Question 1: Find the slopes of the lines which make the following angles with the positive direction of x-axis:

i) $\frac{-\pi}{4}$          ii) $\frac{2\pi}{3}$          iii) $\frac{3\pi}{4}$          iv) $\frac{\pi}{3}$

i)       Angle with the positive direction of x-axis $=$ $\frac{-\pi}{4}$

$\therefore \text{Slope} = m = \tan \theta = \tan \Big($ $\frac{-\pi}{4}$ $\Big) = - 1$

ii)     Angle with the positive direction of x-axis $=$ $\frac{2\pi}{3}$

$\therefore \text{Slope} = m = \tan \theta = \tan \Big($ $\frac{2\pi}{3}$ $\Big) = - \sqrt{3}$

iii)     Angle with the positive direction of x-axis $=$ $\frac{3\pi}{4}$

$\therefore \text{Slope} = m = \tan \theta = \tan \Big($ $\frac{3\pi}{4}$ $\Big) = - 1$

iv)     Angle with the positive direction of x-axis $=$ $\frac{\pi}{3}$

$\therefore \text{Slope} = m = \tan \theta = \tan \Big($ $\frac{\pi}{3}$ $\Big) = \sqrt{3}$

$\\$

Question 2: Find the slope of a line passing through the following points:

i) $(- 3, 2)$ and $( 1, 4)$      ii) $(a{t_1}^2, 2at_1)$ and $( a{t_2}^2, 2 at_2)$      iii) $(3, - 5)$ and $(1, 2)$

i)       The line passing through the following points: $(- 3, 2)$ and $( 1, 4)$

Slope $=$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{4 - 2}{1 - ( -3)}$ $=$ $\frac{2}{4}$ $=$ $\frac{1}{2}$

ii)      The line passing through the following points: $(a{t_1}^2, 2at_1)$ and $( a{t_2}^2, 2 at_2)$

Slope $=$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{2at_2 - 2at_1}{a{t_2}^2 - a{t_1}^2}$ $=$ $\frac{2(t_2-t_2)}{(t_2-t_1)(t_2+t_1)}$ $=$ $\frac{2}{t_2+t_1}$

iii)     The line passing through the following points: $(3, - 5)$ and $(1, 2)$

Slope $=$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{2 - (-5)}{1 - 3}$ $=$ $\frac{7}{-2}$ $=$ $\frac{-7}{2}$

$\\$

Question 3: State whether the two lines in each of the following are parallel, perpendicular or neither:

(i) Through $(5,6)$ and $(2,3)$; through $(9, -2)$ and $(6, -5)$

(ii) Through $(9,5)$ and $(-1,1)$; through $(3, -5)$ and $(8, -3)$

(iii) Through $(5,3)$ and $(1, 1)$; through $(-2,5)$ and $(2, -5)$

(iv) Through $(3, 15)$ and $(16,6$); through $(-5, 3)$ and $(8,2)$

i)       Let $m_1$ be the slope of line joining $(5,6)$ and $(2,3)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{3-6}{2-5}$ $=$ $\frac{-3}{-3}$ $= 1$

Let $m_2$ be the slope of line joining $(9, -2)$ and $(6, -5)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{-5-(-2)}{6-9}$ $=$ $\frac{-3}{-3}$ $= 1$

Since $m_1 = m_2$, the two lines are parallel to each other.

ii)      Let $m_1$ be the slope of line joining $(-1,1)$ and $(9, 5)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{5-1}{9-(-1)}$ $=$ $\frac{4}{10}$ $=$ $\frac{2}{5}$

Let $m_2$ be the slope of line joining $(3, -5)$ and $(8, -3)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{-3-(-5)}{8-3}$ $=$ $\frac{2}{5}$

Since $m_1 = m_2$, the two lines are parallel to each other.

iii)     Let $m_1$ be the slope of line joining $(6,3)$ and $(1,1)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{1-3}{1-6}$ $=$ $\frac{-2}{-5}$ $=$ $\frac{2}{5}$

Let $m_2$ be the slope of line joining $(-2, 5)$ and $(2, -5)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{-5-(-5)}{2-(-2)}$ $=$ $\frac{-10}{4}$ $=$ $\frac{-5}{2}$

Since $m_1 \times m_2 = -1$, the two lines are perpendicular to each other.

iv)     Let $m_1$ be the slope of line joining $(3, 15)$ and $(16,6)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{6-15}{16-3}$ $=$ $\frac{-9}{13}$

Let $m_2$ be the slope of line joining $(-5, 3)$ and $(8, 2)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{2-3}{8-(-5)}$ $=$ $\frac{-1}{13}$

Since $m_1 \neq m_2 = 1$ and neither $m_1 \times m_2 = -1$  the two lines are neither parallel or perpendicular to each other.

$\\$

Question 4: Find the slope of a line (i) which bisects the first quadrant angle (ii) which makes an angle of $30^{\circ}$ with the positive direction of y-axis measured anticlockwise.

i)       We know that the angle between the coordinate axes is $\frac{\pi}{2}$.

The line bisects the first quadrant.

Therefore the inclination of the line with positive x-axis is $\frac{\pi}{4}$

Hence the slope of line $(m) = \tan$ $\frac{\pi}{4}$ $= 1$

ii)      Given the line makes $30^{\circ}$ with positive y-axis.

Therefore the angle with positive x-axis is $30^{\circ}+90^{\circ} = 120^{\circ}$

Hence the slope of line $(m) = \tan 120^{\circ} = -\tan 60^{\circ} = -\sqrt{3}$

$\\$

Question 5: Using the method of slope, show that the following points are collinear:

(i) $A (4, 8), B (5, 12) , C (9 , 28)$           (ii) $A (16, - 18), B (3, - 6), C (- 10,6)$

(i)      Given $A (4, 8), B (5, 12) , C (9 , 28)$

Slope of $AB =$ $\frac{12-8}{5-4}$ $=$ $\frac{4}{1}$ $= 4$

Slope of $BC =$ $\frac{28-12}{9-5}$ $=$ $\frac{16}{4}$ $= 4$

Slope of $CA =$ $\frac{8-28}{4-9}$ $=$ $\frac{-20}{-5}$ $= 4$

Since all the three lines have the same slope, they are parallel to each other. And since they have a common point, they are collinear.

(ii)    Given $A (16, - 18), B (3, - 6), C (- 10,6)$

Slope of $AB =$ $\frac{-6-(-18)}{3-16}$ $=$ $\frac{-12}{13}$

Slope of $BC =$ $\frac{6-(-6)}{10-3}$ $=$ $\frac{-12}{13}$

Slope of $CA =$ $\frac{-18-6}{16-(-10)}$ $=$ $\frac{-24}{26}$ $=$ $\frac{-12}{13}$

Since all the three lines have the same slope, they are parallel to each other. And since they have a common point, they are collinear.

$\\$

Question 6: What is the value of $y$ so that the line through $(3,y)$ and $(2,7)$ is parallel to the line through $(-1,4)$ and $(0, 6)$ ?

Let $m_1$ be the slope of line joining $(-1, 4)$ and $(0,6)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{6-4}{0-(-1)}$ $=$ $\frac{2}{1}$ $= 2$

Let $m_2$ be the slope of line joining $(3, y)$ and $(2, 7)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{7-y}{2-3}$ $=$ $\frac{7-y}{-1}$ $= -7+y$

Since the two lines are parallel to each other $m_1 = m_2$.

$\therefore 2 = -7 + y \Rightarrow y = 9$

$\\$

Question 7: What can be said regarding a line if its slope is (i) zero (ii) positive (iii) negative?

i)       If the slope $= m = \tan \theta = 0 \Rightarrow \theta = 0$

When the slope of a line is $0$, then the line is parallel to x-axis.

ii)      If the slope is positive, then $\tan \theta$ is positive $\Rightarrow \theta$ is acute.

Thus the line makes an acute angle $( 0 < \theta < 90^{\circ})$ with positive x-axis.

iii)     If the slope is negative, then $\tan \theta$ is negative $\Rightarrow \theta$ is obtuse.

Thus the line makes an obtuse angle $( \theta > 90^{\circ})$ with positive x-axis.

$\\$

Question 8: Show that the line joining $(2, - 3)$ and $(-5, 1)$ is parallel to the line joining $(7,-1)$ and $(0,3)$.

Let $m_1$ be the slope of line joining $(2, -3)$ and $(-5, 1)$

Slope $\therefore m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{1-(-3)}{-5-2}$ $=$ $\frac{4}{-7}$ $=$ $\frac{-4}{7}$

Let $m_2$ be the slope of line joining $(7, -1)$ and $(0,3)$

Slope $\therefore m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{3-(-1)}{0-7}$ $=$ $\frac{-4}{7}$

Since $m_1 = m_2$ the two lines are parallel to each other .

$\\$

Question 9: Show that the line joining $(2, - 5)$ and $(- 2,5)$ is perpendicular to the line joining $(5, 3)$ and $(1, 1)$.

Let $m_1$ be the slope of line joining $(2,-5)$ and $(-2,5)$

$\therefore \text{Slope} = m_1 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{5-(-5)}{-2-2}$ $=$ $\frac{10}{-4}$ $=$ $\frac{-5}{2}$

Let $m_2$ be the slope of line joining $(6,3)$ and $(1,1)$

$\therefore \text{Slope} = m_2 =$ $\frac{y_2 - y_1}{x_2-x_2}$ $=$ $\frac{1-3}{1-6}$ $=$ $\frac{-2}{-5}$ $=$ $\frac{2}{5}$

Since $m_1 \times m_2 = -1$ the two lines are perpendicular to each other.

$\\$

Question 10: Without using Pythagoras theorem, show that the points $A (0, 4), B (1, 2)$ and $C ( 3, 3)$ are the vertices of a right angled triangle.

Given $A (0, 4), B (1, 2)$ and $C ( 3, 3)$ are the vertices of a right angled triangle.

Slope of $AB =$ $\frac{2-4}{1-0}$ $= -2$

Slope of $BC =$ $\frac{3-2}{3-1}$ $=$ $\frac{1}{2}$

Slope of $CA =$ $\frac{4-3}{0-3}$ $=$ $\frac{-1}{3}$

Slope of $AB \times$ Slope $BC = -2 \times$ $\frac{1}{2}$ $= -1$

Therefore $AB \perp BC$

Hence $\triangle ABC$ is a right angled triangle.

$\\$

Question 11: Prove that the points $(4, -1), (-2, 4), (4, 0)$ and $(2, 3)$ are the vertices of a rectangle.

Slope of $AB (m_1) =$ $\frac{-4-(-1)}{-2-(-4)}$ $=$ $\frac{-3}{2}$

Slope of $BC (m_2) =$ $\frac{0-(-4)}{4-(-2)}$ $=$ $\frac{4}{6}$ $=$ $\frac{2}{3}$

Slope of $CD (m_3) =$ $\frac{3-0}{2-4}$ $=$ $\frac{-3}{2}$

Slope of $DA (m_4) =$ $\frac{-1-3}{-4-2}$ $=$ $\frac{-4}{-6}$ $=$ $\frac{2}{3}$

Now $m_1 = m_3 \Rightarrow AB \parallel CD$

Similarly, $m_3 = m_4 \Rightarrow BC \parallel DA$

Also $m_1 \times m_2 =$ $\frac{-3}{2}$ $\times$ $\frac{2}{3}$ $= -1 \Rightarrow AB \perp BC$

Similarly, $m_3 \times m_4 =$ $\frac{-3}{2}$ $\times$ $\frac{2}{3}$ $= -1 \Rightarrow CD \perp DA$

Therefore $ABCD$ is a rectangle.

$\\$

Question 12: If the point s $A (h, 0), P (a, b)$ and $B (0, k)$ lie on a line, show that: $\frac{a}{h}$ $+$ $\frac{b}{k}$ $= 1$

Given $A (h, 0), P (a, b)$ and $B (0, k)$ lie on a line i.e. they are collinear.

$\therefore \text{slope of } AP = \text{slope of } PB = \text{slope of } BA$

$\Rightarrow$ $\frac{b-0}{a-h}$ $=$ $\frac{k-b}{0-a}$ $=$ $\frac{0-k}{h-0}$

$\Rightarrow$ $\frac{k-b}{-a}$ $=$ $\frac{-k}{h}$

$\Rightarrow kh - bh = ka$

$\Rightarrow ka + bh = kh$

$\Rightarrow$ $\frac{a}{h}$ $+$ $\frac{b}{k}$ $= 1$

$\\$

Question 13: The slope of a line is double of the slope of another line. If tangents of the angle between them is $1/3$, find the slopes of the other line.

Let $m_1$ and $m_2$ be the slopes of the given lines

Given $m_2 = 2 m_1$

Let $\theta$ be the angle between the lines between the two lines

$\therefore \tan \theta = \Big|$ $\frac{m_2-m_1}{1+ m_1m_2}$ $\Big|$

$\Rightarrow$ $\frac{1}{3}$ $= \Big|$ $\frac{m_2-m_1}{1+ m_1m_2}$ $\Big|$

$\Rightarrow$ $\frac{1}{3}$ $= \Big|$ $\frac{2m_1-m_1}{1+ 2{m_1}^2 }$ $\Big|$

$\Rightarrow$ $\frac{1}{3}$ $= \Big|$ $\frac{m_1}{1+ 2{m_1}^2 }$ $\Big|$

$\Rightarrow$ $\frac{m_1}{1+ 2{m_1}^2 }$ $= \pm$ $\frac{1}{3}$

Case 1: Positive sign

$\frac{m_1}{1+ 2{m_1}^2 }$ $=$ $\frac{1}{3}$

$\Rightarrow 3m_1 = 1 + 2 {m_1}^2$

$\Rightarrow 2{m_1}^2 - 3m_1 + 1 = 0$

$\Rightarrow (2m_1 -1)(m_1-1) = 0$

$\Rightarrow m_1 =$ $\frac{1}{2}$ $, 1$

$\therefore m_2 = 1, 2$

Case 2: Negative sign

$\frac{m_1}{1+ 2{m_1}^2 }$ $= -$ $\frac{1}{3}$

$\Rightarrow 3m_1 = -1 - 2 {m_1}^2$

$\Rightarrow 2{m_1}^2 + 3m_1 + 1 = 0$

$\Rightarrow (2m_1 +1)(m_1+1) = 0$

$\Rightarrow m_1 = -$ $\frac{1}{2}$ $, -1$

$\therefore m_2 = -1, -2$

$\\$

Question 14: Consider the following population and year graph: Find the slope of the line $AB$ and using it, find what will be the population in the year 2010.

From the graph: $A ( 1985, 92), B(1995, 97) , C( 2010, P)$

Slope of line $AB =$ $\frac{97-92}{1995-1985}$ $=$ $\frac{5}{10}$ $=$ $\frac{1}{2}$

Now slope of $BC =$ Slope of $AB$

$\therefore$ $\frac{P-97}{2010-1995}$ $=$ $\frac{1}{2}$

$\Rightarrow$ $\frac{P-97}{15}$ $=$ $\frac{1}{2}$

$\Rightarrow 2P-194 = 15$

$\Rightarrow P =$ $\frac{209}{2}$

$\Rightarrow P = 104.5$ Cr.

$\\$

Question 15: Without using the distance formula, show that points $(-2,-1), (4,0), (3,3)$ and $(- 3,2)$ are the vertices of a parallelogram.

Given points $A(-2,-1), B(4,0), C(3,3)$ and $D(- 3,2)$ are the vertices of a quadrilateral.

Slope of $AB (m_1) =$ $\frac{0-(-1)}{4-(-2)}$ $=$ $\frac{1}{6}$

Slope of $BC (m_2) =$ $\frac{3-0}{4-3}$ $=$ $\frac{3}{-1}$ $= -3$

Slope of $CD (m_3) =$ $\frac{2-3}{-3-3}$ $=$ $\frac{1}{6}$

Slope of $DA (m_4) =$ $\frac{-1-2}{-2-(-3)}$ $=$ $\frac{-3}{1}$ $= -3$

Now $m_1 = m_3 \Rightarrow AB \parallel CD$

Similarly, $m_2 = m_4 \Rightarrow BC \parallel DA$

Therefore $ABCD$ is a parallelogram.

$\\$

Question 16: Find the angle between the $\text{x-axis}$ and the line joining the points $( 3, -1)$ and $( 4, -2)$.

Slope of line joining $( 3, -1)$ and $( 4, -2)$

$\therefore m_1 =$ $\frac{-2-(-1)}{4-3}$ $=$ $\frac{-1}{1}$ $= -1$

Slope of x-axis $= 0$

$\therefore m_2 = 0$

If $\theta$ is the angle between the line and the $\text{x-axis}$, then

$\tan \theta = \Big|$ $\frac{m_1 - m_2}{1+m_1m_2}$ $\Big | = \Big|$ $\frac{-1-0}{1}$ $\Big | = -1$

$\therefore \theta = \tan^{-1} (-1) = 135^{\circ}$

$\\$

Question 17: Line through the points $(- 2, 6)$ and $(4,8)$ is perpendicular to the line through the points $(8, 12)$ and $(x, 24)$. Find the value of $x$.

Given points $A(- 2, 6), B(4,8) , P(8, 12) , Q(x, 24)$

Slope of $AB ( m_1) =$ $\frac{8-6}{4-(-2)}$ $=$ $\frac{2}{6}$ $=$ $\frac{1}{3}$

Slope of $PQ ( m_2) =$ $\frac{24-12}{x-8}$ $=$ $\frac{12}{x-8}$

Given $m_1 \times m_2 = - 1$

$\therefore$ $\frac{1}{3}$ $\times$ $\frac{12}{x-8}$ $= -1$

$\Rightarrow 12 = - 3x + 24$

$\Rightarrow 3x = 12$

$\Rightarrow x = 4$

$\\$

Question 18: Find the value of $x$ for which the points $(x , -1), (2, 1)$  and $( 4, 5)$  are collinear.

Given points $A(x, -1), B(2, 1), C(4, 5)$

Slope of $AB ( m_1) =$ $\frac{1-(-1)}{2-x}$ $=$ $\frac{2}{2-x}$

Slope of $BC ( m_2) =$ $\frac{5-1}{4-2}$ $=$ $\frac{4}{2}$ $= 2$

Given $m_1 = m_2$

$\therefore$ $\frac{2}{2-x}$ $= 2$

$\Rightarrow 2=4-2x$

$\Rightarrow 2x=2$

$\Rightarrow x = 1$

$\\$

Question 19: Find the angle between x-axis and the line joining the points $( -2,-2)$ and $(2,2)$

Slope of line joining $( -2,-2)$ and $(2,2)$

$\therefore m_1 =$ $\frac{2-(-2)}{2-(-2)}$ $=$ $\frac{4}{4}$ $= 1$

Slope of x-axis $= 0$

$\therefore m_2 = 0$

If $\theta$ is the angle between the line and the $\text{x-axis}$, then

$\tan \theta = \Big|$ $\frac{m_1 - m_2}{1+m_1m_2}$ $\Big | = \Big|$ $\frac{1-0}{1}$ $\Big | = 1$

$\therefore \theta = \tan^{-1} (1) = 45^{\circ}$

$\\$

Question 20: By using the concept of slope, show that the points $(- 2, -1), (4,0), (3,3)$ and $( - 3, 2)$  are the vertices of a parallelogram.

Given points $A(-2,-1), B(4,0), C(3,3)$ and $D(- 3,2)$ are the vertices of a quadrilateral.

Slope of $AB (m_1) =$ $\frac{0-(-1)}{4-(-2)}$ $=$ $\frac{1}{6}$

Slope of $BC (m_2) =$ $\frac{3-0}{4-3}$ $=$ $\frac{3}{-1}$ $= -3$

Slope of $CD (m_3) =$ $\frac{2-3}{-3-3}$ $=$ $\frac{1}{6}$

Slope of $DA (m_4) =$ $\frac{-1-2}{-2-(-3)}$ $=$ $\frac{-3}{1}$ $= -3$

Now $m_1 = m_3 \Rightarrow AB \parallel CD$

Similarly, $m_2 = m_4 \Rightarrow BC \parallel DA$

Therefore $ABCD$ is a parallelogram.

$\\$

Question 21: A quadrilateral has vertices $(4,1),(1,7),(-6,0)$ and $(-1,-9)$. Show that the mid-points of the sides of this quadrilateral form a parallelogram.

Given points: $A(4,1), B(1,7), C(-6,0)$ and $D(-1,-9)$

Let $E$ be the mid point of $AB \Rightarrow E = \Big($ $\frac{4+1}{2}$ $,$ $\frac{1+7}{2}$ $\Big) = \Big($ $\frac{5}{2}$ $, 4 \Big)$

Let $F$ be the mid point of $BC \Rightarrow E = \Big($ $\frac{1-6}{2}$ $,$ $\frac{7+0}{2}$ $\Big) = \Big($ $\frac{-5}{2}$ $,$ $\frac{7}{2}$ $\Big)$

Let $G$ be the mid point of $CD \Rightarrow E = \Big($ $\frac{-6-1}{2}$ $,$ $\frac{0-9}{2}$ $\Big) = \Big($ $\frac{-7}{2}$ $,$ $\frac{-9}{2}$ $\Big)$

Let $H$ be the mid point of $DA \Rightarrow E = \Big($ $\frac{4-1}{2}$ $,$ $\frac{1-9}{2}$ $\Big) = \Big($ $\frac{3}{2}$ $, -4 \Big)$

Slope of $EF (m_1) =$ $\frac{\frac{7}{2}-4}{\frac{-5}{2}-\frac{5}{2}}$ $=$ $\frac{1}{10}$

Slope of $FG (m_2) =$ $\frac{\frac{-9}{2}-\frac{7}{2}}{\frac{-7}{2}+\frac{5}{2}}$ $=$ $\frac{-8}{-1}$ $= 8$

Slope of $GH (m_3) =$ $\frac{-4+\frac{9}{2}}{\frac{3}{2}+\frac{7}{2}}$ $=$ $\frac{\frac{1}{2}}{5}$ $=$ $\frac{1}{10}$

Slope of $HF (m_4) =$ $\frac{4-(-4)}{\frac{5}{2}-\frac{3}{2}}$ $=$ $\frac{8}{1}$ $= 8$

Since $m_1 = m_3 \Rightarrow EF \parallel GH$

Also since $m_2 = m_4 \Rightarrow FG \parallel HF$

$\therefore EFGH$ is a parallelogram.