Coordinate Geometry - Straight Line

Find the distance between the points whose polar coordinates are $\left(a, \dfrac{\pi}{2}\right)$ and $\left(3a, \dfrac{\pi}{6}\right)$.

The given polar coordinates are $\;$ $A\left(r_1, \theta_1\right) = \left(a, \dfrac{\pi}{2}\right)$, $\;\;$ $B\left(r_2, \theta_2\right) = \left(3a, \dfrac{\pi}{6}\right)$

$\begin{aligned} \text{Distance } AB & = \sqrt{r_1^2 + r_2^2 -2 r_1 r_2 \cos \left(\theta_1 - \theta_2\right)} \\\\ & = \sqrt{a^2 + \left(3a\right)^2 - 2 \times a \times 3a \cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{6}\right)} \\\\ & = \sqrt{10a^2 - 6a^2 \cos \left(\dfrac{\pi}{3}\right)} \\\\ & = \sqrt{10a^2 - 3a^2} \\\\ & = a \sqrt{7} \; \text{units} \end{aligned}$

Coordinate Geometry - Straight Line

Find the points whose polar coordinates are:

1. $\left(2, 45^\circ\right)$


2. $\left(5, \dfrac{-\pi}{2}\right)$

1. The polar coordinates are $\;$ $\left(r, \theta\right) = \left(2, 45^\circ\right)$
   
   Let the corresponding Cartesian coordinates be $\left(x, y\right)$
   
   Now, $\;$ $x = r \cos \theta = 2 \cos 45^\circ = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
   
   and $\;$ $y = r \sin \theta = 2 \sin 45^\circ = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
   
   $\therefore \;$ The corresponding Cartesian coordinates are $\;$ $\left(x, y\right) = \left(\sqrt{2}, \sqrt{2}\right)$



2. The polar coordinates are $\;$ $\left(r, \theta\right) = \left(5, \dfrac{-\pi}{2}\right)$
   
   Let the corresponding Cartesian coordinates be $\left(x, y\right)$
   
   Now, $\;$ $x = r \cos \theta = 5 \cos \left(\dfrac{-\pi}{2}\right) = 0$
   
   and $\;$ $y = r \sin \theta = 2 \sin \left(\dfrac{-\pi}{2}\right) = -2$
   
   $\therefore \;$ The corresponding Cartesian coordinates are $\;$ $\left(x, y\right) = \left(0, -2\right)$

Coordinate Geometry - Straight Line

Find the area of the triangle, the equations whose sides are $\;$ $y + x = 0$, $\;$ $y = x + 6$ $\;$ and $\;$ $y = 7x + 5$.

The sides of the triangle are

$x + y = 0$ $\;\;\; \cdots \; (1)$,

$y = x + 6$ $\;$ i.e. $\;$ $x - y + 6 = 0$ $\;\;\; \cdots \; (2)$,

$y = 7x + 5$ $\;$ i.e. $\;$ $7x - y + 5 = 0$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Let equations $(3)$ and $(1)$ intersect at point $C$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(-3, 3\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{1}{6}, \dfrac{37}{6}\right)$

Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{-5}{8}, \dfrac{5}{8}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Then, length of altitude $AD = \left|\dfrac{7 \times \left(-3\right) -1 \times 3 + 5}{\sqrt{7^2 + \left(-1\right)^2}}\right| = \dfrac{19}{\sqrt{50}} = \dfrac{19}{5 \sqrt{2}}$ units

Distance $BC = \sqrt{\left(\dfrac{1}{6}+ \dfrac{5}{8}\right)^2 + \left(\dfrac{37}{6} - \dfrac{5}{8}\right)^2} = \sqrt{\left(\dfrac{19}{24}\right)^2 + \left(\dfrac{133}{24}\right)^2} = \dfrac{\sqrt{18050}}{24} = \dfrac{95 \sqrt{2}}{24}$ units

$\therefore \;$ Area of triangle whose sides are the given equations is

$= \dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times \dfrac{19}{5 \sqrt{2}} \times \dfrac{95 \sqrt{2}}{24} = \dfrac{361}{48}$ sq units

Coordinate Geometry - Straight Line

If $\left(x_1, y_1\right)$ be the coordinates of the foot of the perpendicular from the origin to the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$, show that $\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$

Let $A = \left(0, 0\right)$; $\;$ $B = \left(x_1, y_1\right)$ $\;$ be the foot of the perpendicular

Slope of perpendicular $AB = m_1 = \dfrac{y_1}{x_1}$

Equation of given line is $\;$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$

i.e. $\;$ $bx + ya = ab$

$\therefore \;$ Slope of given line $= m_2 = \dfrac{-b}{a}$

$\because \;$ $AB$ is perpendicular to the given line, $\;$ $m_1 = \dfrac{-1}{m_2}$

i.e. $\;$ $\dfrac{y_1}{x_1} = \dfrac{a}{b}$ $\implies$ $y_1 = \dfrac{a x_1}{b}$ $\;\;\; \cdots \; (1)$

$\because \;$ $B$ is the foot of the perpendicular, point $B$ lies on the given line.

$\therefore \;$ We have, $\;$ $\dfrac{x_1}{a} + \dfrac{y_1}{b} = 1$

i.e. $\;$ $bx_1 + ay_1 = ab$

i.e. $\;$ $bx_1 + a \times \dfrac{ax_1}{b} = ab$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $b^2 x_1 + a^2 x_1 = ab^2$

i.e. $\;$ $\left(a^2 + b^2\right) x_1 = ab^2$ $\implies$ $a^2 + b^2 = \dfrac{ab^2}{x_1}$ $\;\;\; \cdots \; (2)$

Now,

$\begin{aligned} \left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) & = \left(x_1^2 + \dfrac{a^2 x_1^2}{b^2}\right) \left(x_1 + \dfrac{ax_1}{b}\right) \\\\ & = \dfrac{\left(b^2 x_1^2 + a^2 x_1^2\right) \left(b x_1 + a x_1\right)}{b^3} \\\\ & = \dfrac{x_1^2 \left(a^2 + b^2\right) x_1 \left(a + b\right)}{b^3} \\\\ & = \dfrac{x_1^3 \left(a^2 + b^2\right) \left(a + b\right)}{b^3} \\\\ & = x_1^3 \times \dfrac{ab^2}{x_1} \times \dfrac{\left(a + b\right)}{b^3} \;\;\; \left[\text{by equation (2)}\right] \\\\ & = \dfrac{a \left(a + b\right)x_1^2}{b} \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \left(a + b\right) x_1 y_1 & = \left(a + b\right) \times x_1 \times \dfrac{ax_1}{b} \\\\ & = \dfrac{a \left(a + b\right) x_1^2}{b} \;\;\; \cdots \; (4) \end{aligned}$

$\therefore \;$ We have from equations $(3)$ and $(4)$

$\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$

Hence proved.

Coordinate Geometry - Straight Line

Find the coordinates of the foot of the perpendicular from the point $\left(-2, 0\right)$ on the line $2x + 3y = 9$.

Given point $= A \left(x_1, y_1\right) = \left(-2, 0\right)$

Let the foot of the perpendicular from point $A$ be at the point $\;$ $B \left(x_2, y_2\right)$

Equation of given line is $\;$ $2x + 3y = 9$ $\;\;\; \cdots \; (1)$

Slope of the given line $\;$ $2x + 3y = 9$ is $\;$ $m = \dfrac{-2}{3}$

$\because \;$ $AB$ is perpendicular to the given line, slope of perpendicular $= m_1 = \dfrac{-1}{m} = \dfrac{3}{2}$

$\therefore \;$ Equation of perpendicular $AB$ passing through the point $A$ and having slope $= m_1$ is

$y - 0 = \dfrac{3}{2} \left(x + 2\right)$

i.e. $\;$ $2y = 3x + 6$

i.e. $\;$ $3x - 2y + 6 = 0$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\left(0, 3\right)$

The point of intersection is the foot of the perpendicular from point $A$ on the line given by equation $(1)$.

$\therefore \;$ $B = \left(0, 3\right)$

Coordinate Geometry - Straight Line

The equations of two sides of a triangle are $\;$ $2x - y = 0$ $\;$ and $\;$ $x + 3y - 7 = 0$. The orthocenter is the point $\left(2,1\right)$. Find the equation of the third side.

The sides of the triangle are

$2x - y = 0$ $\;\;\; \cdots \; (1)$, $\;$ $x + 3y - 7 = 0$ $\;\;\; \cdots \; (2)$

Let the equation of the third side of the triangle be $\;$ $ax + by + c = 0$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(1, 2\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{-7b-3c}{3a-b}, \dfrac{7a+c}{3a-b}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Slope of equation $(3)$ is $m_1 = \dfrac{-a}{b}$

$\therefore \;$ Slope of altitude $AD = \dfrac{-1}{m_1} = \dfrac{b}{a}$

$AD$ passes through the point $A$ and has slope $\dfrac{-1}{m_1}$.

$\therefore \;$ Equation of $AD$ is

$y - 2 = \dfrac{b}{a} \left(x - 1\right)$

i.e. $\;$ $ay - 2a= bx - b$

i.e. $\;$ $bx - ay + 2a-b = 0$ $\;\;\; \cdots \; (4)$

Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.

Slope of equation $(1)$ is $m_2 = 2$

$\therefore \;$ Slope of altitude $BE = \dfrac{-1}{m_2} = \dfrac{-1}{2}$

$BE$ passes through the point $B$ and has slope $\dfrac{-1}{m_2}$.

$\therefore \;$ Equation of $BE$ is

$y - \left(\dfrac{7a + c}{3a - b}\right) = \dfrac{-1}{2} \left[x + \dfrac{7b + 3c}{3a - b}\right]$

i.e. $\;$ $\left(3a - b\right)y - 7a - c = \dfrac{-1}{2} \left[\left(3a - b\right)x + 7b + 3c\right]$

i.e. $\;$ $\left(3a - b\right)x + \left(6a - 2b\right)y - 14a + 7b + c = 0$ $\;\;\; \cdots \; (5)$

Given: Orthocenter of the triangle is $\;$ $O \left(2, 1\right)$.

$\therefore \;$ Point $O$ satisfies both equations $(4)$ and $(5)$.

Substituting $\left(x, y\right) = \left(2,1\right)$ in equation $(4)$ gives,

$2b - a + 2a - b = 0$

i.e. $\;$ $a + b = 0$ $\implies$ $a = -b$ $\;\;\; \cdots \; (6)$

Substituting $\left(x, y\right) = \left(2,1\right)$ in equation $(5)$ gives,

$2 \left(3a - b\right) + 6a - 2b - 14a + 7b + c = 0$

i.e. $\;$ $6a - 2b + 6a - 2b - 14a + 7b + c = 0$

i.e. $\;$ $-2a + 3b + c = 0$ $\;\;\; \cdots \; (7)$

In view of equation $(6)$, equation $(7)$ becomes

$2b + 3b + c = 0$ $\implies$ $c = -5b$ $\;\;\; \cdots \; (8)$

$\therefore \;$ In view of equations $(6)$ and $(8)$ equation $(3)$ becomes,

$-bx + by -5b = 0$

$\therefore \;$ The equation of the third side of the triangle is

$x - y + 5 = 0$

Coordinate Geometry - Straight Line

The equations of the sides of a triangle are $\;$ $x - y - 1 = 0$, $\;$ $2x - 5 = 0$ $\;$ and $\;$ $3x + y -2 = 0$. Find the coordinates of the orthocenter.

The sides of the triangle are

$x - y -1 = 0$ $\;\;\; \cdots \; (1)$, $\;$ $2x - 5 = 0$ $\;\;\; \cdots \; (2)$, $\;$ $3x + y -2 = 0$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Let equations $(3)$ and $(1)$ intersect at point $C$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{5}{2}, \dfrac{-11}{2}\right)$

Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{3}{4}, \dfrac{-1}{4}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Slope of equation $(3)$ is $m_1 = -3$

$\therefore \;$ Slope of altitude $AD = \dfrac{-1}{m_1} = \dfrac{1}{3}$

$AD$ passes through the point $A$ and has slope $\dfrac{-1}{m_1}$.

$\therefore \;$ Equation of $AD$ is

$y - \dfrac{3}{2} = \dfrac{1}{3} \left(x - \dfrac{5}{2}\right)$

i.e. $\;$ $6y - 9 = 2x - 5$

i.e. $\;$ $x - 3y + 2 = 0$ $\;\;\; \cdots \; (4)$

Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.

Slope of equation $(1)$ is $m_2 = 1$

$\therefore \;$ Slope of altitude $BE = \dfrac{-1}{m_2} = -1$

$BE$ passes through the point $B$ and has slope $\dfrac{-1}{m_2}$.

$\therefore \;$ Equation of $BE$ is

$y + \dfrac{11}{2} = -1 \left(x - \dfrac{5}{2}\right)$

i.e. $\;$ $2y + 11 = -2x + 5$

i.e. $\;$ $x + y + 3 = 0$ $\;\;\; \cdots \; (5)$

Solving equations $(4)$ and $(5)$ simultaneously gives the point of intersection as $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$

$\therefore \;$ The orthocenter is $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$.