Straight Line

• 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.

The equation of a straight line (or any curve) is the relation between the x and y (and z) coordinates of all points lying on it.
The general form of the equation of a straight line is: $Ax + By + C = 0$

Slope of a straight line gives you an idea about its inclination with reference to x-axis. Slope is also referred as gradient.

Slope of a Straight Line:

$\mathrm{Slope \ of \ a \ straight \ line} = \dfrac{\mathrm{Rise}}{\mathrm{Run}} = \dfrac{\Delta y}{\Delta x}$

$m = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Equation of a Straight Line:

Slope Intercept Format:

$y = mx + b$

where m = slope and b = y-intercept
For the above line, y-intercept = 1, and
slope = $\dfrac{\mathrm{Rise}}{\mathrm{Run}}$ = $\dfrac{2}{4}$ = $\dfrac{1}{2}$
So the equation will be, $y = 0.5x + 1$ $2y = x + 2$ $x - 2y + 2 = 0$

Slope Point Format:

$y - y_{1} = m (x - x_{1})$

where $x_{1}$ and $y_{1}$ are the coordinates of the point through which the line passes.

Double Intercept Format:

$\dfrac{x}{a}+\dfrac{y}{b} = 1$

where a = x intercept and b = y intercept
For parallel lines slopes are equal i.e. $m_{1} = m_{2}$
For perpendicular lines
$m_{1} * m_{2}$ = -1
or,

$m_2 = \dfrac{-1}{m_1}$

Angle between Two Straight Lines:

Angle between two straight lines is given by: $\alpha = \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right)$

Solved Example:

1-1-01

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

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:

1-1-03

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: $y = mx + b, 0 = mx + b, \dfrac{-b}{m} = x$

Solved Example:

1-1-04

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, $\dfrac{AM}{MP} = \dfrac{BN}{NP}$ $\dfrac{2}{x-1} = \dfrac{3}{5-x}, x = \dfrac{13}{5}$

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,

$\dfrac{x}{a} + \dfrac{y}{b} = 1, \dfrac{x}{6} + \dfrac{y}{4} = 1, 2x+3y =12$

Solved Example:

1-1-06

The image of the point (4, -3) with respect to the line y = x is:

Solution:

Slope of original (red) line = m$_1$ = 1

Since reflection (blue) line is perpendicular to original (red) line,

Slope of reflection (blue) line m$_2$ = $\dfrac{-1}{m_1}$ = -1

Let M(a,a) be the point of intersection as shown in the figure. We have taken both co-ordinates same for M, because it also lies on y=x line.
Now slope of MA = -1
$\dfrac{-3-a}{4-a} = -1, a= 0.5$ But M is the mid-point of AB. Using mid-point formula, $\dfrac{p+4}{2} = 0.5, \mathrm{\ or,\ } p = -3$ $\mathrm{similarly\ }\dfrac{q-3}{2} = 0.5, \mathrm{\ or,\ } q = 4 \mathrm{\ or,\ } B = (-3,4)$

Solved Example:

1-1-07

A line L passes thro' the points (1, 1) and (2, 0) and another line $L_2$ passes through ($\dfrac{1}{2}$,0) and $\perp$ to $L$. Then the area of the triangle formed by the lines $L$, $L_2$ and y-axis, is:

Solution:

Slope of first (red) line = m$_1$ = $\dfrac{y_2 - y_1}{x_2 - x_1}$ = $\dfrac{0-1}{2-1}$ = -1

Equation of first (red) line:
$y - y_1 = m(x- x_1)$ $y- 0 = -1 (x - 2)$ $y = -x + 2$

Slope of second (blue) line = $\dfrac{-1}{m_1}$ = 1

Equation of second (blue) line: $y - y_1 = m(x- x_1), y- 0 = 1 (x - 0.5), y = x -0.5$
Solving these two equations simultaneously, we will get the intersection point as M (1.25,0.75).
\begin{align*} \mathrm{Area\ of\ Triangle} &=\dfrac{1}{2} \times \mathrm{base} \times \mathrm{height}\\ &=\dfrac{1}{2} \times 2.5 \times 1.25\\ &= \dfrac{25}{16} \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:

1-1-10

The area enclosed within the curve |x|+|y|=1 is:

Solution:

• In quadrant 1 (Q1), since both x and y are positive, |x| = x, |y| = y and the line will be: x + y = 1

• In quadrant 2 (Q2), since x is negative and y is positive, |x| = -x, |y| = y the line will be: -x + y = 1

• In quadrant 3 (Q3), since both x and y are negative, |x| = -x, |y| = -y the line will be: -x - y = 1

• In quadrant 4 (Q4), since x is positive and y is negative, |x| = x, |y| = -y the line will be: x - y = 1

Required area = 4 $\times$ Area in the first quadrant = 4 $\times \dfrac{1}{2} \times$ base $\times$ height = 4 $\times \dfrac{1}{2} \times$ 1 $\times$ 1 = 2

Solved Example:

1-1-11

If the equation of base of an equilateral triangle is 2x-y=1 and the vertex is (-1, 2), then the length of the side of the triangle is:

Solution:

\begin{align*} \tan 60^\circ &=\dfrac {h}{\left( \dfrac {x}{2}\right) } \sqrt {3}=\dfrac {h}{\left(\dfrac {x}{2}\right)}\\ h&=\dfrac {\sqrt {3}}{2}x, \mathrm{or},\ x =\dfrac {2}{\sqrt {3}}h \end{align*} \begin{align*} \mathrm{Distance}&=\left\vert \dfrac {2\left( x_1\right) -\left( y_1\right) -1}{\sqrt {2^{2}+\left( -1\right) ^{2}}}\right\vert\\ &=\left| \dfrac {2\left( -1\right) -\left( 2\right) -1}{\sqrt {2^{2}+\left( -1\right) ^{2}}}\right|\\ &=\left| \dfrac {-5}{\sqrt {5}}\right| =\sqrt {5} \end{align*} \begin{align*} x &=\dfrac {2}{\sqrt{3}}h\\ &=\dfrac {2}{\sqrt{3}}\sqrt{5}\\ &=\dfrac {\sqrt {4}\sqrt {5}}{\sqrt{3}}\\ &=\sqrt {\dfrac {20}{3}} \end{align*}

Solved Example:

1-1-12

Let PS be the median of the triangle with vertices P (2,2), Q (6,-1) and R (7,3). The equation of the line passing through (1,-1) and parallel to PS is:

Solution:
S = midpoint of QR=($\dfrac{6+7}{2}$,$\dfrac{-1+3}{2}$)=($\dfrac{13}{2}$,1) $PS=\dfrac{2-1}{2-\dfrac{13}{2}}=\dfrac{-2}{9},$
The required equation is: $y+1 =\dfrac{-2}{9}(x-1), \mathrm{Or,\ }2x+9y+7 =0$.

Solved Example:

1-1-13

The area bounded by the curves $x + 2 \vert y \vert = 1 \mathrm{\ and\ } x = 0$ is:

Solution:

When y < 0, $\vert$y$\vert$ = -y, So when y < 0, the equation becomes, x - 2y = 1

When y > 0, $\vert$y$\vert$ = +y, So when y > 0,the equation becomes, x + 2y = 1

Area of Triangle,

$A = \dfrac{1}{2} \times b \times h = \dfrac{1}{2} \times 1 \times 1 = \dfrac{1}{2} \mathrm{\ sq.\ units}$