We will learn how to find the equation of a straight line in intercept form.
The equation of a line which cuts off intercepts a and b respectively from the x and y axes is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
Let the straight line AB intersects the xaxis at A and the yaxis at B where OA = a and OB = b.
Now we have to find the equation of the straight line AB.
Let P(x, y) be any point on the line AB. Draw PQ perpendicular on OX and PR perpendicular on OX. Then, join the points O and P. Now, PQ = y, OQ = x.
Clearly, we see that
Area of the ∆OAB = Area of the ∆OPA + Area of the ∆OPB
⇒ ½ OA ∙ OB = ½ ∙ OA ∙ PQ + ½ ∙ OB ∙ PR
⇒ ½ a ∙ b = ½ ∙ a ∙ y + ½ ∙ b ∙ x
⇒ ab = ay + bx
⇒ \(\frac{ab}{ab}\) = \(\frac{ay + bx}{ab}\), dividing both sides by ab
⇒ 1 = \(\frac{ay}{ab}\) + \(\frac{bx}{ab}\)
⇒ 1 = \(\frac{y}{b}\) + \(\frac{x}{a}\)
⇒ \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, which is the equation of the line in the intercept form.
The equation \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 is the satisfied by the coordinates of any point P lying on the line AB.
Therefore, \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 represent the equation of the straight line AB.
Solved examples to find the equation of a straight line in intercept form:
1. Find the equation of the line which cuts off an intercept 3 on the positive direction of xaxis and an intercept 5 on the negative direction of yaxis.
Solution:
The equation of a line which cuts off intercepts a and b respectively from the x and y axes is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
Here, a = 3 and b = 5
Therefore, the equation of the straight line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 ⇒ \(\frac{x}{3}\) + \(\frac{y}{5}\) = 1 ⇒ \(\frac{x}{3}\)  \(\frac{y}{5}\) = 1 ⇒ 5x – 3y = 15 ⇒ 5x – 3y – 15 = 0.
2. Find the intercepts of the straight line 4x + 3y = 24 on the coordinate axes.
Solution:
Given equation 4x + 3y = 24.
Now convert the given equation into intercept form.
4x + 3y = 24
⇒ \(\frac{4x + 3y}{24}\) = \(\frac{24}{24}\), Dividing both sides by 24
⇒ \(\frac{4x}{24}\) + \(\frac{3y}{24}\) = 1
⇒ \(\frac{x}{6}\) + \(\frac{y}{8}\) = 1, which is the intercept form.
Therefore, xintercept = 6 and yintercept = 8.
Note: (i) The straight line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 intersects the xaxis at A(a, 0) and the yaxis at B(0, b).
(ii) In \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, a is xintercept and b is y intercept.
These intercept a and b may be positive as well as negative.
(iii) If the straight line AB passing through the origin then, a = 0 and b = 0. If we put a = 0 and b = 0 in the intercept form, then \(\frac{x}{0}\) + \(\frac{y}{0}\) = 1, which is undefined. For this reason the equation of a straight line passing through the origin cannot be expressed in the intercept form.
(iv) A line parallel to the xaxis does not intercept the xaxis at any finite distance and hence, we cannot get any finite x intercept (i.e., a) of such a line. For this reason, a line parallel to xaxis cannot be expressed in the intercept from. In like manner, we cannot get any finite y intercept (i.e., b) of a line parallel to yaxis and hence, such a line cannot be expressed in the intercept form.
11 and 12 Grade Math
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