Question

Line l has intercepts a and b on the coordinate axes. when the axes are rotated through a given angle keeping origin fixed, the same line l has intercepts p and q on the transformed axes. prove that 1/a^2+1/b^2 =1/p^2+1/q^2

Answer 1

Transformation of Co-ordinates

Given:

• Line $l$ has intercepts $a$ and $b$ on the coordinate axes.

• When the axes are rotated through a given angle keeping origin fixed, the same line $l$ has intercepts $p$ and $q$ on the transformed axes.

To prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{p^{2}}+\frac{1}{q^{2}}$

Proof:

The given straight line is $\frac{x}{a}+\frac{y}{b}=1$

Let the axes rotated through an angle $\theta$, so that

• $x=x'\:cos\theta-y'\:sin\theta$

• $y=x'\:sin\theta+y'\:cos\theta$

The given straight line changes into

• $\frac{x'\:cos\theta-y'\:sin\theta}{a}+\frac{x'\:sin\theta+y'\:cos\theta}{b}=1$

• $\Rightarrow (\frac{cos\theta}{a}+\frac{sin\theta}{b})x'+(\frac{cos\theta}{b}-\frac{sin\theta}{a})y'=1$

• $\Rightarrow \frac{x'}{\frac{1}{\frac{cos\theta}{a}+\frac{sin\theta}{b}}}+\frac{y'}{\frac{1}{\frac{cos\theta}{b}-\frac{sin\theta}{a}}}=1$

By the given condition,

• $p=\frac{1}{\frac{cos\theta}{a}+\frac{sin\theta}{b}}$

• $q=\frac{1}{\frac{cos\theta}{b}-\frac{sin\theta}{a}}$

$\Rightarrow$

• $\frac{1}{p}=\frac{cos\theta}{a}+\frac{sin\theta}{b}$

• $\frac{1}{q}=\frac{cos\theta}{b}-\frac{sin\theta}{a}$

Squaring and adding the above equations, we get

$\quad\frac{1}{p^{2}}+\frac{1}{q^{2}}=(\frac{cos\theta}{a}+\frac{sin\theta}{b})^{2}+(\frac{cos\theta}{b}-\frac{sin\theta}{a})^{2}$

$=\frac{cos^{2}\theta}{a^{2}}+\frac{2\:sin\theta\:cos\theta}{ab}+\frac{sin^{2}\theta}{b^{2}}$

$\quad\quad +\frac{cos^{2}\theta}{b^{2}}-\frac{2\:sin\theta\:cos\theta}{ab}+\frac{sin^{2}\theta}{a^{2}}$

$=(sin^{2}\theta+cos^{2}\theta)\:(\frac{1}{a^{2}}+\frac{1}{b^{2}})$

$=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

$\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{p^{2}}+\frac{1}{q^{2}}$

Hence proved.

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Answer 2

Step-by-step explanation:

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