# If p is the length of perpendicular from the origin to the line whose intercepts on the x-axis are a and b, then show that 1/p^{2} = 1/a^{2} + 1/b^{2}

**Solution:**

It is known that the equation of a line whose intercepts on the axis a and b is

x/a + y/b = 1

bx + ay = ab

bx + ay - ab = 0 ....(1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x\(_1\), y\(_1\)) is given by

d = |Ax\(_1\) + By\(_1\) + C|/√A² + B²

On comparing equation (1) to the general equation of line Ax + By + C = 0 , we obtain

A = b, B = a and C = - ab

Therefore, if p is the length of the perpendicular from point (x_{1}, y_{1}) = (0, 0) to line (1), We obtain

p = |A(0) + B(0) - ab|/√a² + b²

= |- ab|/√a² + b²

On squaring both sides, we obtain

⇒ p^{2} = (- ab)^{2}/(a^{2} + b^{2})

⇒ p^{2} (a^{2} + b^{2} ) = a^{2}b^{2}

⇒ (a^{2} + b^{2})/a^{2}b^{2} = 1/p^{2}

⇒ 1/p^{2} = 1/a^{2} + 1/b^{2}

Hence, we showed 1/p^{2} = 1/a^{2} + 1/b^{2}

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 18

## If p is the length of perpendicular from the origin to the line whose intercepts on the x-axis are a and b, then show that 1/p^{2} = 1/a^{2} + 1/b^{2}

**Summary:**

If p is the length of perpendicular from the origin to the line whose intercepts on the x-axis are a and b, then we are asked to show that 1/p^{2} = 1/a^{2} + 1/b^{2} and we have shown it

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