Find the equation of the lines which cut off intercepts on the axes whose sum and product are 1 and $-6$ respectively.

**OR**

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x\cos heta - y\sin heta = k\cos 2 heta$ and $x\sec heta + y\, ex

Question

Find the equation of the lines which cut off intercepts on the axes whose sum and product are 1 and $-6$ respectively.

**OR**

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x\cos	heta - y\sin	heta = k\cos 2	heta$ and $x\sec	heta + y\,	ext{cosec}\,	heta = k$ respectively, then prove that $p^2 + 4q^2 = k^2$.

Accepted Answer

>

> [!TLDR]
> Solve $a+b=1,\ ab=-6$ as roots of a quadratic to get the two intercepts, then write $x/a+y/b=1$ for each pair. [OR alternative: find the perpendicular distances $p,q$ from the origin to each given line in normal form, then verify $p^2+4q^2=k^2$.]
>
> **Main question.** Let the intercepts be $a$ (on $x$-axis) and $b$ (on $y$-axis), with $a+b=1$ and $ab=-6$. So $a,b$ are roots of:
> $$t^2 - (a+b)t + ab = 0 \implies t^2-t-6=0 \implies (t-3)(t+2)=0 \implies t=3,\ -2$$
> So $(a,b) = (3,-2)$ or $(-2,3)$.
>
> Equation in intercept form $\dfrac xa+\dfrac yb=1$:
>
> - For $(a,b)=(3,-2)$: $\dfrac x3-\dfrac y2=1 \implies 2x-3y=6 \implies 2x-3y-6=0$.
> - For $(a,b)=(-2,3)$: $-\dfrac x2+\dfrac y3=1 \implies -3x+2y=6 \implies 3x-2y+6=0$.
>
> ---
> **OR (alternative question):** Prove $p^2+4q^2=k^2$ given $p,q$ are perpendicular distances from the origin to $x\cos	heta-y\sin	heta=k\cos2	heta$ and $x\sec	heta+y\,	ext{cosec}\,	heta=k$ respectively.
>
> For the first line: $(\cos	heta)^2+(-\sin	heta)^2=1$, so it is already in normal form $x\cos\alpha+y\sin\alpha=p$ with $\alpha=-	heta$. So $p = k\cos2	heta$.
> …

Concept understanding — Coordinate Axes Equations

Coordinate Axes Equations

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