Geometry, as a logical system, is a means and even the most powerful means to make children feel the strength of the human spirit that is of their own spirit. - H. FREUDENTHAL

9.1 Introduction

We are familiar with two-dimensional coordinate geometry from earlier classes. Mainly, it is a combination of algebra and geometry. A systematic study of geometry by the use of algebra was first carried out by celebrated French philosopher and mathematician René Descartes, in his book ‘La Géométry, published in 1637. This book introduced the notion of the equation of a curve and related analytical methods into the study of geometry. The resulting combination of analysis and geometry is referred now as analytical geometry. In the earlier classes, we initiated the study of coordinate geometry, where we studied about coordinate axes, coordinate plane, plotting of points in a

plane, distance between two points, section formulae, etc. All these concepts are the basics of coordinate geometry.

Let us have a brief recall of coordinate geometry done in earlier classes. To recapitulate, the location of the points $(6,-4)$ and $(3,0)$ in the XY-plane is shown in Fig 9.1.

We may note that the point $(6,-4)$ is at 6 units distance from the $y$-axis measured along the positive $x$-axis and at 4 units distance from the $x$-axis measured along the negative $y$-axis. Similarly, the point $(3,0)$ is at 3 units distance from the $y$-axis measured along the positive $x$-axis and has zero distance from the $x$-axis. formulae:

We also studied there following important

I. Distance between the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is

$ PQ=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} $

For example, distance between the points $(6,-4)$ and $(3,0)$ is

$ \sqrt{(3-6)^{2}+(0+4)^{2}}=\sqrt{9+16}=5 \text{ units. } $

II. The coordinates of a point dividing the line segment joining the points $(x_1, y_1)$ and $(x_2, y_2)$ internally, in the ratio $m: n$ are $(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n})$.

For example, the coordinates of the point which divides the line segment joining

A $(1,-3)$ and $B(-3,9)$ internally, in the ratio $1: 3$ are given by $x=\frac{1 .(-3)+3.1}{1+3}=0$

$ \text{ and } y=\frac{1.9+3 \cdot(-3)}{1+3}=0 $

III. In particular, if $m=n$, the coordinates of the mid-point of the line segment joining the points $(x_1, y_1)$ and $(x_2, y_2)$ are $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.

IV. Area of the triangle whose vertices are $(x _{1,} y_1),(x_2, y_2)$ and $(x_3, y_3)$ is

$ \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| . $

For example, the area of the triangle, whose vertices are $(4,4),(3,-2)$ and $(-3,16)$ is

$ \frac{1}{2}|4(-2-16)+3(16-4)+(-3)(4+2)|=\frac{|-54|}{2}=27 $

Remark If the area of the triangle $ABC$ is zero, then three points $A, B$ and $C$ lie on a line, i.e., they are collinear.

In the this Chapter, we shall continue the study of coordinate geometry to study properties of the simplest geometric figure - straight line. Despite its simplicity, the line is a vital concept of geometry and enters into our daily experiences in numerous interesting and useful ways. Main focus is on representing the line algebraically, for which slope is most essential.

9.2 Slope of a Line

A line in a coordinate plane forms two angles with the $x$-axis, which are supplementary.

The angle (say) $\theta$ made by the line $l$ with positive direction of $x$-axis and measured anti clockwise is called the inclination of the line. Obviously $0^{\circ} \leq \theta \leq 180^{\circ}$ (Fig 9.2).

We observe that lines parallel to $x$-axis, or coinciding with $x$-axis, have inclination of $0^{\circ}$. The inclination of a vertical line (parallel to or coinciding with $y$-axis) is $90^{\circ}$.

Definition 1 If $\theta$ is the inclination of a line $l$, then $\tan \theta$ is called the slope or gradient of the line $l$.

The slope of a line whose inclination is $90^{\circ}$ is not defined.

The slope of a line is denoted by $m$.

Thus, $m=\tan \theta, \theta \neq 90^{\circ}$

It may be observed that the slope of $x$-axis is zero and slope of $y$-axis is not defined.

9.2.1 Slope of a line when coordinates of any two points on the line are given

We know that a line is completely determined when we are given two points on it. Hence, we proceed to find the slope of a line in terms of the coordinates of two points on the line.

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two points on non-vertical line $l$ whose inclination is $\theta$. Obviously, $x_1 \neq x_2$, otherwise the line will become perpendicular to $x$-axis and its slope will not be defined. The inclination of the line $l$ may be acute or obtuse. Let us take these two cases.

Draw perpendicular $QR$ to $x$-axis and $PM$ perpendicular to $RQ$ as shown in Figs. 9.3 (i) and (ii).

Case 1 When angle $\theta$ is acute:

In Fig 9.3 (i), $\angle MPQ=\theta \quad \quad \quad \quad \quad \quad \ldots (1)$

Therefore, slope of line $l=m=\tan \theta$.

But in $\triangle MPQ$, we have $\tan \theta=\frac{MQ}{MP}=\frac{y_2-y_1}{x_2-x_1} \quad \quad \quad \quad \quad \quad \ldots (2)$

From equations (1) and (2), we have

$ m=\frac{y_2-y_1}{x_2-x_1} $

Case II When angle $\theta$ is obtuse:

In Fig 9.3 (ii), we have

$\angle MPQ=180^{\circ}-\theta$.

Therefore, $\theta=180^{\circ}-\angle MPQ$.

Now, slope of the line $l$

$ \begin{aligned} m & =\tan \theta \\ & =\tan (180^{\circ}-\angle MPQ)=-\tan \angle MPQ \\ & =-\frac{MQ}{MP}=-\frac{y_2-y_1}{x_1-x_2}=\frac{y_2-y_1}{x_2-x_1} . \end{aligned} $

Consequently, we see that in both the cases the slope $m$ of the line through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$.

9.2.2 Conditions for parallelism and perpendicularity of lines in terms of their

slopes In a coordinate plane, suppose that non-vertical lines $l_1$ and $l_2$ have slopes $m_1$ and $m_2$, respectively. Let their inclinations be $\alpha$ and $\beta$, respectively.If the line $\boldsymbol{l_1}$ is parallel to $\boldsymbol{l_2}$ (Fig 9.4), then their inclinations are equal, i.e.,

$ \alpha=\beta, \text{ and hence, } \tan \alpha=\tan \beta $

Therefore $m_1=m_2$, i.e., their slopes are equal.

Conversely, if the slope of two lines $l_1$ and $l_2$ is same, i.e.,

Then

$ m_1=m_2 $

$ \tan \alpha=\tan \beta \text{. } $

By the property of tangent function (between $0^{\circ}$ and $180^{\circ}$ ), $\alpha=\beta$.

Therefore, the lines are parallel.

Hence, two non vertical lines $l_1$ and $l_2$ are parallel if and only if their slopes are equal.

If the lines $ \boldsymbol{l_1 } $ and $\boldsymbol{l_2 } $ are perpendicular (Fig 9.5), then $\beta=\alpha+90^{\circ}$.

Therefore, $\tan \beta=\tan (\alpha+90^{\circ})$

$ =-\cot \alpha=-\frac{1}{\tan \alpha} $

i.e., $\quad m_2=-\frac{1}{m_1}$ or $\quad m_1 m_2=-1$

Conversely, if $m_1 m_2=-1$, i.e., $\tan \alpha \tan \beta=-1$.

Then $\tan \alpha=-\cot \beta=\tan (\beta+90^{\circ})$ or $\tan (\beta-90^{\circ})$

Therefore, $\alpha$ and $\beta$ differ by $90^{\circ}$.

Thus, lines $l_1$ and $l_2$ are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other,

i.e., $\quad m_2=-\frac{1}{m_1}$ or, $m_1 m_2=-1$.

Let us consider the following example.

9.2.3 Angle between two lines

When we think about more than one line in a plane, then we find that these lines are either intersecting or parallel. Here we will discuss the angle between two lines in terms of their slopes.

Let $L_1$ and $L_2$ be two non-vertical lines with slopes $m_1$ and $m_2$, respectively. If $\alpha_1$ and $\alpha_2$ are the inclinations of lines $L_1$ and $L_2$, respectively. Then

$ m_1=\tan \alpha_1 \text{ and } m_2=\tan \alpha_2 . $

We know that when two lines intersect each other, they make two pairs of vertically opposite angles such that sum of any two adjacent angles is $180^{\circ}$. Let $\theta$ and $\phi$ be the adjacent angles between the lines $L_1$ and $L_2$ (Fig 9.6). Then

$ \theta=\alpha_2-\alpha_1 \text{ and } \alpha_1, \alpha_2 \neq 90^{\circ} \text{. } $

Therefore $\tan \theta=\tan (\alpha_2-\alpha_1)=\frac{\tan \alpha_2-\tan \alpha_1}{1+\tan \alpha_1 \tan \alpha_2}=\frac{m_2-m_1}{1+m_1 m_2} \quad(.$ as $.1+m_1 m_2 \neq 0)$ and $\phi=180^{\circ}-\theta$ so that

$\tan \phi=\tan (180^{\circ}-\theta)=-\tan \theta=-\frac{m_2-m_1}{1+m_1 m_2}$, as $1+m_1 m_2 \neq 0$

Now, there arise two cases:

Case I If $\frac{m_2-m_1}{1+m_1 m_2}$ is positive, then $\tan \theta$ will be positive and $\tan \phi$ will be negative, which means $\theta$ will be acute and $\phi$ will be obtuse.

Case II If $\frac{m_2-m_1}{1+m_1 m_2}$ is negative, then $\tan \theta$ will be negative and $\tan \phi$ will be positive, which means that $\theta$ will be obtuse and $\phi$ will be acute.

Thus, the acute angle (say $\theta$ ) between lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$, respectively, is given by

$ \tan \theta=|\frac{m_2-m_1}{1+m_1 m_2}|, \text{ as } 1+m_1 m_2 \neq 0 \quad \quad \quad \quad \quad \quad \quad \quad \ldots(1) $

The obtuse angle ( say $\phi$ ) can be found by using $\phi=180^{\circ}-\theta$.

9.3 Various Forms of the Equation of a Line

We know that every line in a plane contains infinitely many points on it. This relationship between line and points leads us to find the solution of the following problem:

How can we say that a given point lies on the given line? Its answer may be that for a given line we should have a definite condition on the points lying on the line. Suppose $P(x, y)$ is an arbitrary point in the XY-plane and $L$ is the given line. For the equation of $L$, we wish to construct a statement or condition for the point $P$ that is true, when $P$ is on $L$, otherwise false. Of course the statement is merely an algebraic equation involving the variables $x$ and $y$. Now, we will discuss the equation of a line under different conditions.

9.3.1 Horizontal and vertical lines

If a horizontal line $L$ is at a distance $a$ from the $x$ axis then ordinate of every point lying on the line is either $a$ or $-a$ [Fig 9.8 (a)]. Therefore, equation of the line $L$ is either $y=a$ or $y=-a$. Choice of sign will depend upon the position of the line according as the line is above or below the $y$-axis. Similarly, the equation of a vertical line at a distance $b$ from the $y$-axis is either $x=b$ or $x=-b$ [ Fig 9.8(b)].

9.3.2 Point-slope form

Suppose that $P_0(x_0, y_0)$ is a fixed point on a non-vertical line $L$, whose slope is $m$. Let $P(x, y)$ be an arbitrary point on L (Fig 9.10).

Then, by the definition, the slope of $L$ is given by

$ m=\frac{y-y_0}{x-x_0} \text{, i.e., } y-y_0=m(x-x_0) \quad \quad \quad \quad \quad \quad \ldots (1) $

Since the point $P_0(x_0, y_0)$ along with all points $(x, y)$ on $L$ satisfies (1) and no other point in the plane satisfies (1). Equation (1) is indeed the equation for the given line $L$.

Thus, the point $(x, y)$ lies on the line with slope $m$ through the fixed point $(x_0, y_0)$, if and only if, its coordinates satisfy the equation

$ y-y_0=m(x-x_0) $

9.3.3 Two-point form

Let the line $L$ passes through two given points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$. Let $P(x, y)$ be a general point on $L$ (Fig 9.11).

The three points $P_1, P_2$ and $P$ are collinear, therefore, we have

Fig 9.11 slope of $P_1 P=$ slope of $P_1 P_2$

$ \text{ i.e., } \quad \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}, \quad \text{ or } \quad y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) \text{. } $

Thus, equation of the line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by

$ y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) \quad \quad \quad \quad \quad \quad \quad \quad \ldots (2) $

9.3.4 Slope-intercept form

Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines.

Case I Suppose a line $L$ with slope $m$ cuts the $y$-axis at a distance $c$ from the origin (Fig 9.12). The distance $c$ is called the $y$ intercept of the line L. Obviously, coordinates of the point where the line meet the $y$-axis are $(0, c)$. Thus, $L$ has slope $m$ and passes through a fixed point $(0, c)$. Therefore, by point-slope form, the equation of $L$ is

$ y-c=m(x-0) \text{ or } y=m x+c $

Thus, the point $(x, y)$ on the line with slope $m$ and $y$-intercept $c$ lies on the line if and only if

$ y=m x+c \quad \quad \quad \quad \quad \quad \ldots(3) $

Note that the value of $c$ will be positive or negative according as the intercept is made on the positive or negative side of the $y$-axis, respectively.

Case II Suppose line $L$ with slope $m$ makes $x$-intercept $d$. Then equation of $L$ is

$ y=m(x-d) \quad \quad \quad \quad \quad \quad \ldots(4) $

Students may derive this equation themselves by the same method as in Case I.

9.3.5 Intercept - form

Suppose a line L makes $x$-intercept $a$ and $y$-intercept $b$ on the axes. Obviously $L$ meets $x$-axis at the point $(a, 0)$ and $y$-axis at the point $(0, b)$ (Fig .9.13). By two-point form of the equation of the line, we have

$ \begin{aligned} & y-0=\frac{b-0}{0-a}(x-a) \text{ or } a y=-b x+a b, \\ & \text{ i.e., } \quad \frac{x}{a}+\frac{y}{b}=1 . \end{aligned} $

Thus, equation of the line making intercepts $a$ and $b$ on $x$-and $y$-axis, respectively, is

$ \frac{x}{a}+\frac{y}{b}=1 \quad \quad \quad \quad \quad \quad \quad \ldots (5) $

9.4 Distance of a Point From a Line

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let $L: A x+By+C=0$ be a line, whose distance from the point $P(x_1, y_1)$ is $d$. Draw a perpendicular PM from the point $P$ to the line $L$ (Fig 9.14). If the

line meets the $x$-and $y$-axes at the points $Q$ and $R$, respectively. Then, coordinates of the points are $Q(-\frac{C}{A}, 0)$ and $R(0,-\frac{C}{B})$. Thus, the area of the triangle $P Q R$ is given by

$ \text{ area }(\Delta PQR)=\frac{1}{2} PM \cdot QR \text{, which gives } PM=\frac{2 area(\Delta PQR)}{QR} \quad \quad \quad \ldots (1) $

Also, area $(\Delta PQR)=\frac{1}{2}\left|x_1(0+\frac{C}{B})+(-\frac{C}{A})(-\frac{C}{B}-y_1)+0(y_1-0)\right|$

$ =\frac{1}{2}|x_1 \frac{C}{B}+y_1 \frac{C}{A}+\frac{C^{2}}{AB}| $

or 2 area $(\Delta PQR)=|\frac{C}{AB}| \cdot|A _{x_1}+B y_1+C|$, and

$ QR=\sqrt{(0+\frac{C}{A})^{2}+(\frac{C}{B}-0)^{2}}=|\frac{C}{AB}| \sqrt{A^{2}+B^{2}} $

Substituting the values of area $(\triangle PQR)$ and $QR$ in (1), we get

$ PM=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}} $

or

$ d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}} . $

Thus, the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$ d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}} . $

9.4.1 Distance between two parallel lines

We know that slopes of two parallel lines are equal.Therefore, two parallel lines can be taken in the form

$\quad \quad \quad\quad y=m x+c_1 \quad \quad \quad \quad \quad \ldots (1)$

and $\quad \quad \quad y=m x+c_2 \quad \quad \quad \quad \quad \ldots (2)$

Line (1) will intersect $x$-axis at the point $A(-\frac{c_1}{m}, 0)$ as shown in Fig 9.15.

Distance between two lines is equal to the length of the perpendicular from point A to line (2). Therefore, distance between the lines (1) and (2) is

$ \frac{|(-m)(-\frac{c_1}{m})+(-c_2)|}{\sqrt{1+m^{2}}} \text{ or } d=\frac{|c_1-c_2|}{\sqrt{1+m^{2}}} \text{. } $

Thus, the distance $d$ between two parallel lines $y=m x+c_1$ and $y=m x+c_2$ is given by

$ d=\frac{|c_1-c_2|}{\sqrt{1+m^{2}}} $

If lines are given in general form, i.e., $A x+B y+C_1=0$ and $A x+B y+C_2=0$,

then above formula will take the form $d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}$

Students can derive it themselves.

Summary

• Slope $(m)$ of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}, \quad x_1 \neq x_2$.

• If a line makes an angle a with the positive direction of $x$-axis, then the slope of the line is given by $m=\tan \alpha, \alpha \neq 90^{\circ}$.

• Slope of horizontal line is zero and slope of vertical line is undefined.

• An acute angle (say $\theta$ ) between lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$ is given by $\tan \theta=|\frac{m_2-m_1}{1+m_1 m_2}|, 1+m_1 m_2 \neq 0$.

• Two lines are parallel if and only if their slopes are equal.

• Two lines are perpendicular if and only if product of their slopes is -1 .

• Three points A, B and C are collinear, if and only if slope of $AB=$ slope of $BC$.

• Equation of the horizontal line having distance $a$ from the $x$-axis is either $y=a$ or $y=-a$.

• Equation of the vertical line having distance $b$ from the $y$-axis is either $x=b$ or $x=-b$.

• The point $(x, y)$ lies on the line with slope $m$ and through the fixed point $(x_o, y_o)$, if and only if its coordinates satisfy the equation $y-y_0=m(x-x_0)$.

• Equation of the line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$.

• The point $(x, y)$ on the line with slope $m$ and $y$-intercept $c$ lies on the line if and only if $y=m x+c$.

• If a line with slope $m$ makes $x$-intercept $d$. Then equation of the line is $y=m(x-d)$.

• Equation of a line making intercepts $a$ and $b$ on the $x$-and $y$-axis, respectively, is $\frac{x}{a}+\frac{y}{b}=1$.

• Any equation of the form $A x+By+C=0$, with $A$ and $B$ are not zero, simultaneously, is called the general linear equation or general equation of a line.

• The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by $d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

• Distance between the parallel lines $A x+B y+C_1=0$ and $A x+B y+C_2=0$, is given by $d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}$.