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:
I. 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
II. 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)
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