Textbook - Coordinate Geometry Questions in English

(N/A) Taking $A$ as the origin $(0,0)$,we consider $AD$ as the $x$-axis and $AB$ as the $y$-axis.
By observing the grid provided in the figure:
- The point $P$ is located at $4$ units along the $x$-axis and $6$ units along the $y$-axis,so its coordinates are $P(4, 6)$.
- The point $Q$ is located at $3$ units along the $x$-axis and $2$ units along the $y$-axis,so its coordinates are $Q(3, 2)$.
- The point $R$ is located at $6$ units along the $x$-axis and $5$ units along the $y$-axis,so its coordinates are $R(6, 5)$.
Thus,the coordinates of the vertices of the triangle are $P(4, 6)$,$Q(3, 2)$,and $R(6, 5)$.

(N/A) Taking $C$ as the origin $(0,0)$,$CB$ as the $x$-axis,and $CD$ as the $y$-axis:
$1$. The coordinates of the vertices are determined by counting the units from the origin $C$ along the axes.
$2$. The coordinates of the vertices $P, Q,$ and $R$ are $P(12, 2), Q(13, 6),$ and $R(10, 3)$.
$3$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$4$. Substituting the values:
$\text{Area} = \frac{1}{2} |12(6 - 3) + 13(3 - 2) + 10(2 - 6)|$
$\text{Area} = \frac{1}{2} |12(3) + 13(1) + 10(-4)|$
$\text{Area} = \frac{1}{2} |36 + 13 - 40|$
$\text{Area} = \frac{1}{2} |9| = 4.5 \text{ square units}$.

(1:16) Given that,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$.
Since $\frac{AD}{AB} = \frac{AE}{AC}$,by the Converse of Thales Theorem (Basic Proportionality Theorem),$DE \parallel BC$.
Therefore,$\Delta ADE \sim \Delta ABC$ by $AA$ similarity criterion.
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus,$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Now,calculate the area of $\Delta ABC$ using the coordinates $A(4, 6)$,$B(1, 5)$,and $C(7, 2)$:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |4(5 - 2) + 1(2 - 6) + 7(6 - 5)|$
Area $= \frac{1}{2} |4(3) + 1(-4) + 7(1)| = \frac{1}{2} |12 - 4 + 7| = \frac{1}{2} |15| = 7.5$ square units.
Area of $\Delta ADE = \frac{1}{16} \times \text{Area}(\Delta ABC) = \frac{1}{16} \times 7.5 = \frac{7.5}{16} = \frac{15}{32}$ square units.
The ratio of the area of $\Delta ADE$ to the area of $\Delta ABC$ is $1:16$.

(N/A) The median $AD$ of the triangle divides the side $BC$ into two equal parts.
Therefore,$D$ is the mid-point of side $BC$.
The coordinates of the mid-point of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Here,$B = (6, 5)$ and $C = (1, 4)$.
Coordinates of $D = \left(\frac{6+1}{2}, \frac{5+4}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right)$ or $(3.5, 4.5)$.

(N/A) $1$. First,find the coordinates of the midpoint $D$ of the side $BC$. Since $D$ is the midpoint of $BC$ with $B(6, 5)$ and $C(1, 4)$,the coordinates of $D$ are $\left(\frac{6+1}{2}, \frac{5+4}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right)$.
$2$. Now,we need to find the coordinates of point $P$ that divides the line segment $AD$ in the ratio $m:n = 2:1$,where $A(4, 2)$ and $D(\frac{7}{2}, \frac{9}{2})$.
$3$. Using the section formula,the coordinates of $P$ are given by $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
$4$. Substituting the values: $P = \left(\frac{2 \times \frac{7}{2} + 1 \times 4}{2+1}, \frac{2 \times \frac{9}{2} + 1 \times 2}{2+1}\right) = \left(\frac{7+4}{3}, \frac{9+2}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.

(N/A) The median $BE$ of the triangle divides the side $AC$ into two equal parts. Therefore,$E$ is the midpoint of side $AC$.
Coordinates of $E = \left(\frac{4+1}{2}, \frac{2+4}{2}\right) = \left(\frac{5}{2}, 3\right)$.
Point $Q$ divides the side $BE$ in a ratio $2 : 1$. Using the section formula:
Coordinates of $Q = \left(\frac{2 \times \frac{5}{2} + 1 \times 6}{2+1}, \frac{2 \times 3 + 1 \times 5}{2+1}\right) = \left(\frac{5+6}{3}, \frac{6+5}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.
The median $CF$ of the triangle divides the side $AB$ into two equal parts. Therefore,$F$ is the midpoint of side $AB$.
Coordinates of $F = \left(\frac{4+6}{2}, \frac{2+5}{2}\right) = \left(5, \frac{7}{2}\right)$.
Point $R$ divides the side $CF$ in a ratio $2 : 1$. Using the section formula:
Coordinates of $R = \left(\frac{2 \times 5 + 1 \times 1}{2+1}, \frac{2 \times \frac{7}{2} + 1 \times 4}{2+1}\right) = \left(\frac{10+1}{3}, \frac{7+4}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.

(N/A) The coordinates of the vertices are $A(4, 2)$,$B(6, 5)$,and $C(1, 4)$.
First,find the midpoint $D$ of side $BC$ using the midpoint formula: $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{6+1}{2}, \frac{5+4}{2}) = (3.5, 4.5)$.
Now,point $P$ divides the median $AD$ in the ratio $2:1$. Using the section formula $P = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$:
$P = (\frac{2(3.5) + 1(4)}{2+1}, \frac{2(4.5) + 1(2)}{2+1}) = (\frac{7+4}{3}, \frac{9+2}{3}) = (\frac{11}{3}, \frac{11}{3})$.
Next,calculate the centroid $G$ of $\Delta ABC$ using the formula $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$:
$G = (\frac{4+6+1}{3}, \frac{2+5+4}{3}) = (\frac{11}{3}, \frac{11}{3})$.
Observation: The point $P$ is the same as the centroid $G$ of the triangle.

(N/A) The coordinates of the centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are given by the formula:
Centroid $= \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Given vertices are $A(4, 2), B(6, 5),$ and $C(1, 4)$.
Here,$x_1 = 4, y_1 = 2, x_2 = 6, y_2 = 5, x_3 = 1, y_3 = 4$.
Substituting these values into the formula:
Centroid $= \left( \frac{4 + 6 + 1}{3}, \frac{2 + 5 + 4}{3} \right)$
$= \left( \frac{11}{3}, \frac{11}{3} \right)$
Thus,the coordinates of the centroid are $\left( \frac{11}{3}, \frac{11}{3} \right)$.

(C) $P$ is the mid-point of side $AB$. Therefore,the coordinates of $P$ are $\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right) = \left(-1, \frac{3}{2}\right)$.
Similarly,the coordinates of $Q, R$,and $S$ are $(2, 4)$,$\left(5, \frac{3}{2}\right)$,and $(2, -1)$ respectively.
Length of $PQ = \sqrt{(-1-2)^2 + \left(\frac{3}{2}-4\right)^2} = \sqrt{(-3)^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $QR = \sqrt{(2-5)^2 + \left(4-\frac{3}{2}\right)^2} = \sqrt{(-3)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $RS = \sqrt{(5-2)^2 + \left(\frac{3}{2}-(-1)\right)^2} = \sqrt{3^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $SP = \sqrt{(2-(-1))^2 + \left(-1-\frac{3}{2}\right)^2} = \sqrt{3^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of diagonal $PR = \sqrt{(-1-5)^2 + \left(\frac{3}{2}-\frac{3}{2}\right)^2} = \sqrt{(-6)^2 + 0^2} = 6$.
Length of diagonal $QS = \sqrt{(2-2)^2 + (4-(-1))^2} = \sqrt{0^2 + 5^2} = 5$.
Since all sides are equal $(PQ = QR = RS = SP = \sqrt{\frac{61}{4}})$ and the diagonals are not equal $(PR \neq QS)$,the quadrilateral $PQRS$ is a rhombus.