EQUATION OF A STRAIGHT LINE IN PARAMETRIC FORM

Consider a line AB makes an angle \alpha with the positive direction of x – axis and a point on the line is Q\left( {{x_1},\,{y_1}} \right).

Let a general point P on the line AB and the distance between P and Q is r.

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Draw a line QL, PM perpendicular to x-axis and drawQN \bot PM.

In a right angled triangle PQN,

\cos \alpha  = \frac{{QN}}{{QP}},\sin \alpha  = \frac{{PN}}{{QP}}

\begin{array}{l}
QN = LM = OM - OL = x - {x_1}\\
PN = PM - MN = PM - QL = y - {y_1}
\end{array}

Therefore,

\cos \alpha  = \frac{{x - {x_1}}}{r},\sin \alpha  = \frac{{y - {y_1}}}{r}

Therefore, the equation of a straight line in parametric form is:

{r = \frac{{x - {x_1}}}{{\cos \alpha }} = \frac{{y - {y_1}}}{{\sin \alpha }}}

 

Example:

Determine the equation of a line passes through the point (–1, – 2) and makes an angle of 30o with the positive direction of x-axis, in parametric form. Find the coordinates of a point at distance of 2 units.

The equation of the line is:

\frac{{x - {x_1}}}{{\cos \alpha }} = \frac{{y - {y_1}}}{{\sin \alpha }} = r

{x_1} =  - 1,{y_1} =  - 2{\rm{ and }}\alpha  = 30^\circ ,r = 2

 \frac{{x + 1}}{{\cos 30^\circ }} = \frac{{y + 2}}{{\sin 30^\circ }} = 2

\frac{{x + 1}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{y + 2}}{{\frac{1}{2}}} = 2

x + 1 = \sqrt 3 ,y + 2 = 1

x + 1 = \sqrt 3 ,y = 1 - 2

x = \sqrt 3  - 1,y =  - 1

The coordinates of the point are{\left( {\sqrt 3  - 1, - 1} \right)}.

Post Author: E-Maths