Introduction to Straight Lines

STRAIGHT LINE

GENERAL EQUATION OF A STRAIGHT LINE

Any equation of first degree of the form ax + by + c = 0, where a, b, c, are constants always represents a straight line (at least one out of a and b is non-zero).

Slope or gradient of a line:

If $\theta$ is the angle at which a straight line is inclined to the positive direction of x-axis, then tan$\theta$ is called the slope or gradient of the line, where 0 £ $\theta$ < 180° ($\theta$ ¹ 90°).

Note: (i) If a line is parallel to x-axis, then its slope is equal to tan0° = 0.

(ii) Slope of a line perpendicular to x-axis is not defined. Whenever we say that the slope of a line is not defined, we mean that the line is perpendicular to x-axis.

INTERCEPT OF A STRAIGHT LINE ON THE AXES

Intercept of a line on x-axis:

If a line cut the axis at (a, 0), then 'a' is called the intercept of a line on x-axis or the x-intercept. If a line intersect the x-axis at (–2, 0), then the length of intercept is |–2| = 2.

Intercept of a line on y-axis:

If a line cuts y-axis at (0, b) then b is called the intercept of the line on y-axis or y intercept.

DIFFERENT FORMS OF THE STRAIGHT LINES

Slope intercept form:

Equation of a line whose slope is m (m = tan$\theta$) and which cut an intercept c on the y- axis (i.e. which passes through the point (0, c) is given by y = mx + c.

Intercept form:

The equation of the line which cuts off intercepts a and b on x-axis and y-axis respectively is given by $\frac{\text{x}}{a}+\frac{y}{b}=1$.

Thus intercept of a straight line on x-axis can be found by putting y = 0 in the equation of the line and then finding the value of x. Similarly intercept on y-axis can be found by putting x = 0 in the equation of the line and then finding the value of y.

Normal Form:

The equation of a straight line upon which the length of perpendicular from the origin is p and the perpendicular makes an angle $\alpha$ with the positive direction of x-axis is given by

x cos$\alpha$ + y sin$\alpha$ = p

Note: In normal form of equation of a straight line p is always taken as positive and $\alpha$is measured from positive direction of x-axis in anticlockwise direction between 0 and 2$\pi$.

Note: In the normal form x cos$\alpha$+y sin$\alpha$=p, p is always taken as positive.

Point-slope form: The equation of a line passing through the point (x₁, y₁) and having slope m is given by $\left(y-y_1\right)=m\left(x-x_1\right)$.

Two points form: The equation of a straight line passing through two given points (x₁, y₁) and (x₂, y₂) is given by $\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Note: Slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by $\frac{y_2-y_1}{x_2-x_1}.$

Parametric form:

The equation of line passes through a given point P(x₁, y₁) and makes a given angle $\theta$ with the positive direction of the x-axis is given by $\frac{x-x_1}{\text{cos }\theta }=\frac{y-y_1}{\sin \theta }=r$

where r is the distance between the variable point Q (x, y) and the fixed point P(x₁, y₁).

$\bullet$ Any point on the line will be of the form (x₁ + r cos$\theta$, y₁ + r sin$\theta$). For different values of r, we will get different points on the line

$\bullet$Here |r| will gives the distance of the point Q from the fixed point P(x₁, y₁).

$\bullet$ If P(x₁, y₁) is any point on the line which makes an angle$\theta$ with the positive direction of x axis, then there will be two points on the line at a distance r from P(x₁, y₁). One will be relatively upward if r is taken positive and other will be relatively downward if r is taken negative.

Illustration 1: Find the points on the line y = x – 2, which are lying at a unit distance from the point (5, 3).

Key concept 1: In general case if in any question there is something related to distance, we use the parametric form of line.

Solution: Clearly (5, 3) is lying on the given line. Now we will write the parametric equation of the given line passing through (5, 3), which is

$\frac{x-5}{\frac{1}{\sqrt{2}}}=\frac{y-3}{\frac{1}{\sqrt{2}}}=r$

Here |r| will represent the distance between (5, 3) and (x, y)

Any point on the line is $\left(5+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}}\right)$. Now if we take r = 1, we get the coordinate of the point which is at a unit distance from (5, 3) and relatively upward, and if we take r = –1, we will get the point which is also at a unit distance from (5, 3) but relatively downward. Hence the points are

${\text{P}}^{\prime }\left(5+\frac{1}{\sqrt{2}},3+\frac{1}{\sqrt{2}}\right)\text{and}{P}^{\prime \prime }\left(5-\frac{1}{\sqrt{2}},3-\frac{1}{\sqrt{2}}\right)$

Key concept 2: Take any point on the given line in terms of a single variable and use the distance formula.

Solution: Any point on the given line can be taken as ($\alpha$, $\alpha$ - 2). Given that the distance between the points ($\alpha$, $\alpha$ - 2) and (5, 3) is 1.

\begin{align}  \Rightarrow \sqrt{{{(\alpha \text{ - }5)}^{2}}\text{ + }{{(\alpha \text{ - }5)}^{2}}}=\text{ 1} \\  \Rightarrow \text{ }\alpha \text{ = 5}\pm \text{ }\dfrac{\text{1}}{\sqrt{2}} \\ \end{align}

Hence the required points are $\left(5+\frac{1}{\sqrt{2}},3+\frac{1}{\sqrt{2}}\right)\text{and}\left(5-\frac{1}{\sqrt{2}},3-\frac{1}{\sqrt{2}}\right)$

ANGLE BETWEEN TWO STRAIGHT LINES

If $\theta$ is the acute angle between two lines, then

tan$\theta$ = $\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$ where m₁ and m₂ are the slopes of the two lines and are finite.

Notes:

$\bullet$If the two lines are perpendicular to each other then m₁m₂ = -1.

$\bullet$Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0.

$\bullet$If the two lines are parallel or are coincident, then m₁ = m₂.

$\bullet$Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0.

$\bullet$If any of the two lines is perpendicular to x-axis, then the slope of that line is not define (infinite).

Let m₁ = $\infty$, then $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$ = $\left|\frac{1-\frac{m_2}{m_1}}{\frac{1}{m_1}+m_2}\right|=\left|\frac{1}{m_2}\right|$

or $\theta$ = |90° - $\alpha$|, where tan$\alpha$ = m₂ i.e. angle $\theta$ is the complimentary to the angle which the oblique line makes with the x-axis.

Illustration 2: Find the equation to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is x - 2y = 3 and the opposite vertex is the point (2, 2).

Solution: Clearly other two sides of the triangle will be making an angle of 45^o with the given line. That means we have to find the equation of line passing through the point (3, 2) and making an angle of 45^o with the given line. Slope of the given line x – 2y = 3 is m₁ = 1/2.

$\Rightarrow$ tan 45° = $\pm$$\frac{m-m_1}{1+m_{1}m},\text{ i}\text{.e}\text{., 1=}\pm \frac{\text{m-1/2}}{1+\frac{1}{2}m}$$\Rightarrow \text{ m = 3, -}\frac{\text{1}}{\text{3}}.$Hence the required equations of the two lines are 3x – y – 7 = 0 and x + 3y – 9 = 0THE DISTANCE BETWEEN TWO PARALLEL LINESThe distance between two parallel lines:

ax + by + c₁ = 0 and ax + by + c₂ = 0 is $\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}$.CONCURRENCY OF THREE LINESThe condition for 3 lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 to be concurrent is\left| \begin{align} {{a}_{1}}{{b}_{1}}{{c}_{1}} \\ {{a}_{2}}{{b}_{2}}{{c}_{2}} \\ {{a}_{3}}{{b}_{3}}{{c}_{3}} \\ \end{align} \right|=0BISECTORS OF ANGLE BETWEEN TWO GIVEN LINESLet a₁x+b₁y+c₁ = 0……..(1) a₂x + b₂y + c₂ = 0….(2)are two intersecting lines.Let any point p(x, y) be any point on either of the two bisectors of angles of (1) and (2).Then p is equidistance from (1) and (2)$\Rightarrow \left|\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}\right|=\left|\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}\right|$ $\Rightarrow \frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$which are the required equations of the two bisectors of angles between (1) and (2).If the two given lines are not perpendicular i.e. a₁ a₂ + b₁ b₂ $\ne$ 0, then one of these equation is the equation of the bisector of acute angle and the other that of the obtuse angle.The equation of acute and the obtuse angle bisectors:Method 1Step 1:Take one of the given lines and let its slope be m₁ and take one of the bisectors and let it’s slope be m₂.Step 2: If $\theta$ be the acute angle between them, then find $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$Step 3: If tan$\theta$ > 1then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle.If tan$\theta$ < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle. Method 2:If the constant term c₁ and c₂ in the two equations a₁x+b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0 are of the same sign, then Case 1: if $a_1{a}_2+b_1{b}_2>0,$then $\frac{a_{1}x+b_{1}y+c}{\sqrt{{a_1}^2+{b_1}^2}}=\frac{a_{2}x+b_{2}y+c}{\sqrt{{a_2}^2+{b_2}^2}}$will give the equation of obtuse angle bisector and$\frac{a_{1}x+b_{1}y+c}{\sqrt{{a_1}^2+{b_1}^2}}=-\frac{a_{2}x+b_{2}y+c}{\sqrt{{a_2}^2+{b_2}^2}}$will give the equation of acute angle bisector.Case 2: if ${a_1}{a_2} + {b_1}{b_2} < 0,$then $\dfrac{{{a_1}x + {b_1}y + c}}{{\sqrt {{a_1}^2 + {b_1}^2} }} = \dfrac{{{a_2}x + {b_2}y + c}}{{\sqrt {{a_2}^2 + {b_2}^2} }}$will give the equation of acute angle bisector and $\dfrac{{{a_1}x + {b_1}y + c}}{{\sqrt {{a_1}^2 + {b_1}^2} }} =- \dfrac{{{a_2}x + {b_2}y + c}}{{\sqrt {{a_2}^2 + {b_2}^2} }}$ will give the equation of obtuse angle bisector.Note: Whether both the lines are perpendicular or not but the angle bisectors of these lines will always be mutually perpendicular.The equation of the bisector of the angle which contain a given point:The equation of the bisector of the angle between the two lines containing the point ($\alpha ,\beta$) is $\frac{a_{1}x+b_{1}y+c}{\sqrt{{a_1}^2+{b_1}^2}}=\frac{a_{2}x+b_{2}y+c}{\sqrt{{a_2}^2+{b_2}^2}}$if $a_1\alpha +b_1\beta +c_{1}and{a}_2\alpha +b_2\beta +c_2$ are of the same signsor$\frac{a_{1}x+b_{1}y+c}{\sqrt{{a_1}^2+{b_1}^2}}=-\frac{a_{2}x+b_{2}y+c}{\sqrt{{a_2}^2+{b_2}^2}}$ if $a_1\alpha +b_1\beta +c_{1}and{a}_2\alpha +b_2\beta +c_2$ are of the opposite signsThe equation of the bisector of the angle containing the origin:Write the equations of the two lines so that the constants c₁ and c₂ are positive. Then the equation $\frac{a_{1}x+b_{1}y+c}{\sqrt{{a_1}^2+{b_1}^2}}=\frac{a_{2}x+b_{2}y+c}{\sqrt{{a_2}^2+{b_2}^2}}$ is the equation ofthe bisector containing the origin.Note:if ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}<0,$, then the origin will lie in the acute angle and if${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}>0,$ then origin will lie in the obtuse angle.Illustration 3:For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 find equation of the bisector of the(a)acute angle(b)angle which contain (1, 2)(c)angle which contain originSolution:(a)The equation of given lines are 4x + 3y – 6 = 0…(1)and 5x + 12y + 9 = 0 …(2)equation of angle bisectors between these two lineare$\frac{4x+3y-6}{\sqrt{4^2+3^2}}=\pm \frac{5x+12y+9}{\sqrt{5^2+{12}^2}}$….(3)9x – 7y – 41 = 0 and 7x + 9y – 3 = 0Here we will get two equation of bisectors. Two find the acute angle bisector take one equation of the given line and one equation of bisector. Given line be 4x + 3y – 6 = 0 and one bisector be 7x + 9y – 3 = 0. Now $\tan \text{ }\!\!\theta\!\!\text{ }\,\,\text{=}\,\,\left| \dfrac{-\dfrac{4}{3}+\dfrac{7}{9}}{1+\dfrac{4}{3}.\dfrac{7}{9}} \right|<1$Hence bisector 7x + 9y – 3 = 0 will be acute angle bisector.(b)Equation of bisector which contain the point (1, 2) since4.1 + 3.2 – 6 > 0 and 5.0 + 12.2 + 9 > 0Hence equation of bisector which bisect the angle which contains the point (1, 2) is$\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}$$\Rightarrow$ 9x – 7y – 41 = 0;(c);;;;;Since 4.0 + 3.0 – 6 < 0 and 5.0 + 12.0 +9 > 0.Hence the equation of the bisector which contain the origin is $-\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}$ $\Rightarrow$ 7x + 9y – 3 = 0;FAMILY OF LINES;Suppose L₁= a₁x+b₁y+c₁ = 0 and L₂=a₂x + b₂y + c₂ = 0 are two intersecting lines and let the point of their intersection be ($\alpha ,\beta$). Now if we write these two equations in this form $L_1+\lambda L_2=0$ (where $\lambda$ is a parameter)$\Rightarrow (a_{1}x+b_{1}y+c_1)+\lambda (a_{2}x+b_{2}y+c_2)=0$………(1);;;

then for different values of $\lambda$, (1) will give different straight lines.Now $\Rightarrow (a_1\alpha +b_1\beta +c_1)+\lambda (a_2\alpha +b_2\beta +c_2)=0$$\Rightarrow 0+\lambda .0=0$$\Rightarrow$($\alpha ,\beta$) always lies on (1) whatever be the value of $\lambda$.Hence (1) represent a family of straight lines passing through the point of intersection of a₁x+b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0.Note: $\bullet$Whenever we have to show that a line always passes through a fixed point, we use the concept of family of lines.$\bullet$Family of lines perpendicular to a given line ax+by+c = 0 is given by bx-ay+k=0, where k is a parameter. $\bullet$Family of lines parallel to a given line ax+by+c = 0 is given by ax+by+k=0, where k is a parameter.Illustration 4:Find the equation of the straight line which belongs to both of the following family of lines 5x + 3y – 2 + $\lambda$₁ (3x – y – 4) = 0 and x - y + 1 + $\lambda$₂ (2x – y – 2) = 0Solution.Lines of first family are concurrent at (1, -1) and that of second at (3, 4) $\therefore$ Required line passes through both of these points $\therefore$ Equation is 5x – 2y – 7 = 0PAIR OF STRAIGHT LINESThe general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines if\Delta \,=\,\left| \begin{align} ahg \\ hbf \\ gfc \\ \end{align} \right|\,=0and h²$⩾$ab.$\Rightarrow$ abc + 2fgh - af² - bg² - ch² = 0 and h² $⩾$ ab.The homogeneous second degree equation ax² + 2hxy + by² = 0 represents a pair of straight lines through the origin if h² ³ ab.If the lines through the origin whose joint equation is ax² + 2hxy + by² = 0, are y = m₁x and y = m₂x, theny² - (m₁ + m₂)xy + m₁m₂x² = 0 andy² + $\frac{2h}{b}$xy + $\frac{a}{b}{x}^2$ = 0 are identical, so that$m_1+m_2=-\frac{2h}{b},m_1{m}_2=\frac{a}{b}$.If $\theta$ be the angle between two lines, through the origin, then$\tan \theta =\pm \frac{\sqrt{{\left(m_1+m_2\right)}^2-4m_1{m}_2}}{1+m_1{m}_2}$ = $\pm$$\frac{2\sqrt{h^2-ab}}{a+b}$.

The lines are perpendicular if a + b = 0 and coincident if h² = ab.

Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Curve and a Line:

If the line lx + my + n = 0, ((n ¹ 0) i.e. the line does not pass through origin) cut the curve ax² + 2hxy + by² + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e.

ax² + 2hxy + by² + (2gx + 2fy) $\left(\frac{\ell x+my}{-n}\right)+c{\left(\frac{\ell x+my}{-n}\right)}^2=0$

is the equation of the lines OA and OB

Illustration 5: If the lines joining origin to the points of intersection of the curve

2x² + 3y² + m = 0 and the line y = 2x + 3 are perpendicular, then find the value of m.

Solution. Homogenise the given curve with line, we get the required line

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

Since, the lines are perpendicular

$\Rightarrow$ Coefficient of x² + coefficient of y² = 0

$\Rightarrow$ m = -9

ROTATION OF CO-ORDINATE AXES

Let OX, OY be the original axes and OX' and OY' be the new axes obtained after rotating OX and OY through an angle $\theta$ in the anticlockwise direction. Let P be any point in the plane having coordinates (x, y) with respect to axes OX and OY and (x', y') with respect to axes OX' and OY'. Then

x = x' cos$\theta$ – y' sin$\theta$, y = x' sin$\theta$ + y' cos$\theta$ ...(1)

and

x' = x cos$\theta$ + y sin$\theta$, y' = – x sin$\theta$ + y cos$\theta$ …(2)

Note: The above transformation can also be displaced by a table.

$\bullet$If f(x, y) = 0 is the equation of a curve then it's transformed equation is

f( x' cos$\theta$ – y' sin$\theta$, x' sin$\theta$ + y' cos$\theta$) = 0