Question Statement

Q1. Find the lines represented by each of the following joint equations:

(i) $x^2+5xy+6y^2=0$

(ii) $7x^2-2xy-9y^2=0$

(iii) $x^2+6xy=0$

(iv) $x^2-8xy+12y^2=0$

(v) $5x^2+3xy-2y^2=0$

(vi) $x^2-3xy-y^2=0$

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Background and Explanation

A joint equation of the form $a{x}^2+2hxy+b{y}^2=0$ represents two straight lines passing through the origin. To find these individual lines, we factorize the homogeneous quadratic expression into two linear factors. When simple factorization by grouping is not possible, we treat the equation as a quadratic in one variable and apply the quadratic formula.

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Solution

Part (i)

Given the equation:

$x^2+5xy+6y^2=0$

To factorize, we split the middle term $5xy$ into $2xy+3xy$:

$x^2+2xy+3xy+6y^2=0$

Group the terms to extract common binomial factors:

$x(x+2y)+3y(x+2y)=0$

Factor out $(x+2y)$:

$(x+3y)(x+2y)=0$

Setting each factor equal to zero gives the required lines:

$x+3y=0\text{and}x+2y=0$

Part (ii)

Given the equation:

$7x^2-2xy-9y^2=0$

Factorize by grouping. We look for terms that give a common binomial factor:

$7x(x+y)-9y(x+y)=0$

Verify: $7x(x+y)=7x^2+7xy$ and $-9y(x+y)=-9xy-9y^2$, which sum to $7x^2-2xy-9y^2$.

Factor out $(x+y)$:

$(7x-9y)(x+y)=0$

Therefore, the required lines are:

$7x-9y=0\text{and}x+y=0$

Part (iii)

Given the equation:

$x^2+6xy=0$

Factor out the common term $x$:

$x(x+6y)=0$

Setting each factor to zero:

$x=0\text{and}x+6y=0$

These are the required lines (the y-axis and a line through the origin with slope $-\frac{1}{6}$).

Part (iv)

Given the equation:

$x^2-8xy+12y^2=0$

Split $-8xy$ as $-6xy-2xy$ to enable grouping:

$x^2-6xy-2xy+12y^2=0$

Group terms:

$x(x-6y)-2y(x-6y)=0$

Factor out $(x-6y)$:

$(x-2y)(x-6y)=0$

Therefore, the required lines are:

$x-2y=0\text{and}x-6y=0$

Part (v)

Given the equation:

$5x^2+3xy-2y^2=0$

Split $3xy$ as $5xy-2xy$:

$5x^2+5xy-2xy-2y^2=0$

Group terms:

$5x(x+y)-2y(x+y)=0$

Factor out $(x+y)$:

$(x+y)(5x-2y)=0$

Therefore, the required lines are:

$x+y=0\text{and}5x-2y=0$

Part (vi)

Given the equation:

$x^2-3xy-y^2=0$

Since this cannot be factorized easily by rational coefficients (inspection method fails), we treat it as a quadratic equation in $x$:

$x^2-(3y)x+\left(-y^2\right)=0$

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with $a=1$, $b=-3y$, and $c=-y^2$:

$x=\frac{-(-3y)\pm \sqrt{(3y{)}^2-4(1)(-y^2)}}{2(1)}$

Simplify the discriminant:

$x=\frac{3y\pm \sqrt{9y^2+4y^2}}{2}=\frac{3y\pm \sqrt{13y^2}}{2}=\frac{3y\pm \sqrt{13}y}{2}$

Factor out $y$:

$x=\left(\frac{3\pm \sqrt{13}}{2}\right)y$

This yields two separate equations. Multiply both sides by 2 and rearrange:

$2x=(3+\sqrt{13})y\Rightarrow 2x-(3+\sqrt{13})y=0$

$2x=(3-\sqrt{13})y\Rightarrow 2x-(3-\sqrt{13})y=0$

Therefore, the required lines are:

$2x-(3+\sqrt{13})y=0\text{and}2x-(3-\sqrt{13})y=0$

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Key Formulas or Methods Used

• Homogeneous Quadratic Equation: General form $a{x}^2+2hxy+b{y}^2=0$ representing two lines through the origin
• Factorization by Grouping: Splitting the middle term $2hxy$ to create two groups with common binomial factors
• Quadratic Formula: For equation $a{x}^2+bx+c=0$, solutions are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Linear Factor Interpretation: Each linear factor $(lx+my)$ corresponds to a line $lx+my=0$

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Summary of Steps

1. Identify the equation as a homogeneous quadratic (joint equation) in $x$ and $y$
2. Attempt factorization by splitting the middle term and grouping to find common binomial factors
3. If factorizable: Set each linear factor equal to zero to obtain the two lines
4. If not easily factorable: Treat as quadratic in one variable (say $x$), treating the other variable ($y$) as constant, and apply the quadratic formula
5. Simplify the resulting expressions and rearrange into standard line form $lx+my=0$