Subject: Optional Mathematics

This note contents various method or way to measure the slope intercept.

**To find the equation of a straight line in form of y = mx + c**

Let a straight line XY make an intercept c on Y-axis. then OY = c.

Let m be the slope of the line and \(\theta\) be its inclination.

Then, m = tanΘ

Let P(x, y) be any point on the XY. Draw perpendicular TR from P to X-axis.

Then, OR = x and RT = y.

Again, draw perpendicular YS from Y to the line segment TR, then

YS = OR = x

ST = RT - RS = RT - OY = y - c

Also,

∠TYS = ∠YXO = \(\theta\)

From right angled ΔYTS,

tanΘ = \(\frac{ST}{YS}\)

or, m = \(\frac{y - c}{x}\)

or, y - c = mx

or, **y = mx + c, **which is the equation of a straight line in the required form.

**To find the equation of a straight line in the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1**

Let the straight line EF cut the axis at E and F. Let OE = a and OF = b. These are the intercept on the X-axis and the Y-axis respectively. Obviously, the coordinates of E and F are respectively (a, 0) and (0, b).

Let G(x, y) be any point on the line EF. Then,

Slope of the line segment EG = \(\frac{y - 0}{x - a}\)

Slope of the line segment GF= \(\frac{b -y}{0 - x}\)

But EG and GF are the segments of the same straight line.

So, \(\frac{y - 0}{x - a}\) = \(\frac{b - y}{0 - x}\)

or, -xy = bx - ab - xy + ay

Dividing both sides by ab we get**\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1,** which bis the equation of straight line in the required form.

**To find the equation of a straight line in the form x cos∝ + y sin∝ = p**

Let a straight line XY cut the axes at X and Y. Then,

X-intercept = OX and Y-intercept =OY

Then, equation of the line XY is given by

\(\frac{x}{OX}\) + \(\frac{y}{OY}\) = 1 . . . . . . . . . . . . . . . . . (i)

Draw perpendicular OZ from origin to the line XY.

Let OZ = p and∠XOZ = \(\alpha\).

Then, ∠YOZ = 90° - \(\alpha\) and

∠OYZ =90° - ∠YOZ = 90° - (90° - ∝) = ∝

From right angled triangle OXZ,

sec∝ = \(\frac{OX}{OZ}\)

∴ OX = OZ sec\(\alpha\) = p sec\(\alpha\)

From right angled triangle OYZ,

cosec∝ = \(\frac{OY}{OZ}\)

∴ OY = OZ cosec\(\alpha\) = p cosec\(\alpha\)

Now,

putting the values of OA and OB in (i) we get.

\(\frac{x}{p \: secα} + \frac{y}{p \: cosecα}\) = 1

or, \(\frac{x \: cosα}{p} + \frac{y \: sinα}{p}\) = 1

or, **x cosα + y sinα = p,** which is the equation of a straight line in the required form.

- y = mx+ c
- \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
- x cos∝ + y sin∝ = p

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Find out the equation of a straight line whose inclination is 45° and y-intercept is 2.

Solution:

Here,

Inclination of the line (θ) = 45°

∴ Slope of the line (m) = tanθ

= tan 45°

= 1

y-intercept (c) = 2

∴ Equation of the line in slope intercept form is given by, y = mx + c

So,

y = 1 . x + 2

or, y - x - 2 = 0 is the required equation.

Find the equation of a line passing through the point (0, 5) and inclination at an angle of 60° with the x-axis.

Solution:

Here,

Inclination of the line (θ) = 60°

∴ Slope of the line (m) = tan 60° = \(\sqrt{3}\)

∴ Equation of the line is given by, y = mx + c

or, y = \(\sqrt{3}\) x + c . . . . . . . . . . . . (1)

Since the line passes though (0, 5)

5 = c

∴ c = 5

Putting the value of c in (1)

y = \(\sqrt{3}\) x + 5 is the required equation.

Find the equation of a straight line whose X-intercept and Y-intercept are 3 and -4 respectively.

Here,

X-intercept (a) = 3

Y-intercept (b) = -4

Equation of a staright line is given by \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

or, \(\frac{x}{3}\) - \(\frac{y}{4}\) = 1

or, 4x - 3y = 12

4x - 3y - 12 = 0 is the required equation.

A straight line passes through the point (6, 5) and cuts off intercepts on the axes equal in magnitude but opposite in sign. Find its equation.

Solution:

Here,

The line cuts off, equal; intercepts on the axes equal in magnitude but opposite in sign.

So, X-intercept = -Y-intercept, i.e. a = -b.

So, its equation is given by \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

or, x - y = a . . . . . . . . . (i)

Since it passes through the point (6, 5)

or, 6 - 5 = a

∴ a = 1

Putting the value of a in equation (i)

x - y - 1 = 0 is the required equation.

Find the equation of a straight line which passes through the point (-2, 3) and is such that its portion between the axes is divided by the point in the ratio 3 : 4.

Solution:

Let AB be the required line that cuts X-axis at A and Y-axis at B such that OA = a and OB = b.

By the questions, C(-2, 3) divides the line AB in the ratio 3 : 4.

Using section formula

or, x = \(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}\), y = \(\frac{m_1y_2 + m_2y_1}{m_1 + m_2}\)

or, -2 = \(\frac{3.0 + 4.a}{3 + 4}\), 3 = \(\frac{3.b + 4.0}{3 + 4}\)

∴ a = \(\frac{-7}{2}\), ∴ b = 7

The equation of line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

\(\frac{x}{\frac{7}{-2}}\) + \(\frac{y}{7}\) = 1

∴ 2x - y + 7 = 0 is the required equation.

The length of perpendicular drawn from the origin on a straight line is 3 units and the perpendicular is inclined at an angle of 120° to X-axis. Find its equation.

Solution:

Here,

perpendicular length (p) = 3

Angle made by perpendicular in the positive direction with X-axis (α) = 120°

Equation of the line is given by x cosα + y sinα = p

or, x . cos 120° + y sin 120° = 3

or, x . (\(\frac{-1}{2}\)) + y . \(\frac{\sqrt{3}}{2}\) = 3

or, -x + \(\sqrt{3}\) y = 6

∴ x - \(\sqrt{3}\) y + 6 = 0 is the required equation.

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