
The formula for the centroid is given by:
$G = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
Therefore,option $A$ is the correct answer.

(A) The formula for the coordinates of the midpoint of 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 $(8, 10)$ and $(4, 8)$.
Here,$x_1 = 8, y_1 = 10, x_2 = 4, y_2 = 8$.
Substituting these values into the formula:
Midpoint $= \left(\frac{8 + 4}{2}, \frac{10 + 8}{2}\right) = \left(\frac{12}{2}, \frac{18}{2}\right) = (6, 9)$.
Therefore,the required coordinates of the midpoint are $(6, 9)$.

(C) The midpoint formula for a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Given $A(4, 3)$ and $B(8, 9)$,we have $x_1 = 4, y_1 = 3, x_2 = 8, y_2 = 9$.
Substituting these values into the formula:
Midpoint $= \left( \frac{4 + 8}{2}, \frac{3 + 9}{2} \right)$
Midpoint $= \left( \frac{12}{2}, \frac{12}{2} \right)$
Midpoint $= (6, 6)$.

(A) The midpoint $M(x, y)$ of a line segment with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by the formula:
$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Given $A(3, 5)$ and $B(7, 5)$,we have $x_1 = 3, y_1 = 5$ and $x_2 = 7, y_2 = 5$.
Substituting these values into the formula:
$x = \frac{3 + 7}{2} = \frac{10}{2} = 5$
$y = \frac{5 + 5}{2} = \frac{10}{2} = 5$
Therefore,the midpoint is $(5, 5)$.

(A) The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
Midpoint $= (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$
Given points are $(x_1, y_1) = (12, 10)$ and $(x_2, y_2) = (0, 8)$.
Substituting these values into the formula:
Midpoint $= (\frac{12 + 0}{2}, \frac{10 + 8}{2})$
Midpoint $= (\frac{12}{2}, \frac{18}{2})$
Midpoint $= (6, 9)$
Therefore, the correct option is $A$.

(C) Let the $Y-$ axis divide the line segment joining $A(-3, -4)$ and $B(1, -2)$ in the ratio $m:n$ at point $M(0, y)$.
Since the point lies on the $Y-$ axis,its $x-$ coordinate is $0$.
Using the section formula for the $x-$ coordinate: $x = \frac{mx_2 + nx_1}{m + n}$.
Substituting the values: $0 = \frac{m(1) + n(-3)}{m + n}$.
This implies $m - 3n = 0$,which gives $m = 3n$.
Therefore,the ratio $\frac{m}{n} = \frac{3}{1}$,or $m:n = 3:1$.

(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 the points $P(3, 4)$ and $Q(a, 8)$,the midpoint $M$ is $\left(\frac{3 + a}{2}, \frac{4 + 8}{2}\right)$.
We are given that the midpoint is $M(5, 6)$.
Equating the coordinates,we get $\frac{3 + a}{2} = 5$ and $\frac{4 + 8}{2} = 6$.
From the first equation,$3 + a = 10$,which implies $a = 10 - 3 = 7$.
Thus,the value of $a$ is $7$.

(B) According to the section formula,if a point $P(x, y)$ divides the line segment joining points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ in the ratio $m:n$,the coordinates of $P$ are given by:
$x = \frac{mx_{2} + nx_{1}}{m + n}$
$y = \frac{my_{2} + ny_{1}}{m + n}$
Here,the ratio is given as $\lambda : 1$,so $m = \lambda$ and $n = 1$.
Substituting these values into the formula:
$x = \frac{\lambda x_{2} + (1)x_{1}}{\lambda + 1} = \frac{\lambda x_{2} + x_{1}}{\lambda + 1}$
$y = \frac{\lambda y_{2} + (1)y_{1}}{\lambda + 1} = \frac{\lambda y_{2} + y_{1}}{\lambda + 1}$
Thus,the coordinates are $\left(\frac{\lambda x_{2}+x_{1}}{\lambda+1}, \frac{\lambda y_{2}+y_{1}}{\lambda+1}\right)$.