We will discuss here about the method of finding the equation of a straight line in the slope-intercept form.

Let the straight line AB intersect x-axis at C and y-intersect at D.

Let ∠ACX = θ and OD = c.

Then, tan θ = m(say).

We have to find the equation of the straight line AB.

Now take any point P (x, y) on the line. Let PM ⊥ OX.

Then, OM = x and PM = y.

Draw DE ⊥ PM. Clearly, DE ∥ OX.

Also, PE = PM – EM = PM - OD = y - c, and DE = OM = x.

As DE ∥ OX, ∠PDE = ∠PCX = θ. Therefore, in the right-angled triangle PED,

tan θ = \(\frac{PE}{DE}\) = \(\frac{y - c}{x}\)

⟹ m = \(\frac{y - c}{x}\)

⟹ y – c = mx

⟹ y = mx + c

This is the relation between the x-coordinate and y-coordinate of any point on the line AB.

y = mx + c is the equation of the straight line whose slope is m and which cuts off an intercept c on the y-axis.

Solved examples of finding the equation of a straight line in the slope-intercept form:

**1.** The equation of the straight line inclined at 30° with the positive
direction of the x-axis and cuts an intercept 5 units on the positive direction
of the y-axis is

y = tan 30° ∙ x + 5, (since m = tan 30° and c = +5)

⟹ y = \(\frac{√3}{3}\)x + 5

**2.** The equation of the straight line inclined at 45° with
the positive direction of the x-axis and cuts an intercept 7 units on the
positive direction of the y-axis is

y = tan 45° ∙ x + (-7), (since m = tan 45° and c = -7)

⟹ y = x – 7

**Notes:** **1.** The x-axis is inclined at 0° with the positive
direction of the x-axis i.e. m = tan 0and cuts at intercept 0 unit on the
y-axis i.e. c = 0. So, the equation of the x-axis is y = tan 0° ∙ x + 0, (since
m = tan 0° and c = 0)

⟹ y = x + 0 or x

Therefore, the equation of the x-axis is y = 0

**2.** If a line parallel to x-axis and at a distance a from
the x-axis then the slope m = tan 0 and the intercept on the y-axis c = a. So,
the equation of the parallel line is y = tan 0 ∙ x + a, (since m = tan 0° and c
= a)

__From Slope-intercept Form____ to HOME__

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