##### Straight Line Important Formulae
Slope: $m = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Slope-Intercept form: $y = mx + b$ Slope Point Format: $y - y_{1} = m (x - x_{1})$ Angle between two lines: $\alpha = \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right)$
After successful completion of this topic, you should be able to:

• Find and interpret equation of a straight line in various forms.
• Perform slope calculations including parallel and perpendicular lines.
• Find angle between two coplanar, non-parallel lines.

###### Solved Example: 1-1-01 Page 35

The angle between lines 2x- 9y + 16=0 and x + 4y + 5 = 0 is given by:

Solution:
First let's find out slopes of each line. $2x- 9y + 16 =0 \mathrm{\ Or, \ } y = \dfrac{2}{9}x+\dfrac{16}{9}$ $\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}$ Similarly, slope of the second line, $m_2 = \dfrac{-1}{4}$ The angle between two lines is given by: \begin{align*} \alpha &= \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right )\\ &= \tan^{-1}\left(\dfrac{\dfrac{-1}{4}-\dfrac{2}{9}}{1+ \dfrac{2}{9}\times \dfrac{-1}{4}}\right )\\ &=\tan^{-1} \left( {\dfrac{ \dfrac{-17}{36}}{\dfrac{34}{36}}}\right )\\ &=\tan^{-1}\left(\dfrac{-1}{2}\right) \end{align*}

###### Solved Example: 1-1-02 Page 35

What is the equation of the straight line which passes through the point (1, 2) and cuts off equal intercepts from the axes ?

Solution:
Method I: Use double intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ But both intercepts are equal. $\dfrac{x}{a} + \dfrac{y}{a} = 1$ But the line passes through (1,2) $\dfrac{1}{a} + \dfrac{2}{a} = 1$ $1 + 2 = a$ $3 = a$ Substituting, $\dfrac{x}{3} + \dfrac{y}{3} = 1$ $x + y = 3$ Method II: Since it cuts equal intercepts, the slope will be m = -1. It passes through (1,2). \begin{align*} y- y_1 &= m(x- x_1)\\ y- 2 &= (-1)(x- 1)\\ y - 2 &= -x + 1\\ x + y -3 &= 0\\ x + y &= 3\\ \end{align*}

###### Solved Example: 1-1-03 Page 35

Given the slope-intercept form of a line as y = mx + b, which one of the following is true?

Solution:
For the straight line y = mx + b, m is the slope and b is the y-intercept.
To get x-intercept,
We will have to substitute y=0: \begin{align*} y &= mx + b\\ 0 &= mx + b\\ \dfrac{-b}{m} &= x \end{align*}

###### Solved Example: 1-1-04 Page 35

What is the equation of the straight line which is perpendicular to y = x and passes through (3, 2)?

Solution:
Slope of the given line y = x: m$_1$ = 1
Slope of the perpendicular line: m$_2$ = $\dfrac{-1}{m_1}$ = -1
$y - y_1 = m(x - x_1)$ But it passes through (3,2) \begin{align*} y - 2 &= (-1)(x - 3)\\ y - 2 &= -x + 3)\\ x + y &= 5\\ \end{align*}

###### Solved Example: 1-1-05

If the co-ordinates of the middle point of the portion of a line intercepted between coordinate axes (3,2), then the equation of the line will be: Solution: As shown in the figure, the coordinates of intersection point of the straight line with x- axis will be:

B (2$\times$3, 0)= B (6,0)

and y-axis will be:

A(0, 2$\times$2) = A(0,4).

So, x-intercept = a = 6, y-intercept = b= 4
Using double -intercept form of straight line equation, \begin{align*} \dfrac{x}{a} + \dfrac{y}{b} &= 1\\ \dfrac{x}{6} + \dfrac{y}{4} &= 1\\ 2x+3y &= 12 \end{align*}

###### Solved Example: 1-1-08

The area of triangle formed by the lines $x =0, y =0\ \mathrm{and} \dfrac{x}{a} + \dfrac{y}{b} =1$, is:

Solution: The equation of the line $\dfrac{x}{a} + \dfrac{y}{b} =1$ is in double intercept form. This line makes an intercept 'a' on the x-axis and 'b' on the y-axis.

Area of the right angled triangle is = $\dfrac{1}{2}$ $\times$ (base) $\times$ (Height) = $\dfrac{1}{2}$ $\times$ a $\times$ b.

###### Solved Example: 1-1-09

The diagonals of a parallelogram PQRS are along the lines x+3y=4 and 6x-2y=7. Then PQRS must be a:

Solution:
$m_1=-\dfrac{1}{3}$ and $m_2=3$. Hence lines x +3y =4 and 6x -2y =7 are perpendicular to each other. Therefore the parallelogram is rhombus.

###### Solved Example: 1-1-14

Consider the point $P (1,2)$ and the line L: $x - 2y + 2 = 0$. Which of the following is a correct statement?

Solution: The line L: $x - 2y + 2 = 0$ can be brought to slope-intercept format as: $y =0.5x + 1$

A rough sketch can be drawn beginning y-intercept point (0,1).

The line has a positive slope, means uphill (from left to right).

The slope is 0.5 or $\dfrac{1}{2}$ means every 2 units it travels horizontally (run), it travels 1 unit along y-axis (rise).

From this information, it can be found out that the point P is towards the non-origin side.
Alternate method:
Substitute (0,0) in the given equation of L. We get 0-0+2 = 2>0 for Origin.
Substitute (1,2) in the given equation of L. We get 1-4+2 = -1<0 for (1,2).
Since both answers have opposite sign, the point P is on the opposite side of the origin.

###### Solved Example: 1-1-15

A ray of light coming from the point A(1, 2) is reflected at a point P(x,0) on the x-axis and then passes through the point B(5, 3). The coordinates of the point A are: (BITSAT 2016) Solution: From similarity of triangles, $\dfrac{AM}{MP} = \dfrac{BN}{NP}$ $\dfrac{2}{x-1} = \dfrac{3}{5-x}, x = \dfrac{13}{5}$