Question

A pair of straight lines are drawn through the origin form with the line $2x + 3y = 6$ an isosceles triangle right angled at the origin. Find the equation of the pair of straight lines & the area of the triangle correct to two places of decimals.

• A. Equation: $5x^2 - 24xy - 5y^2 = 0$, Area: $2.77$
• B. Equation: $x^2 - 24xy - y^2 = 0$, Area: $3.14$
• C. Equation: $5x^2 + 24xy - 5y^2 = 0$, Area: $2.77$
• D. Equation: $5x^2 - 24xy - 5y^2 = 0$, Area: $3.00$

Answer 1

Answer: (A)

Equation: $5x^2 - 24xy - 5y^2 = 0$, Area: $2.77$

Answer 2

Let the pair of straight lines through the origin be $y = m_1x$ and $y = m_2x$. For the triangle to be right-angled at the origin, the lines must be perpendicular, so $m_1m_2 = -1$. Let $m_1 = m$ and $m_2 = -1/m$. The combined equation of these lines is $(y-mx)(y+x/m) = 0$, which simplifies to $mx^2 - (1-m^2)xy - my^2 = 0$.

The vertices of the triangle are the origin $(0,0)$, and the intersection points of the pair of lines with $2x+3y=6$. Let these intersection points be P and Q. For the triangle to be isosceles, $OP = OQ$.

The intersection of $y=mx$ with $2x+3y=6$ gives $2x+3mx=6$, so $x_P = \frac{6}{2+3m}$ and $y_P = \frac{6m}{2+3m}$. $OP^2 = x_P^2 + y_P^2 = \left(\frac{6}{2+3m}\right)^2 + \left(\frac{6m}{2+3m}\right)^2 = \frac{36(1+m^2)}{(2+3m)^2}$.

The intersection of $y=-x/m$ with $2x+3y=6$ gives $2x - 3x/m = 6$, so $x_Q = \frac{6m}{2m-3}$ and $y_Q = -\frac{6}{2m-3}$. $OQ^2 = x_Q^2 + y_Q^2 = \left(\frac{6m}{2m-3}\right)^2 + \left(\frac{-6}{2m-3}\right)^2 = \frac{36(m^2+1)}{(2m-3)^2}$.

Setting $OP^2 = OQ^2$: $\frac{36(1+m^2)}{(2+3m)^2} = \frac{36(m^2+1)}{(2m-3)^2}$ This implies $(2+3m)^2 = (2m-3)^2$. $4 + 12m + 9m^2 = 4m^2 - 12m + 9$ $5m^2 + 24m - 5 = 0$ Factoring gives $(5m-1)(m+5) = 0$. So, $m = 1/5$ or $m = -5$.

If $m=1/5$, the slopes are $1/5$ and $-5$. If $m=-5$, the slopes are $-5$ and $1/5$. Both yield the same pair of lines: $y = \frac{1}{5}x$ and $y = -5x$. The equations are $x - 5y = 0$ and $5x + y = 0$. The combined equation is $(x-5y)(5x+y) = 0$, which is $5x^2 - 24xy - 5y^2 = 0$.

The area of the triangle formed by the pair of lines $Ax^2 + 2Hxy + By^2 = 0$ and the line $lx+my+n=0$ is given by $\frac{n^2 \sqrt{H^2-AB}}{|Bl^2 - 2Hlm + Am^2|}$. Here, $A=5$, $H=-12$, $B=-5$, $l=2$, $m=3$, $n=-6$. Area = $\frac{(-6)^2 \sqrt{(-12)^2 - (5)(-5)}}{|(-5)(2^2) - 2(-12)(2)(3) + (5)(3^2)|}$ Area = $\frac{36 \sqrt{144 + 25}}{| -20 + 144 + 45 |}$ Area = $\frac{36 \sqrt{169}}{|169|} = \frac{36 \times 13}{169} = \frac{36}{13}$.

As a decimal, $\frac{36}{13} \approx 2.76923...$. Rounded to two decimal places, the area is $2.77$.