# Straight Line, Rectangular Coordinates

THE STRAIGHT LINE:

RECTANGULAR COORDINATES:

THEOREM1:

To find the equation of a straight line which is parallel to one of the coordinate axes.

Let CL be any line parallel to the axis of y and passing through a point C on the axis such that OC = c.

Let P be any point on this line whose coordinates are x and y.

Then the abscissa of the point P is always c, which implies

x = c          (1)

This is true for each and every point on the line CL and is the equation of the line.

It is clear that equation (1) does not contain the coordinate y.

In the same way the equation of a straight line parallel to the axis of x is y = d

Corollary:

The equation of the x axis is y = 0.

The equation of the y axis is x = 0.

THEOREM 2:

To find the equation of a straight line which cuts off a given intercept on the axis of y and is inclined at a given angle to the axis of x.

Let C be a point on the axis of y such that OC = c.

Through C draw a straight line LCT inclined at an angle ∝ = tan-1 m to the axis of x, so that tan ∝ = m.

The straight line LCT is hence the required straight line and we have to find the relation between the coordinates of any point P lying on it.

Let us draw a perpendicular PM to OX and a line CN parallel to OX which meets PM at N.

Let the coordinates of P be x and y, so that OM = x and MP = y.

Then MP = NP + MN = CN tan ∝ + OC = m.x + c

x = c       or y = mx + c

This equation is true for any point on the given straight line, hence it is the equation of a straight line.

Corollary:

The equation of a straight line passing through the origin (that is which cuts off a zero intercept from the y axis) is y = mx.

Example:

Find the equation of the straight line cutting off an intercept 5 from the positive direction of y axis and inclined at 120 degree to the x axis.

Solution:

Given that ∝ = 120° and c = 5

Hence y = tan 120°.x + 5

Or y = - √ x + 5

Or y + x √ - 5 = 0 (Answer)

THEOREM 3:

To find the equation of a straight line which cuts off given intercepts a and b from the axes.

Let A and B be two points on OX and OY such that OA = a and OB = b.

Join AB. Let P(x,y) be any point on AB. Draw a perpendicular PQ on OX.

We know that (by geometry),

Example:

Find the equation of a straight line which passes through the point (3, -2) and cutting off intercepts equal but of opposite signs, from the two axes.

Solution:

Let the intercepts cut off from the two axes be of lengths a and –a.

Putting b = -a in equation + = 1 we get,

+ y/(-a) = 1

x – y = a            (1)

Given that the line passes through the point (3, -2), so putting (3, -2) in equation (1) we get,

3 – (-2) = a

a = 5

Hence the required equation is x – y = 5.                 (Answer)

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