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.
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
Putting the value of x in eq. (ii), we get
Hence, the point of intersection P(x1, y1) is
Now, the equation of line in intercept form is:
where a and b are the intercepts on the axis.
Given that: a = b
⇒ x + y = a …(i)
Putting the value of ‘a’ in eq. (i), we get
⇒ 23x + 23y = 11
Hence, the required line is 23x + 23y = 11