Question

Which of the following second-degree equation represents a pair of straight lines?

• A. ${\mathrm{x}}^2–\mathrm{xy}–{\mathrm{y}}^2=1$
• B. $–{\mathrm{x}}^2+\mathrm{xy}–{\mathrm{y}}^2=1$
• C. $4{\mathrm{x}}^2–4\mathrm{xy}+{\mathrm{y}}^2=4$
• D. ${\mathrm{x}}^2+{\mathrm{y}}^2=4$

Answer 1

The correct option is C

$4{\mathrm{x}}^2–4\mathrm{xy}+{\mathrm{y}}^2=4$

Explanation for the correct option: c) $4{\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=4$

The general equation that represents a pair of straight lines is $\mathrm{abc}+2\mathrm{fgh}–{\mathrm{af}}^2–{\mathrm{bg}}^2–{\mathrm{ch}}^2=0$,

where the second-degree equation that represents a pair of straight lines is in the form ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

Compare the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ with the equation $4{\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=4$.

From the second degree equation, the values of $\mathrm{a}=4,\mathrm{b}=1,\mathrm{c}=-4,\mathrm{h}=-2,\mathrm{g}=0,\mathrm{f}=0$

Put the values in the general equation of straight lines.

$\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0$

$\Rightarrow \left(4\right)\left(1\right)\left(-4\right)+2\left(0\right)\left(0\right)\left(-2\right)-\left(4\right){\left(0\right)}^2-\left(1\right){\left(0\right)}^2-\left(-4\right){\left(-2\right)}^2=0$

$\Rightarrow$ $-16+0-0-0+4\left(4\right)=0$

$\Rightarrow$ $0=0$

Since both sides are equal, the second-degree equation $4{\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=4$, represents a pair of straight lines.

Explanation for the incorrect options:

a) ${\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2=1$.

Compare the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ with the equation ${\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2=1$.

The values of $\mathrm{a}=1,\mathrm{b}=-1,\mathrm{c}=-1,\mathrm{f}=0,\mathrm{g}=0,\mathrm{h}=-\frac{1}{2}$.

Put the values in the general equation of straight lines.

$\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0$

$\Rightarrow$$\left(1\right)\left(-1\right)\left(-1\right)+2\left(0\right)\left(0\right)\left(-\frac{1}{2}\right)-\left(1\right)\left(0\right)-\left(-1\right)\left(0\right)-\left(-1\right){\left(-\frac{1}{2}\right)}^2=0$

$\Rightarrow$ $1+0-0+0+\frac{1}{4}=0$

$\Rightarrow$ $\frac{5}{4}\ne 0$

Since both sides are not equal, the second-degree equation ${\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2=1$, does not represent a pair of straight lines.

b)$–{\mathrm{x}}^2+\mathrm{xy}–{\mathrm{y}}^2=1$

Compare the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ with the equation $–{\mathrm{x}}^2+\mathrm{xy}–{\mathrm{y}}^2=1$.

The values of $\mathrm{a}=-1,\mathrm{b}=-1,\mathrm{c}=-1,\mathrm{f}=0,\mathrm{g}=0,\mathrm{h}=\frac{1}{2}$

Put the values in the general equation of straight lines.

$\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0$

$\Rightarrow$$\left(-1\right)\left(-1\right)\left(-1\right)+2\left(0\right)\left(0\right)\left(\frac{1}{2}\right)-\left(-1\right)\left(0\right)-\left(-1\right)\left(0\right)-\left(-1\right)\left(\frac{1}{2}\right)=0$

$\Rightarrow$ $-1+0+0+0+\frac{1}{2}=0$

$\Rightarrow$ $-\frac{1}{2}\ne 0$

Since both sides are not equal, the second-degree equation $–{\mathrm{x}}^2+\mathrm{xy}–{\mathrm{y}}^2=1$, does not represent a pair of straight lines.

d)${\mathrm{x}}^2+{\mathrm{y}}^2=4$

Compare the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ with the equation ${\mathrm{x}}^2+{\mathrm{y}}^2=4$.

The values of $\mathrm{a}=1,\mathrm{b}=1,\mathrm{c}=-4,\mathrm{f}=0,\mathrm{g}=0,\mathrm{h}=0$
Put the values in the general equation of straight lines.

$\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0$

$\Rightarrow$$\left(1\right)\left(1\right)\left(-4\right)+2\left(0\right)\left(0\right)\left(0\right)-\left(1\right)\left(0\right)-\left(1\right)\left(0\right)-\left(-4\right)\left(0\right)=0$

$\Rightarrow$ $-4+0-0-0+0=0$

$\Rightarrow$ $-4\ne 0$

Since both sides are not equal, the second-degree equation ${\mathrm{x}}^2+{\mathrm{y}}^2=4$, does not represent a pair of straight lines.

Hence, the correct answer is option c) $4{\mathrm{x}}^2-4\mathrm{xy}+{\mathrm{y}}^2=4$.