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4.5 Q-2

Question Statement

Solve the quadratic equation 3x2+7xy+2y2=03x^2 + 7xy + 2y^2 = 0 to determine the lines it represents and the angle between them.


Background and Explanation

This question involves solving a homogeneous quadratic equation in xx and yy. To determine the lines it represents, we rewrite the equation in terms of yx=m\frac{y}{x} = m, which corresponds to the slopes of the lines. We then use the formula for the angle between two lines:
tan⁑θ=∣m1βˆ’m2∣1+m1m2,\tan\theta = \frac{|m_1 - m_2|}{1 + m_1 m_2},
or equivalently for a homogeneous equation:
tan⁑θ=2h2βˆ’aba+b,\tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b},
where aa, hh, and bb are coefficients of x2x^2, xyxy, and y2y^2, respectively.


Solution

Step 1: Rewrite the equation

Rewrite the quadratic equation in terms of yx=m\frac{y}{x} = m:

2(yx)2+7(yx)+3=02\left(\frac{y}{x}\right)^2 + 7\left(\frac{y}{x}\right) + 3 = 0

This becomes:

2m2+7m+3=02m^2 + 7m + 3 = 0

Step 2: Solve for mm

Using the quadratic formula, m=βˆ’BΒ±B2βˆ’4AC2Am = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}:

m=βˆ’7Β±72βˆ’4(2)(3)2(2)m = \frac{-7 \pm \sqrt{7^2 - 4(2)(3)}}{2(2)}

Simplify step by step:

  1. Calculate the discriminant:
49βˆ’24=25=5 \sqrt{49 - 24} = \sqrt{25} = 5
  1. Substitute into the formula:
m=βˆ’7Β±54 m = \frac{-7 \pm 5}{4}
  1. Find the two roots:
m=βˆ’7+54=βˆ’24=βˆ’12 m = \frac{-7 + 5}{4} = -\frac{2}{4} = -\frac{1}{2} m=βˆ’7βˆ’54=βˆ’124=βˆ’3 m = \frac{-7 - 5}{4} = -\frac{12}{4} = -3

Thus, m1=βˆ’12m_1 = -\frac{1}{2} and m2=βˆ’3m_2 = -3.

Step 3: Find the equations of the lines

The lines are given by:

y=m1xandy=m2xy = m_1x \quad \text{and} \quad y = m_2x

Substitute m1m_1 and m2m_2:

  1. Line 1: y=βˆ’12xβ‡’x+2y=0y = -\frac{1}{2}x \quad \Rightarrow \quad x + 2y = 0
  2. Line 2: y=βˆ’3xβ‡’3x+y=0y = -3x \quad \Rightarrow \quad 3x + y = 0

Step 4: Find the angle between the lines

The formula for the angle between two lines in a homogeneous quadratic equation is:

tan⁑θ=2h2βˆ’aba+b\tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b}

Substitute a=3a = 3, h=72h = \frac{7}{2}, and b=2b = 2:

tan⁑θ=2(72)2βˆ’(3)(2)3+2\tan\theta = \frac{2\sqrt{\left(\frac{7}{2}\right)^2 - (3)(2)}}{3 + 2}

Simplify step by step:

  1. Square hh:
(72)2=494 \left(\frac{7}{2}\right)^2 = \frac{49}{4}
  1. Subtract abab:
494βˆ’6=254 \frac{49}{4} - 6 = \frac{25}{4}
  1. Take the square root:
254=52 \sqrt{\frac{25}{4}} = \frac{5}{2}
  1. Multiply by 2 and divide by a+ba + b:
2(52)5=55=1 \frac{2\left(\frac{5}{2}\right)}{5} = \frac{5}{5} = 1

Thus:

tan⁑θ=1β‡’ΞΈ=tanβ‘βˆ’1(1)=45∘\tan\theta = 1 \quad \Rightarrow \quad \theta = \tan^{-1}(1) = 45^\circ

Key Formulas or Methods Used

  1. Quadratic Formula:
    m=βˆ’BΒ±B2βˆ’4AC2Am = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}

  2. Angle Between Two Lines in Homogeneous Form:
    tan⁑θ=2h2βˆ’aba+b\tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b}


Summary of Steps

  1. Rewrite the quadratic equation in terms of m=yxm = \frac{y}{x}.
  2. Solve the quadratic equation for mm using the quadratic formula.
  3. Write the equations of the lines using the roots of mm.
  4. Apply the formula for the angle between lines to calculate ΞΈ\theta.
  5. Conclude that the angle between the lines is 45∘45^\circ.