Question

Find the equation of the line passing through the intersection of the lines 3x – 4y + 1 = 0 and 5x + y – 1 = 0 and which cuts off equal intercepts from the axes.

Answer 1

Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

3x – 4y + 1 = 0 …(i)

5x + y – 1 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 4, we get

20x + 4y – 4 = 0 …(iii)

On adding eq. (iii) and (i), we get

20x + 4y – 4 + 3x – 4y + 1 = 0

⇒ 23x – 3 = 0

⇒ 23x = 3

⇒ x = 3/23

Putting the value of x in eq. (ii), we get

Hence, the point of intersection P(x₁, y₁) is

\((\frac{3}{23},\frac{8}{23})\)

Now, the equation of line in intercept form is:

\(\frac{x}{a}+\frac{y}{b}=1\)

where a and b are the intercepts on the axis.

Given that: a = b

⇒ x + y = a …(i)

If eq. (i) passes through the point \((\frac{3}{23},\frac{8}{23})\), we get

Putting the value of ‘a’ in eq. (i), we get

x+y = 11/23

⇒ 23x + 23y = 11

Hence, the required line is 23x + 23y = 11