Mix Examples - Coordinate Geometry Questions in English

(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 $3:2$.

(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.
Therefore,$\frac{AM}{AB} = \frac{MN}{BC}$.
Given $MN = 3$ and $BC = 7$,we have $\frac{AM}{AB} = \frac{3}{7}$.
Since $AB = AM + MB$,we can write $\frac{AM}{AM + MB} = \frac{3}{7}$.
By cross-multiplication,$7 AM = 3 AM + 3 MB$.
$4 AM = 3 MB$.
Therefore,$\frac{AM}{MB} = \frac{3}{4}$.
Thus,$M$ divides $\overline{AB}$ from $A$ in the ratio $3:4$.

(C) 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 $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
This formula is derived from the shoelace formula or by calculating the area of trapezoids formed by dropping perpendiculars from the vertices to the $x$-axis.
Comparing this with the given options,option $C$ is the correct representation.

(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.

(B) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Here,$(x_1, y_1) = (7, 5)$ and $(x_2, y_2) = (2, 5)$.
Substituting the values into the formula:
$d = \sqrt{(2 - 7)^2 + (5 - 5)^2}$
$d = \sqrt{(-5)^2 + (0)^2}$
$d = \sqrt{25 + 0}$
$d = \sqrt{25}$
$d = 5$
Therefore,the distance between the two points is $5$ units.

(C) The distance of any point $P(x, y)$ from the $X$-axis is given by the absolute value of its $y$-coordinate,which is $|y|$.
Given the point $P(2, -3)$,the $x$-coordinate is $2$ and the $y$-coordinate is $-3$.
Therefore,the distance from the $X$-axis is $|-3| = 3$ units.
Thus,the correct option is $C$.

(A) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(2, 4)$ and $B(-3, 4)$.
Here,$x_1 = 2, y_1 = 4$ and $x_2 = -3, y_2 = 4$.
Substituting these values into the formula:
$d = \sqrt{(-3 - 2)^2 + (4 - 4)^2}$
$d = \sqrt{(-5)^2 + (0)^2}$
$d = \sqrt{25 + 0}$
$d = \sqrt{25} = 5$.
Therefore,the distance between the points is $5$ units.

(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 $(5, 5)$.

(B) 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 $B$.

(A) The distance between points $A(3, a)$ and $B(4, 1)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $d = \sqrt{10}$,we have $\sqrt{(4 - 3)^2 + (1 - a)^2} = \sqrt{10}$.
Squaring both sides,we get $(4 - 3)^2 + (1 - a)^2 = 10$.
$1^2 + (1 - 2a + a^2) = 10$.
$1 + 1 - 2a + a^2 = 10$.
$a^2 - 2a + 2 = 10$.
$a^2 - 2a - 8 = 0$.
Factoring the quadratic equation: $(a - 4)(a + 2) = 0$.
Thus,$a = 4$ or $a = -2$.
Since $4$ is provided in option $A$,the correct value is $4$.

(B) The given vertices are $A(0, 0)$,$B(3.1, 0)$,and $C(0, 4.5)$.
Point $A(0, 0)$ is the origin.
Point $B(3.1, 0)$ lies on the $X$-axis.
Point $C(0, 4.5)$ lies on the $Y$-axis.
Since the $X$-axis and $Y$-axis are perpendicular to each other at the origin,the angle formed at vertex $A(0, 0)$ is $90^{\circ}$.
Therefore,the triangle formed by these vertices is a right-angled 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 $A(-4, -6)$ and $B(6, b)$ with distance $d = 10$.
Substituting the values into the formula:
$10 = \sqrt{(6 - (-4))^2 + (b - (-6))^2}$
$10 = \sqrt{(6 + 4)^2 + (b + 6)^2}$
$10 = \sqrt{10^2 + (b + 6)^2}$
Squaring both sides:
$100 = 100 + (b + 6)^2$
$(b + 6)^2 = 100 - 100$
$(b + 6)^2 = 0$
Taking the square root of both sides:
$b + 6 = 0$
$b = -6$.

(A) To determine the type of quadrilateral $ABCD$,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 $AB = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{2^2 + 0^2} = 2$.
$2$. Length of $BC = \sqrt{(2 - 2)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = 2$.
$3$. Length of $CD = \sqrt{(0 - 2)^2 + (2 - 2)^2} = \sqrt{(-2)^2 + 0^2} = 2$.
$4$. Length of $DA = \sqrt{(0 - 0)^2 + (0 - 2)^2} = \sqrt{0^2 + (-2)^2} = 2$.
Since all sides are equal $(AB = BC = CD = DA = 2)$,the quadrilateral is either a square or a rhombus.
Next,we check the diagonals $AC$ and $BD$:
Diagonal $AC = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Diagonal $BD = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Since all sides are equal and the diagonals are equal,$\square ABCD$ is a square.

(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 $m:n = 3:1$.

(A) The formula for the coordinates of the centroid of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by:
Centroid $= \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
Given vertices are $A(3, 2)$,$B(7, 5)$,and $C(2, 2)$.
Substituting the values into the formula:
Centroid $= \left(\frac{3 + 7 + 2}{3}, \frac{2 + 5 + 2}{3}\right)$
Centroid $= \left(\frac{12}{3}, \frac{9}{3}\right)$
Centroid $= (4, 3)$

(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 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 $\left(\frac{\lambda x_{2}+x_{1}}{\lambda+1}, \frac{\lambda y_{2}+y_{1}}{\lambda+1}\right)$.

(D) To determine the correct matching,we analyze the properties of the diagonals of quadrilaterals:
$1$. For a rhombus $(1)$,the diagonals bisect each other at right angles $(b)$.
$2$. For a parallelogram $(2)$,the diagonals bisect each other $(a)$.
$3$. For a rectangle $(3)$,the diagonals are congruent and bisect each other $(d)$.
$4$. For a square $(4)$,the diagonals are congruent and bisect each other at right angles $(c)$.
Thus,the correct matching is $(1-b), (2-a), (3-d), (4-c)$.

(C) $1$. Given coordinates: $A(3, 0)$,$B(0, 0)$,$C(0, -4)$.
$2$. Calculate lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sqrt{9} = 3$.
$BC = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{16} = 4$.
$AC = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$3$. Check for right-angled triangle: $AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 = AC^2$. Since $AB^2 + BC^2 = AC^2$,$\Delta ABC$ is a right-angled triangle at $\angle B$. (Option $A$ is true).
$4$. Midpoint of $\overline{AC} = (\frac{3+0}{2}, \frac{0-4}{2}) = (\frac{3}{2}, -2)$. (Option $B$ is true).
$5$. Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6$. (Option $C$ states area is $12$,which is false).
$6$. $AB=3, BC=4, AC=5$ is true. (Option $D$ is true).
Therefore,the false statement is $C$.