Coordinate Geometry: Reflection

Use a graph sheet for this question. Take $1 \; cm = 1$ unit on both the axes.

1. Plot the points $A \left(0, 5\right)$, $B \left(3,0\right)$, $C \left(1, 0\right)$ and $D \left(1, -5\right)$.
2. Reflect the points $B$, $C$ and $D$ on the $Y$ axis and name them as $B'$, $C'$ and $D'$ respectively.
3. Write the coordinates of $B'$, $C'$ and $D'$.
4. Join the points $ABCDD'C'B'A$, in order, and give a name to the closed figure formed.

The points $A \left(0, 5\right)$, $B \left(3,0\right)$, $C \left(1, 0\right)$ and $D \left(1, -5\right)$ are plotted.

Points $B$, $C$ and $D$ are reflected in the $Y$ axis giving the points $B' \left(-3,0\right)$, $C' \left(-1, 0\right)$ and $D' \left(-1, -5\right)$ respectively.

Points $ABCDD'C'B'A$ joined in order give an arrow.

Coordinate Geometry: Equation of a Line

The coordinates of $P \left(2, 6\right)$ and $Q \left(-3, 5\right)$ are given. Find

1. the gradient of $PQ$;
2. the equation of $PQ$;
3. the coordinates of the point where $PQ$ intersects the $X$ axis.

Given: $\;\;$ $P \left(x_1, y_1\right) = \left(2, 6\right)$, $\;\;$ $Q \left(x_2, y_2\right) = \left(-3, 5\right)$

1. Gradient of $PQ$ $= m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - 6}{-3 - 2} = \dfrac{-1}{-5} = \dfrac{1}{5}$



2. Equation of $PQ$ is
   
   $y - y_1 = m \left(x - x_1\right)$
   
   i.e. $\;$ $y - 6 = \dfrac{1}{5} \left(x - 2\right)$
   
   i.e. $\;$ $5y - 30 = x - 2$
   
   i.e. $\;$ $x - 5y + 28 = 0$



3. When $PQ$ intersects the $X$ axis, the $Y$ coordinate of the point of intersection is $0$.
   
   Let $PQ$ intersect the $X$ axis at the point $A\left(a, 0\right)$.
   
   Substituting the coordinates of point $A$ in the equation of $PQ$ gives,
   
   $a - 5\times 0 + 28 = 0$ $\implies$ $a = -28$
   
   $\therefore \;$ $PQ$ intersects the $X$ axis at the point $\left(-28, 0\right)$.

Linear Inequations

Solve the inequation $\;\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}, \; x \in R$
Write the solution set and represent it on a number line.

Consider $\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{x}{2} + \dfrac{x}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{5x}{6}$

i.e. $\;$ $8 \leq 5x$

i.e. $\;$ $\dfrac{8}{5} \leq x$

i.e. $\;$ $1.6 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{1}{6} + \dfrac{4}{3}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{9}{6}$

i.e. $\;$ $x < 3$ $\;\;\; \cdots \; (2)$

$\therefore \;$ From equations $(1)$ and $(2)$, the solution set of the given inequation is

$\left\{x \; \bigg | \; 1.6 \leq x < 3, \; \; x \in R \right\}$

Commercial Mathematics - Banking

A person deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is $8 \%$ per annum and the person gets ₹ $8088$ from the bank after $3$ years, find the value of the monthly installment.

Let money deposited per month $= P = $ ₹ $x$

Time for which money is deposited $= n = 3$ years $= 36$ months

Rate of interest $= r = 8\%$ per annum

Interest $= I = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100}$

i.e. $\;$ $I =$ ₹ $\left[\dfrac{x \times 36 \times 37}{2 \times 12} \times \dfrac{8}{100}\right]$

i.e. $\;$ $I =$ ₹ $\left[\dfrac{111 x}{25}\right] = $ ₹ $4.44 x$

Total money deposited in $36$ months $= $ ₹ $36 x$

Maturity value (MV) $=$ Total money deposited $+ $ Interest

i.e. $\;$ $MV = $ ₹ $\left(36 x + 4.44 x\right) = $ ₹ $40.44 x$

Given: $\;$ $MV =$ ₹ $8088$

$\implies$ $40.44x = 8088$ $\implies$ $x = 200$

i.e. $\;$ money deposited per month $= $ ₹ $200$

Quadratic Equations

A shopkeeper purchases certain number of books for ₹ $960$. If the cost per book was ₹ $8$ less, the number of books that could be purchased for ₹ $960$ would be $4$ more. Write an equation, taking the original cost of each book to be ₹ $x$. Solve the equation to find the original cost of the books.

Let the original cost of each book $= $ ₹ $x$

Amount for which books purchased $= $ ₹ $960$

$\therefore \;$ Original number of books purchased $= \dfrac{960}{x}$

New cost of each book $= $ ₹ $\left(x - 8\right)$

New number of books that can be bought for ₹ $960$ $= \dfrac{960}{x - 8}$

As per question,

new number of books bought $= $ original number of books bought $+ \; 4$

i.e. $\;$ $\dfrac{960}{x - 8} = \dfrac{960}{x} + 4$

i.e. $\;$ $240 \left[\dfrac{1}{x - 8} - \dfrac{1}{x}\right] = 1$

i.e. $\;$ $240 \left[\dfrac{x - x + 8}{x^2 - 8x}\right] = 1$

i.e. $\;$ $x^2 - 8x - 1920 = 0$

i.e. $\;$ $x^2 - 48x + 40x - 1920 = 0$

i.e. $\;$ $x \left(x - 48\right) + 40 \left(x - 48\right) = 0$

i.e. $\;$ $\left(x - 48\right) \left(x + 40\right) = 0$

i.e. $\;$ $x - 48 = 0$ $\;$ or $\;$ $x + 40 = 0$

i.e. $\;$ $x = 48$ $\;$ or $\;$ $x = -40$

$\because \;$ cost of books cannot be negative,

$\therefore \;$ original cost of each book $= $ ₹ $x =$ ₹ $48$

Coordinate Geometry: Section Formula

Calculate the ratio in which the line joining $A \left(-4, 2\right)$ and $B \left(3, 6\right)$ is divided by the point $P \left(x, 3\right)$.

Also find the coordinates of point $P$ and the length of $AP$.

Given: $\;$ $A \left(-4, 2\right)$, $\;$ $B \left(3, 6\right)$

Let the point $P \left(x,3\right)$ divide the line joining the points $A$ and $B$ in the ratio $k : 1$

Then, by section formula [for $Y$ coordinate of point $P$]

$\dfrac{6k + 2 \times 1}{k + 1} = 3$

i.e. $\;$ $6k + 2 = 3k +3$

i.e. $\;$ $3k = 1$ $\implies$ $k = \dfrac{1}{3}$

i.e. $\;$ the point $P$ divides $AB$ in the ratio $1 : 3$

Again, by section formula [for $X$ coordinate of point $P$]

$\dfrac{-4 \times 3 + 3 \times 1}{1 + 3} = x$

i.e. $\;$ $x = \dfrac{-12 + 3}{4} = \dfrac{-9}{4}$

$\therefore \;$ Point $\;$ $P = \left(x, 3\right) = \left(\dfrac{-9}{4}, 3\right)$

By distance formula,

$\begin{aligned} \text{length of AP} & = \sqrt{\left[-4 - \left(\dfrac{-9}{4}\right)\right]^2 + \left[2 - 3\right]^2} \\\\ & = \sqrt{\left(\dfrac{-7}{4}\right)^2 + \left(-1\right)^2} \\\\ & = \sqrt{\dfrac{49}{16} + 1} \\\\ & = \dfrac{\sqrt{65}}{4} \end{aligned}$

Geometric Progression

Find the geometric progression (GP) for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.

Let the first term of GP $= t_1 = a$; $\;$ common ratio $= r$

$n^{th}$ term of GP $= t_n = ar^{n-1}$

$2^{nd}$ term of GP $= t_2 = ar^{2 - 1} = ar$

$3^{rd}$ term of GP $= t_3 = ar^{3 - 1} = ar^2$

$5^{th}$ term of GP $= t_5 = ar^{5 - 1} = ar^4$

Now, $\;$ $t_1 + t_2 = -4$ $\;\;\;$ [given]

i.e. $\;$ $a + ar = -4$ $\implies$ $a = \dfrac{-4}{1 + r}$ $\;\;\; \cdots \; (1)$

$t_5 = 4 \times t_3$ $\;\;\;$ [given]

i.e. $\;$ $a r^4 = 4 \times ar^2$

i.e. $\;$ $r^2 = 4$ $\implies$ $r = \pm 2$

Substituting $r = +2$ in equation $(1)$ gives

$a = \dfrac{-4}{1 + 2} = \dfrac{-4}{3}$

Then the GP is $\;$ $\dfrac{-4}{3}, \; \dfrac{-8}{3}, \; \dfrac{-16}{3}, \; \dfrac{-32}{3}, \; \dfrac{-64}{3}, \cdots$

Substituting $r = -2$ in equation $(1)$ gives

$a = \dfrac{-4}{1 - 2} = 4$

Then the GP is $\;$ $4, \; -8, \; 16, \; -32, \; 64, \cdots$