Coordinate Geometry - Cartesian Coordinates

If the point $D$ divides the base $BC$ of a $\triangle ABC$ in the ratio $n : m$, prove that

$m \cdot \left(AB\right)^2 + n \cdot \left(AC\right)^2 = \left(m + n\right) \left(AD\right)^2 + m \cdot \left(BD\right)^2 + n \cdot \left(DC\right)^2$

Let $B$ be at the origin. Then the coordinates of point $B = \left(0,0\right)$

Let $BC$ be along the $X$ axis and the line perpendicular to $BC$ be the $Y$ axis.

Let $BC = a$. $\;$ Then $\;$ $C = \left(a, 0\right)$

Let $\;$ $a = \left(x, y\right)$

Since point $D$ divides the base $BC$, point $D$ is also on the $X$ axis.

Let $\;$ $D = \left(p, 0\right)$

The point $D$ divides $BC$ in the ratio $n : m$

$\therefore \;$ By section formula for internal division, the $x$ coordinate of point $D$ is

$p = \dfrac{n \times a + m \times 0}{m + n} = \dfrac{na}{m + n}$

$\therefore \;$ $D = \left(\dfrac{na}{m + n}, 0\right)$

Now, by distance formula,

$\left(AB\right)^2 = \left(x - 0\right)^2 + \left(y - 0\right)^ = x^2 + y^2$

$\left(AC\right)^2 = \left(x - a\right)^2 + \left(y - 0\right)^2 = x^2 + a^2 - 2ax + y^2$

$\left(AD\right)^2 = \left(x - \dfrac{na}{m + n}\right)^2 + \left(y - 0\right)^2 = x^2 + \dfrac{a^2 n^2}{\left(m + n\right)^2} - \dfrac{2xna}{m + n} + y^2$

$\left(BD\right)^2 = \left(0 - \dfrac{na}{m + n}\right)^2 + 0^2 = \dfrac{a^2 n^2}{\left(m + n\right)^2}$

$\left(DC\right)^2 = \left(a - \dfrac{na}{m + n}\right)^2 + 0^2 = \left(\dfrac{am + an - an}{m + n}\right)^2 = \dfrac{a^2 m^2}{\left(m + n\right)^2}$

Now,

$\begin{aligned} LHS & = m \cdot \left(AB\right)^2 + n \cdot \left(AC\right)^2 \\\\ & = m \left(x^2 + y^2\right) = n \left(x^2 + a^2 - 2ax + y^2\right) \\\\ & = x^2 \left(m + n\right) + y^2 \left(m + n\right) + a^2 n - 2anx \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + an \left(a - 2x\right) \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} RHS & = \left(m + n\right) \left(AD\right)^2 + m \cdot \left(BD\right)^2 + n \cdot \left(DC\right)^2 \\\\ & = \left(m + n\right) \left[x^2 + \dfrac{a^2 n^2}{\left(m + n\right)^2} - \dfrac{2xna}{m + n} + y^2\right] + m \times \dfrac{a^2 n^2}{\left(m + n\right)^2} + n \times \dfrac{a^2 m^2}{\left(m + n\right)^2} \\\\ & = \left(m + n\right) x^2 + \left(m + n\right) y^2 + \dfrac{a^2 n^2}{m + n} - 2xna + \dfrac{a^2}{\left(m + n\right)^2} \left(mn^2 + nm^2\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + \dfrac{a^2 n^2}{m + n} - 2xna + \dfrac{a^2 mn}{\left(m + n\right)^2} \left(n + m\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + \dfrac{a^2 n^2}{m + n} + \dfrac{a^2 mn}{m + n} \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + a^2 n \left(\dfrac{n}{n + m} + \dfrac{m}{m + n}\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + a^2 n \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + an \left(a - 2x\right) \;\;\; \cdots \; (2) \end{aligned}$

$\therefore \;$ From equations $(1)$ and $(2)$ we have $LHS = RHS$.

Hence proved.

Coordinate Geometry - Cartesian Coordinates

The centroid of triangle $ABC$ is at the point $\left(2, 3\right)$. The coordinates of $A$ and $B$ are $\left(5, 6\right)$ and $\left(-1, 4\right)$. Find the coordinates of point $C$.

Given: $\;$ $A \left(x_1, y_1\right) = \left(5, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-1, 4\right)$, $\;$ centroid $G \left(p, q\right) = \left(2, 3\right)$

Let the coordinates of point $C = \left(x, y\right)$

Let $M$ be the midpoint of $AB$.

Then, by midpoint formula, the coordinates of point $M$ are

$M = \left(\dfrac{5 - 1}{2}, \dfrac{6 + 4}{2}\right) = \left(2, 5\right)$

$CM$ is the median of $\triangle ABC$.

The centroid $G$ divides the median $CM$ internally in the ratio $2 : 1$.

$\therefore \;$ By section formula for internal division, we have,

for the $x$ coordinate: $\;$ $2 = \dfrac{2 \times 2 + 1 \times x}{2 + 1}$

i.e. $\;$ $6 = 4 + x$ $\implies$ $x = 2$

for the $y$ coordinate: $\;$ $3 = \dfrac{2 \times 5 + 1 \times y}{2 + 1}$

i.e. $\;$ $9 = 10 + y$ $\implies$ $y = -1$

$\therefore \;$ $C \left(x, y\right) = \left(2, -1\right)$

Coordinate Geometry - Cartesian Coordinates

Find the area of the quadrilateral whose angular points are $\left(-1, 6\right)$, $\left(-3, -9\right)$, $\left(5, -8\right)$ and $\left(3, 9\right)$.

Let $\;$ $A \left(x_1, y_1\right) = \left(-1, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-3, -9\right)$, $\;$ $C \left(x_3, y_3\right) = \left(5, -8\right)$, $\;$ $D \left(x_4, y_4\right) = \left(3, 9\right)$

Area of quadrilateral $ABCD$ is

$\begin{aligned} \Delta & = \dfrac{1}{2} \left(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - y_1 x_2 - y_2 x_3 - y_3 x_4 - y_4 x_1\right) \\\\ & = \dfrac{1}{2} \left[-1 \times \left(-9\right) - 3 \times \left(-8\right) + 5 \times 9 + 3 \times 6 -6 \times \left(-3\right) + 9 \times 5 + 8 \times 3 - 9 \times \left(-1\right) \right] \\\\ & = \dfrac{1}{2} \left[9 + 24 + 45 + 18 + 18 + 45 + 24 + 9\right] \\\\ & = \dfrac{1}{2} \times 192 = 96 \end{aligned}$

Coordinate Geometry - Cartesian Coordinates

Show that the points $\left(a, b + c\right)$, $\left(b, c + a\right)$ and $\left(c, a + b\right)$ are collinear.

Let $A \left(x_1, y_1\right) = \left(a, b + c\right)$, $\;$ $B \left(x_2, y_2\right) = \left(b, c + a\right)$ and $C \left(x_3, y_3\right) = \left(c, a + b\right)$

Area of triangle formed by the points $ABC$ is

$\begin{aligned} \Delta & = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right] \\\\ & = \dfrac{1}{2} \left[a \left(c + a - a - b\right) + b \left(a + b - b - c\right) + c \left(b + c -c - a\right)\right] \\\\ & = \dfrac{1}{2} \left[a \left(c - b\right) + b \left(a - c\right) + c \left(b - a\right)\right] \\\\ & = \dfrac{1}{2} \left[ac - ab + ab - bc + bc - ac\right] \\\\ & = 0 \end{aligned}$

$\because \;$ The area of triangle formed by the points $A$, $B$ and $C$ is zero,

$\implies$ the points $A$, $B$ and $C$ are collinear.

Coordinate Geometry - Cartesian Coordinates

Find the area of the triangle whose vertices are $\left(a \cos \theta , b \sin \theta\right)$, $\left(-a \sin \theta, b \cos \theta\right)$ and $\left(-a \cos \theta, -b \sin \theta\right)$

Let $\;$ $A \left(x_1, y_1\right) = \left(a \cos \theta , b \sin \theta\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-a \sin \theta, b \cos \theta\right)$, $\;$ $C \left(x_3, y_3\right) = \left(-a \cos \theta, -b \sin \theta\right)$

Area of $\triangle ABC = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right]$

$= \dfrac{1}{2} \left[a \cos \theta \left(b \cos \theta + b \sin \theta\right) - a \sin \theta \left(-b \sin \theta - b \sin \theta\right) - a \cos \theta \left(b \sin \theta - b \cos \theta\right)\right]$

$= \dfrac{1}{2} \left[ab \cos^2 \theta + ab \sin \theta \cos \theta + 2ab \sin^2 \theta -ab \sin \theta \cos \theta + ab \cos^2 \theta\right]$

$= \dfrac{1}{2} \left[2ab \cos^2 \theta + 2 ab \sin^2 \theta\right]$

$= \dfrac{1}{2} \left[2ab \left(\sin^2 \theta + \cos^2 \theta\right)\right]$

$= \dfrac{1}{2} \times 2ab \times 1 = ab$

Coordinate Geometry - Cartesian Coordinates

$P \left(3,7\right)$ is a point on the line joining $A \left(1,1\right)$ and $B \left(6,16\right)$. Find the harmonic conjugate of $P$ with respect to $A$ and $B$.

Let the point $P \left(3,7\right)$ divide the line joining $A \left(1,1\right)$ and $B \left(6,16\right)$ internally in the ratio $k : 1$.

Then, by section formula,

$3 = \dfrac{1 \times 1 + 6 \times k}{k + 1}$

i.e. $\;$ $3k + 3 = 1 + 6k$

i.e. $\;$ $3k = 2$ $\implies$ $k = \dfrac{2}{3}$

i.e. $\;$ the point $P$ divides the line $AB$ internally in the ratio $2:3$

Let $Q \left(x,y\right)$ be the harmonic conjugate of $P$ with respect to $A$ and $B$.

Then the point $Q$ divides $AB$ externally in the ratio $2 : 3$

Then, by section formula for external division,

$x = \dfrac{2 \times 6 - 3 \times 1}{2 - 3} = \dfrac{12 - 3}{-1} = -9$

$y = \dfrac{2 \times 16 - 3 \times 1}{2 - 3} = \dfrac{32 - 3}{-1} = -29$

$\therefore \;$ $Q \left(x, y\right) = \left(-9, -29\right)$

Coordinate Geometry - Cartesian Coordinates

The point $R \left(22,23\right)$ divides the line join of $P \left(7,5\right)$ and $Q$ externally in the ratio $3 : 5$. Find $Q$.

Let $Q = \left(p,q\right)$

The point $R \left(22,23\right)$ divides the line joining $P \left(7,5\right)$ and $Q \left(p,q\right)$ externally in the ratio $3:5$.

Then, by section formula,

$22 = \dfrac{p \times 3 - 7 \times 5}{3 - 5}$

i.e. $\;$ $22 = \dfrac{3p - 35}{-2}$

i.e. $\;$ $3p = -44 + 35$ $\implies$ $p = \dfrac{-9}{3} = -3$

and $\;$ $23 = \dfrac{q \times 3 - 5 \times 5}{3 - 5}$

i.e. $\;$ $23 = \dfrac{3q - 25}{-2}$

i.e. $\;$ $3q = - 46 + 25$ $\implies$ $q = \dfrac{-21}{3} = -7$

$\therefore \;$ $Q = \left(-3, -7\right)$

Coordinate Geometry - Cartesian Coordinates

Find the coordinates of the points which divide internally and externally the line joining $\left(-4, 4\right)$ and $(1,7)$ in the ratio $2 : 1$.

Let $A = \left(-4,4\right)$ and $B = \left(1,7\right)$

Let $P \left(p,q\right)$ divide the line joining $AB$ internally in the ratio $2:1$.

Then, by section formula,

$p = \dfrac{1 \times 2 + \left(-4\right) \times 1}{2 + 1} = \dfrac{2 - 4}{3} = \dfrac{-2}{3}$

$q = \dfrac{7 \times 2 + 4 \times 1}{2 + 1} = \dfrac{14 + 4}{3} = \dfrac{18}{3} = 6$

$\therefore \;$ $P = \left(\dfrac{-2}{3}, 6\right)$

Let $Q \left(x,y\right)$ divide the line joining $AB$ externally in the ratio $2:1$.

Then, by section formula,

$x = \dfrac{1 \times 2 - \left(-4\right) \times 1}{2 - 1} = \dfrac{2 + 4}{1} = 6$

$y = \dfrac{7 \times 2 - 4 \times 1}{2 - 1} = \dfrac{14 - 4}{1} = 10$

$\therefore \;$ $Q = \left(6, 10\right)$