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Let us find the equation of the straight line whose intercepts are a and b on the coordinate axes, respectively.
 
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Let PQ be a straight line that meets the x - axis at the point A and the y axis at the point B. The coordinates of points A and B are (a,0) and (0,b), respectively. Then, OA = a and OB = b.
 
Here, (x_1,y_1) = (a,0) and (x_2,y_2) = (0,b).
 
Substituting the known values in two point form equation, we get:
 
\frac{y - 0}{b - 0} = \frac{x - a}{0 - a}
 
\frac{y}{b} = \frac{x - a}{-a}
 
\frac{y}{b} = \frac{-x}{a} + 1
 
\frac{x}{a} + \frac{y}{b} = 1
 
Hence, the equation of the intercept of the line is \frac{x}{a} + \frac{y}{b} = 1.
Example:
Find the intercepts of the line -3x + 4y + 12 = 0 on the coordinate axes.
 
Solution:
 
The given equation is -3x + 4y + 12 = 0.
 
Let us write the equation in the normal form.
 
-3x + 4y = -12
 
\frac{-3}{-12}x + \frac{4}{-12}y = \frac{-12}{-12} [Dividing by -12 on both sides so that the RHS is equal to 1]
 
\frac{x}{4} + \frac{y}{-3} = 1
 
Comparing the above equation with the equation of the intercept of the line \frac{x}{a} + \frac{y}{b} = 1, we get:
 
a = 4 and b = -3
 
Therefore, the intercept of the line is a = 4 and b = -3.