Equation of Straight Line - Practice Questions & MCQ

Equation of Straight Line

(a) Slope-Intercept form

Consider the given figure

$A B$ is a straight line with slope $m$ and intercept $c$ on Y -axis. $\mathrm{P}(\mathrm{x}, \mathrm{y})$ any point on the straight line. PL is perpendicular to X -axis and MQ is perpendicular to Y -axis

$
\begin{aligned}
& \angle \mathrm{PRL}=\angle \mathrm{PQM}=\theta, \quad \mathrm{OQ}=\mathrm{c} \\
& \mathrm{PM}=\mathrm{PL}-\mathrm{ML}=\mathrm{PL}-\mathrm{OQ}=\mathrm{y}-\mathrm{c} \\
& \mathrm{QM}=\mathrm{OL}=\mathrm{x} \\
& \operatorname{In} \triangle \mathrm{PQM}, \tan \theta=\frac{\mathrm{PM}}{\mathrm{QM}}=\frac{\mathrm{y}-\mathrm{c}}{\mathrm{x}} \\
& \tan \theta=\mathrm{m}=\frac{\mathrm{y}-\mathrm{c}}{\mathrm{x}} \\
& \Rightarrow \mathrm{y}=\mathrm{mx}+\mathrm{c}
\end{aligned}
$
The equation of a straight line whose slope is given as $m$ and making $y$-intercept of length $c$ unit is $y=m x+c$.
If the straight line passing through the origin, then equation of straight line become $y=m x$
If Equation of straight line is $\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0$, then
We can write $\mathrm{By}=-\mathrm{Ax}-\mathrm{C}$

$
\mathrm{y}=\left(-\frac{\mathrm{A}}{\mathrm{~B}}\right) \mathrm{x}+\left(-\frac{\mathrm{C}}{\mathrm{~B}}\right)
$

compare with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$

$
\begin{aligned}
& \text { slope }=m=-\frac{A}{B} \\
& y-\text { intercept }=c=-\frac{C}{B}
\end{aligned}
$

(b) Point-Slope form

Let the equation of give line I with slope ' $m$ ' is

$
y=m x+c
$

$\left(x_1, y_1\right)$ lies on the line i

$
y_1=m x_1+c
$
From (i) and (ii) [(ii) - (i)]

$
y-y_1=m\left(x-x_1\right)
$
The equation of a straight line whose slope is given as ' $m$ ' and passes through the point $\left(x_1, y_1\right)$ is $\mathbf{y}-\mathbf{y}_1=\mathbf{m}\left(\mathbf{x}-\mathbf{x}_1\right)$

Equation of Straight Line

(c) Two-point form

The equation of a straight line passing through the two given points (x₁,y₁) and  (x₂,y₂) is  given by

$
\mathbf{y}-\mathbf{y}_1=\left(\frac{\mathbf{y}_2-\mathbf{y}_1}{\mathbf{x}_2-\mathbf{x}_1}\right)\left(\mathbf{x}-\mathbf{x}_1\right)
$
Proof:
Let the equation of straight line I with slope ' $m$ ' be

$
y=m x+c
$
Points $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\left(\mathrm{x}_2, \mathrm{y}_2\right)$ pass through the given line I , then

$
\begin{aligned}
& \mathrm{y}_1=\mathrm{mx}_1+\mathrm{c} \\
& \mathrm{y}_2=\mathrm{mx}_2+\mathrm{c}
\end{aligned}
$

$\qquad$
Subtract eq (ii) from eq (i)

$
y-y_1=m\left(x-x_1\right)
$
Subtract eq (iii) from eq (i)

$
\mathrm{y}-\mathrm{y}_2=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_2\right)
$
Divide eq (iv) by eq ( $\mathbf{v}$ )

$
\begin{aligned}
& \frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1} \\
& \Rightarrow \mathrm{y}-\mathrm{y}_{\mathbf{1}}=\left(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\right)\left(\mathrm{x}-\mathrm{x}_{\mathbf{1}}\right) \\
& \text { also here } \mathrm{m}=\left(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\right)
\end{aligned}
$

Determinant Form:
If $\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{Q}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{R}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ are collinear, then area of $\triangle \mathrm{PQR}=0$
i.e. $\left.\quad \frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1\end{array}\right| \right\rvert\,=0$
(d) Intercept form of line

Equation of a straight line which makes intercepts ' $a$ ' and 'b' on X -axis and Y -axis respectively is given by

$
\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{~b}}=1
$
Proof:
A straight line which cut $X$-axis at $A(a, 0)$ and $Y$-axis at $B(0, b)$
Using the concept of two points form of a line
Equation of a straight line through the two-point $A(a, 0)$ and $B(0, b)$

$
\begin{aligned}
& y-0=\frac{b-0}{0-a}(x-a) \\
& \Rightarrow-a y=b x-a b \\
& \Rightarrow b x+a y=a b
\end{aligned}
$

divide LHS and RHS by ab

$
\frac{x}{a}+\frac{y}{b}=1
$
Note:

For the general equation $A x+B y+C=0$
If $\mathrm{C} \neq 0$, then $\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0$ can be written as

$
\begin{aligned}
& A x+B y=-C \\
& \frac{x}{-\frac{C}{A}}+\frac{y}{-\frac{C}{B}}=1 \text { or } \frac{x}{a}+\frac{y}{b}=1
\end{aligned}
$

where, $\mathrm{a}=-\frac{\mathrm{C}}{\mathrm{A}}$ and $\mathrm{b}=-\frac{\mathrm{C}}{\mathrm{B}}$
Hence, $x$-intercept is $-\frac{\mathrm{C}}{\mathrm{A}}$ and $y$-intercept is $-\frac{\mathrm{C}}{\mathrm{B}}$

Normal and Parametric form of a line

Normal form of line

Equation of straight line on which the length of the perpendicular from the origin is $p$ and this normal makes an angle $\theta$ with the positive direction of X -axis is given by

$
\mathbf{x} \cos \theta+\mathbf{y} \sin \theta=\mathbf{p}
$

Proof:
$A B$ is the straight line and length of perpendicular from origin to the line is p (i.e. $O N=p$ ).
Line $A B$ cuts $X$-axis and $Y$-axis at point $Q$ and $R$ respectively

$
\begin{aligned}
& \angle \mathrm{NOX}=\theta \\
& \angle \mathrm{NQO}=90^{\circ}-\theta \\
& \therefore \angle \mathrm{NQX}=180^{\circ}-\left(90^{\circ}-\theta\right)=90^{\circ}+\theta \\
& \text { Slope } m=\tan \left(90^{\circ}+\theta\right)=-\cot (\theta)
\end{aligned}
$
In triangle NOL

$
O L=x=p \cdot \cos \theta, N L=y=p \cdot \sin \theta
$
Point $N(p \cdot \cos \theta, p \cdot \sin \theta)$
Using Slope-point form, equation of line $A B$ is

$
\begin{aligned}
& y-p \cdot \sin \theta=-\frac{\cos \theta}{\sin \theta}(x-p \cdot \cos \theta) \\
& x \cdot \cos \theta+y \cdot \sin \theta=p
\end{aligned}
$

Parametric form of a line

The equation of a straight line passing through the point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and making an angle $\theta$ with the positive direction of $X$-axis is

$
\frac{\mathrm{x}-\mathrm{x}_1}{\cos \theta}=\frac{\mathrm{y}-\mathrm{y}_1}{\sin \theta}=\mathrm{r}
$
Where $r$ is the directed distance between the points $(x, y)$ and $\left(x_1, y_1\right)$.
Proof:
$A B$ is a straight line passing through the point $P\left(x_1, y_1\right)$ and meets $X$-axis at $R$ and makes an angle $\theta$ with the positive direction of $X$-axis.

Let $Q(x, y)$ be any point on the line $A B$ at a distance ' $r$ ' from $P$
As from the figure

$
\begin{aligned}
& \mathrm{PN}=\mathrm{ML}=\mathrm{OL}-\mathrm{OM}=\mathrm{x}-\mathrm{x}_1 \\
& \mathrm{QN}=\mathrm{QL}-\mathrm{NL}=\mathrm{QL}-\mathrm{PM}=\mathrm{y}-\mathrm{y}_1 \\
& \mathrm{In} \triangle \mathrm{NPQ} \\
& \cos \theta=\frac{\mathrm{PN}}{\mathrm{PQ}}=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{r}} \\
& \sin \theta=\frac{\mathrm{QN}}{\mathrm{PQ}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{r}}
\end{aligned}
$
From the above two equation

$
\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r
$
Also,

$
\begin{aligned}
& \mathrm{x}=\mathrm{x}_1+\mathrm{r} \cos \theta \\
& \mathrm{y}=\mathrm{y}_1+\mathrm{r} \sin \theta
\end{aligned}
$

Parametric equations of straight line AB