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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 θ. Let the line l meet the X-axis at A and the Y-axis at B.
Let Q(x1,y1) 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.

NQP= θ

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

In ΔPQN, we have:

cosθ = QNPQ=x-x1r         ...isinθ= PNPQ=y-y1r           ...ii           

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

x-x1cosθ=y-y1sinθ=r

This is the parametric form of equation of a line.

From the above equation, we have:

x-x1=rcosθ and y-y1= rsinθ x=x1+rcosθ and y=y1+ rsinθ

Thus, the coordinates of any point on the line at a distance r from the given point (x1, y1) are (x1+rcosθ, y1+ rsinθ).

Note: If Q(x1, y1) is a point on a line l which makes an angle θ with the positive direction of the x-axis, then there will be two points on line l at a distance r from Q(x1, y1) and their coordinates will be

x1+r cosθ, y1+r sinθ  and  x1-r cosθ, y1-r sinθ.

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), θ=60° and r = 2

We know that the parametric form of the equation is

x-x1cosθ=y-y1sinθ=r.

x-1cos60°=y-4sin60°

x-112=y-432

3x-1=y-4

i.e. 3x-y=3-4

Also, the coordinates of any point on the line at a distance 2 units from the given point (1, 4) are 1±2 cos60°, 4±2 sin60° or 1±1, 4±3, i.e., 2, 4+3 and 0, 4-3.

Point of Intersection of Two Intersecting Lines

Theorem: If a1x+b1y=c1 and a2x+b2y=c2 are two intersecting lines, then the coordinates of their point of intersection are

c1b2-c2b1a1b2-a2b1,a1c2-a2c1a1b2-a2b1, where a1b2-a2b1 0

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

a1x+b1y=c1       ...ia2x+b2y=c2       ...ii

Solving the equations (i) and (ii), to eliminate b2i - b1ii gives

a1b2x+b1b2y=c1b2     a2b1x+b1b2y=c2b1 -      -            -a1b2 - a2b1x = c1b2 - c2b1

  x = c1b2 - c2b1a1b2 - a2b1 provided a1b2a2b1 ≠ 0      ...(iii)

Similarly to eliminate x, a2i - a1ii gives

a1a2x+a2b1y=a2c1     a1a2x+a1b2y=a1c2 -      -            -a2b1 - a1b2y = a2c1 - a1c2

  y = a1c1 - a1c2a2b1 - a1b2 provided a1b2a2b1 ≠ 0

      = -a1c2 - a2c1-a1b2 - a2b1= a1c2 - a2c1a1b2 - a2b1                ... iv


Therefore, the coordinates of the point of intersection of lines (i) and (ii) are b2c1-b1c2a1b2-a2b1,a1c2-a2c1a1b2-a2b1.


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

i.e., -a1b1-a2b2 a1a2b1b2a1b2-a2b10

Note: By using Cramer's Rule (determinant form) as denoted by  =a1b1a2b2, we get:
 x=c1b1c2b2,  y=a1b1a2b2

From (iii) and (iv), the coordinates of the point of intersection are given by x,y.

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:

    2x - 2y - 2 = 0    2x - 3y + 1 = 0-     +     -                 y - 3 = 0

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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