NCERT Solutions
Class 11 Maths
Straight Lines

Ex.10.3 Q.18
If p is the length of perpendicular from the origin to the line whose intercepts on the
axes are a and b, then show that: (1 ÷ p2) = (1 ÷ a2) + (1 ÷ b2)
It is known that the equation of a line whose intercepts on the axes are a and b is
+
= 1
=> bx + ay = ab
=> bx + ay – ab = 0 ………... (1)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
d = |Ax1 + By1 + C| ÷ √ (A2 + B2)
On comparing equation 1 to the general equation of line Ax + By + C = 0, we get
A = b, B = a, and C = –ab
Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line 1, we get
p = |A × 0 + B × 0 - ab| ÷ √ (b2 + a2) = |-ab| ÷ √ (b2 + a2)
On squaring both sides, we obtain
p2 = (-ab)2 ÷ (b2 + a2)
=> p2 = a2 b2 ÷ (b2 + a2)
=> p2(b2 + a2) = a2 b2
=> (b2 + a2) ÷ a2 b2 = (1 ÷ p2) => (1 ÷ p2) = (1 ÷ a2) + (1 ÷ b2)
Hence, we showed that (1 ÷ p2) = (1 ÷ a2) + (1 ÷ b2)