Question Statement

Find the joint equation of lines through the origin and perpendicular to the lines:

(i) $2x^2-7xy+6y^2=0$

(ii) $x^2+17xy+60y^2=0$

(iii) $3x^2-13xy-10y^2=0$

(iv) $x^2+3xy-28y^2=0$

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Background and Explanation

A homogeneous equation of degree 2 (of the form $a{x}^2+2hxy+b{y}^2=0$) represents a pair of straight lines passing through the origin. To find lines perpendicular to these through the origin, we determine the individual lines, find their slopes, calculate the negative reciprocal slopes for perpendicularity, and then combine the new line equations into a single joint equation.

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Solution

Part (i): For $2x^2-7xy+6y^2=0$

Step 1: Factorize the given equation to find the individual lines.

$\begin{array}{rl}2x^2-7xy+6y^2& =0\\ 2x^2-4xy-3xy+6y^2& =0\\ 2x(x-2y)-3y(x-2y)& =0\\ (2x-3y)(x-2y)& =0\end{array}$

Thus, the two lines are: $2x-3y=0\text{and}x-2y=0$

Step 2: Find the slopes of the given lines.

For $2x-3y=0$: $\text{Slope }{m}_1=\frac{2}{3}$

For $x-2y=0$: $\text{Slope }{m}_2=\frac{1}{2}$

Step 3: Determine slopes of perpendicular lines.

The slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$.

Perpendicular to first line: $m_1^{\prime }=-\frac{1}{2/3}=-\frac{3}{2}$

Perpendicular to second line: $m_2^{\prime }=-\frac{1}{1/2}=-2$

Step 4: Find equations of perpendicular lines through the origin.

Using point-slope form $y-y_1=m(x-x_1)$ with $(0,0)$:

For slope $-\frac{3}{2}$: \begin{align*} y-0 &= -\frac{3}{2}(x-0) \\ 2y &= -3x \\ 3x+2y &= 0 \end{align*}

For slope $-2$: \begin{align*} y-0 &= -2(x-0) \\ y &= -2x \\ 2x+y &= 0 \end{align*}

Step 5: Form the joint equation.

Multiply equations (i) and (ii): $\begin{array}{rl}(3x+2y)(2x+y)& =0\\ 6x^2+3xy+4xy+2y^2& =0\\ 6x^2+7xy+2y^2& =0\end{array}$

Required joint equation: $6x^2+7xy+2y^2=0$

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Part (ii): For $x^2+17xy+60y^2=0$

Step 1: Factorize the given equation.

$\begin{array}{rl}{x}^2+17xy+60y^2& =0\\ x^2+12xy+5xy+60y^2& =0\\ x(x+12y)+5y(x+12y)& =0\\ (x+5y)(x+12y)& =0\end{array}$

The lines are: $x+5y=0\text{and}x+12y=0$

Step 2: Find slopes and their perpendicular counterparts.

For $x+5y=0$: Slope $=-\frac{1}{5}$, so perpendicular slope $=5$

For $x+12y=0$: Slope $=-\frac{1}{12}$, so perpendicular slope $=12$

Step 3: Find equations of perpendicular lines through origin.

For slope $5$: \begin{align*} y-0 &= 5(x-0) \\ y &= 5x \\ 5x-y &= 0 \end{align*}

For slope $12$: \begin{align*} y-0 &= 12(x-0) \\ y &= 12x \\ 12x-y &= 0 \end{align*}

Step 4: Form the joint equation.

$\begin{array}{rl}(5x-y)(12x-y)& =0\\ 60x^2-5xy-12xy+y^2& =0\\ 60x^2-17xy+y^2& =0\end{array}$

Required joint equation: $60x^2-17xy+y^2=0$

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Part (iii): For $3x^2-13xy-10y^2=0$

Step 1: Find the individual lines using the quadratic formula.

Treating the equation as quadratic in $x$: $3x^2-(13y)x-10y^2=0$

Using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $\begin{array}{rl}x& =\frac{13y\pm \sqrt{(-13y{)}^2-4(3)(-10y^2)}}{2(3)}\\ & =\frac{13y\pm \sqrt{169y^2+120y^2}}{6}\\ & =\frac{13y\pm \sqrt{289y^2}}{6}\\ & =\frac{13y\pm 17y}{6}\end{array}$

This gives two solutions: $x=\frac{30y}{6}=5y\text{and}x=\frac{-4y}{6}=-\frac{2}{3}y$

Thus, the lines are: $x-5y=0\text{and}3x+2y=0$

Step 2: Find perpendicular slopes.

For $x-5y=0$ (slope $\frac{1}{5}$): Perpendicular slope $=-5$

For $3x+2y=0$ (slope $-\frac{3}{2}$): Perpendicular slope $=\frac{2}{3}$

Step 3: Find equations of perpendicular lines through origin.

For slope $-5$: \begin{align*} y-0 &= -5(x-0) \\ y &= -5x \\ 5x+y &= 0 \end{align*}

For slope $\frac{2}{3}$: \begin{align*} y-0 &= \frac{2}{3}(x-0) \\ 3y &= 2x \\ 2x-3y &= 0 \end{align*}

Step 4: Form the joint equation.

$\begin{array}{rl}(5x+y)(2x-3y)& =0\\ 10x^2-15xy+2xy-3y^2& =0\\ 10x^2-13xy-3y^2& =0\end{array}$

Required joint equation: $10x^2-13xy-3y^2=0$

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Part (iv): For $x^2+3xy-28y^2=0$

Step 1: Factorize the given equation.

$\begin{array}{rl}{x}^2+3xy-28y^2& =0\\ x^2+7xy-4xy-28y^2& =0\\ x(x+7y)-4y(x+7y)& =0\\ (x-4y)(x+7y)& =0\end{array}$

The lines are: $x-4y=0\text{and}x+7y=0$

Step 2: Find perpendicular slopes.

For $x-4y=0$ (slope $\frac{1}{4}$): Perpendicular slope $=-4$

For $x+7y=0$ (slope $-\frac{1}{7}$): Perpendicular slope $=7$

Step 3: Find equations of perpendicular lines through origin.

For slope $-4$: \begin{align*} y-0 &= -4(x-0) \\ y &= -4x \\ 4x+y &= 0 \end{align*}

For slope $7$: \begin{align*} y-0 &= 7(x-0) \\ y &= 7x \\ 7x-y &= 0 \end{align*}

Step 4: Form the joint equation.

$\begin{array}{rl}(4x+y)(7x-y)& =0\\ 28x^2-4xy+7xy-y^2& =0\\ 28x^2+3xy-y^2& =0\end{array}$

Required joint equation: $28x^2+3xy-y^2=0$

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Key Formulas or Methods Used

• Factorization of homogeneous equations: Splitting the middle term to factorize $a{x}^2+2hxy+b{y}^2=0$ into two linear factors
• Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (used in part iii when factorization was difficult)
• Slope from line equation: For $ax+by+c=0$, slope $m=-\frac{a}{b}$
• Perpendicular slope condition: If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1\cdot m_2=-1$, or $m_2=-\frac{1}{m_1}$
• Point-slope form: $y-y_1=m(x-x_1)$ for line through $(x_1,y_1)$ with slope $m$
• Joint equation formation: If individual lines are $L_1=0$ and $L_2=0$, their joint equation is $L_1\cdot L_2=0$

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Summary of Steps

1. Factorize the given homogeneous equation to find the two individual lines $L_1=0$ and $L_2=0$ (or use quadratic formula if needed)
2. Calculate slopes of both lines using $m=-\frac{\text{coefficient of }x}{\text{coefficient of }y}$
3. Find perpendicular slopes using $m_{\perp }=-\frac{1}{m}$ for each line
4. Write new line equations through the origin $(0,0)$ using the perpendicular slopes and point-slope form $y=mx$
5. Multiply the two new line equations together and expand to get the required joint equation in the form $a{x}^2+2hxy+b{y}^2=0$