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Straight Line

Understand the concept of slope of a line and extend it to the concepts of conditions for parallelism and perpendicularity of lines and collinearity of three points

Parametric form

Let P(x, y) be any point on the line l having inclination $\mathrm{\theta }$. Let the line l meet the X-axis at A and the Y-axis at B.
Let Q(${x}_{1},{y}_{1}$) be a point on the line at a distance r from P(x, y), i.e., PQ = r

Draw perpendiculars PL and QM on the X-axis and QN on the line segment PL.

$\angle$NQP= $\mathrm{\theta }$

Now, QN = ML = OL − OM
= x x1
Also, PN = PL − NL = PL − QM
= y y1

In $\mathrm{\Delta }$PQN, we have:

From (i) and (ii), we get:

$\frac{x-{x}_{1}}{\mathrm{cos}\theta }=\frac{y-{y}_{1}}{\mathrm{sin}\theta }=r$

This is the parametric form of equation of a line.

From the above equation, we have:

Thus, the coordinates of any point on the line at a distance r from the given point () are ().

Note: If Q() is a point on a line l which makes an angle $\mathrm{\theta }$ with the positive direction of the x-axis, then there will be two points on line l at a distance r from Q() and their coordinates will be

.

The two points will be on the both sides of point Q.

Example: A line with inclination 60$°$ is drawn through the point (1, 4). Find the coordinates of two points on this line which are at a distance of 2 units from the given point.

Solution:
Here, (x1, y1) = (1, 4), $\mathrm{\theta }=60°$ and r = 2

We know that the parametric form of the equation is

$\frac{x-{x}_{1}}{\mathrm{cos}\theta }=\frac{y-{y}_{1}}{\mathrm{sin}\theta }=r$.

$\frac{x-1}{\mathrm{cos}60°}=\frac{y-4}{\mathrm{sin}60°}$

$\frac{x-1}{\left(1}{2}\right)}=\frac{y-4}{\left(\sqrt{3}}{2}\right)}$

$\sqrt{3}\left(x-1\right)=y-4$

i.e. $\sqrt{3}x-y=\sqrt{3}-4$

Also, the coordinates of any point on the line at a distance 2 units from the given point (1, 4) are , i.e., .

Point of Intersection of Two Intersecting Lines

Theorem: If are two intersecting lines, then the coordinates of their point of intersection are

$\left(\frac{{c}_{1}{b}_{2}-{c}_{2}{b}_{1}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}},\frac{{a}_{1}{c}_{2}-{a}_{2}{c}_{1}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right)$, where

Proof:
The equations of the two intersecting lines are as follows:

Solving the equations (i) and (ii), to eliminate gives

provided a1b2a2b1 ≠ 0      ...(iii)

Similarly to eliminate x, gives

provided a1b2a2b1 ≠ 0

Therefore, the coordinates of the point of intersection of lines (i) and (ii) are $\left(\frac{{b}_{2}{c}_{1}-{b}_{1}{c}_{2}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}},\frac{{a}_{1}{c}_{2}-{a}_{2}{c}_{1}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right)$.

Remark: As the lines intersect, they are not parallel.

Note: By using Cramer's Rule (determinant form) as denoted by , we get:

From (iii) and (iv), the coordinates of the point of intersection are given by $\left(\frac{∆\mathrm{x}}{∆},\frac{∆\mathrm{y}}{∆}\right)$.

Example: Show that the lines $x-y-1=0$ and $2x-3y+1=0$ intersect at the point (4, 3).

Solution:

The given equation are:

$x-y-1=0$                  ... (i)

$2x-3y+1=0$             ... (ii)

Multiplying equation (i) by 2 and solving, we get:

y = 3

Substituting the value of y in equation (i), we get:

x = 3 + 1 = 4

x = 4, y = 3

i.e., the point of intersection is (4, 3).

Thus, the lines $x-y-1=0$ and $2x-3y+1=0$ intersect at the point (4, 3).…

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