# Straight Line – Intercept Form: Examples

This lesson will be focused on examples involving the intercept form of the equation of a straight line.

Keep in mind that the problems can be solved using any form of the equation. The form to be used will depend upon the conditions given in the problem. As you solve more problems, you’ll get the general idea of which equations will provide a good starting point.

Example 1. Find the equation of the line passing through (4, -5) cutting off equal intercepts, but of the opposite sign, from the axes.

Solution The information provided in the problem involves intercepts. So it will be better to start with the intercept form of the equation. (Although any other form will lead to the same answer, but we should prefer efficient methods).

Alright. The equation can be assumed to be $$\frac{x}{a}+\frac{y}{b}=1$$.

Given that, the intercepts are equal but opposite in sign. So, a = – b. (I)

And, the line passes through the point (4, -5). So, $$\frac{4}{a}+\frac{-5}{b}=1$$ (II)

Again, two conditions lead to two equations in two variables. Solving I and II, we get the value of a as 9 and b as -9.
Therefore, the required equation is x – y = 9, and the line would look something like this..

Example 2 A line intersects the X-axis and the Y-axis at the point A and B respectively. It also passes through a point C(4, 3) which divides AB in the ratio 1:2 internally. Find the equation of the line.

Solution First, the two conditions – (i) passes through a point, (ii) the point divides AB in the ratio 1:2.

Okay, now, we have to assume an equation of the line, and convert the above conditions into equations to be solved.

The first condition is straightforward. Whatever the equation we assume, we just have to substitute the point’s coordinates for x and y in that equation.

What about the second one? This is how the line might look like..

If we assume the equation of the line to be $$\frac{x}{a}+\frac{y}{b}=1$$, then the second condition will be very easy to apply.

Why? Because the coordinates of A and B are known to be (a, 0) and (0, b) respectively. (Go back in case you forgot why.)

Hence, we can apply the section formula as $$4=\frac{2\times a + 1\times 0}{2+1}$$ which gives a = 6

And, similarly $$3=\frac{2\times 0+1\times b}{2+1}$$ which gives b = 9. And we’re done !

The required equation is $$\frac{x}{6}+\frac{y}{9}=1$$

Note that we didn’t have to explicitly use the first condition (i.e. substituting the coordinates (4, 3) in the equation) because the section formula ensures that A, B and C lie on the same line.

One more..

Example 3 Find the equation to the line, sum of whose intercepts is 10 and
(i) makes equal angles with the axes
(ii) makes a triangle of area 12 with the axes

Solution We can assume the equation to be $$\frac{x}{a}+\frac{y}{b}=1$$ again. The common condition for both the parts is a + b = 10.

(i) The line could look like one of the many below..

If the angles are equal, the intercepts must be equal in magnitude (though they could be of same or opposite signs).

Geometry explains this. For instance, in triangle OAB, the equal angles OAB and OBA, (both measuring 45o) make the sides opposite to them (which are the intercepts) also equal.

What does that mean? a = b, or a = -b. But the latter would mean a+b = 0, which violates the other condition.

Thus, a = b is the only possibility. And using the previous condition (i.e. a+b=10), both a and b must equal 5.

That rules out all the grey lines, and the blue line is the only possibility.

Therefore, the required equation becomes $$\frac{x}{5}+\frac{y}{5}=1$$ or x + y = 5

(ii) We’ve come across a similar problem before. But this time we’ll use the intercept form. The area of the triangle in terms of the intercepts a and b will be |1/2 ab|.

Using the given information we have, |1/2 ab| = 12 or |ab|=24. Also a+b=10.

Solving these, we get four sets of solutions, (a=6, b=4) , (a=4, b=6), (a=12,b=-2) and (a=-2, b=12).

We therefore get four possible lines : $$\frac{x}{6}+\frac{y}{4}=1$$ , $$\frac{x}{12}+\frac{y}{-2}=1$$, $$\frac{x}{4}+\frac{y}{6}=1$$ and $$\frac{x}{-2}+\frac{y}{12}=1$$

That’s it for this lesson. In the next lesson, I’ll discuss another equation of the straight line known as the point-slope form.