Question

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
⠀
⠀⠀⠀☆ $\bf{\dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}}$
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Answer 1

Step-by-step explanation:

Given:-

p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b.

To find:-

show that 1/p²=1/a²+1/b²

Solution:-

The intercepts of the axes =a and b

The intercept of x-axis =a

The intercept of y-axis =b

Then The equation whose intercepts of x, y -axies a and b is x/a+y/b=1

=>(1/a)x + (1/b)y= 1

=>(1/a)x +(1/b)y -1 = 0

On comparing with Ax+By+C=0, then

A=1/a

B=1/b

C= -1

The coordinates of the origin=(0,0)

The perpendicular distance from the Origin =p

Here distance =p

x1=0

x2=0

We know that the perpendicular distance from the origin to the line x/a +y/b =1

= | Ax1+By1+C |/√(A²+B²)

=>p= | (1/a)(0)+(1/b)(0)+(-1) | /√[(1/a)²+(1/b)²]

=>p= |0+0-1 |/ √[(1/a²)+(1/b²)]

=>p= | -1 | /√[(1/a²)+(1/b²)]

=>p=1/√[(1/a²)+(1/b²)]

On squaring both sides then

=>p²=(1/a²)+(1/b²)

Answer:-

⠀If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then

1/p² = 1/a² + 1/b²

Used formula:-

• The equation whose intercepts of x, y -axies a and b is x/a+y/b=1.
• the perpendicular distance from the origin to the line x/a +y/b =1 is

| Ax1+By1+C |/√(A²+B²)