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If the lines 3x2 - kxy - 3y2 = 0 and x + 2y - 8 = 0 form an isosceles triangle then k = ?
1. 8
2. 4
3. -4
4. -8

1 Answer

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Correct Answer - Option 1 : 8

Concept:

  • For a pair of lines ax2 + 2hxy + by2 = 0:
    • Sum of the slopes is: \(\rm -\frac{2h}{b}\).
    • Product of the slopes is: \(\rm \frac{a}{b}\).
  • The angle θ between a pair of lines y = mx + c and y = nx + d, is given by: \(\rm \tan \theta =\frac{|m-n|}{1+mn}\).
  • The product of the slopes of a pair of perpendicular lines is -1.
  • The slopes of parallel lines are equal.

 

Calculation:

Comparing the equation 3x2 - kxy - 3y2 = 0 with the general equation of a pair of lines ax2 + 2hxy + by2 = 0, we can say that: a = 3, b = -3 and 2h = -k.

Product of the slopes of the pair of lines = \(\rm \frac{a}{b}=\frac3{-3}\) = -1.

∴ The pair of lines are perpendicular to each other. It means that the angles made by these two lines with the line x + 2y - 8 = 0, must be equal and 45 each.

If the slope of one of the lines is m, the slope of the other line will be \(\rm -\frac{1}m\).

Also, the slope of the line x + 2y - 8 = 0 is \(\rm n=-\frac{1}{2}\).

Using the formula for angle between a pair of lines, we have:

For slopes m and \(\rm n=-\frac{1}{2}\):

\(\rm \tan 45^\circ =\frac{\left|m-\frac12\right|}{1+m\left(\frac12\right)}\)

Using tan 45 = 1, we get:

⇒ \(\rm 1+m\left(\frac12\right)=\left|m-\frac12\right|\)

⇒ 2 + m = |2m - 1|

Squaring both sides, we get:

⇒ 4 + 4m + m2 = 4m2 - 4m + 1

⇒ 3m2 - 8m - 3 = 0

⇒ 3m2 - 9m + m - 3 = 0

⇒ 3m(m - 3) + (m - 3) = 0

⇒ (m - 3)(3m + 1) = 0

⇒ m - 3 = 0 OR 3m + 1 = 0

⇒ m = 3 OR \(\rm m=-\frac{1}{3}\)

For slopes \(\rm -\frac{1}m\) and \(\rm n=-\frac{1}{2}\):

\(\rm \tan 45^\circ =\frac{\left|-\frac{1}m-\frac12\right|}{1+\left(-\frac{1}m\right)\left(\frac12\right)}\)

⇒ \(\rm 1-\frac{1}{2m}=\left|-\frac{1}{m}-\frac12\right|\)

⇒ 2m - 1 = |-2 - m|

⇒ 4m2 - 4m + 1 = 4 + 4m + m2

Which is the same equation as the previous one.

∴ m = 3 OR \(\rm m=-\frac{1}{3}\).

In either case, the slopes of the pair of lines are 3 and \(\rm -\frac{1}{3}\).

Now, using the formula for the sum of the slopes, we get:

\(\rm -\frac{2h}{b}=3+\left(-\frac13\right)\)

⇒ \(\rm \frac{k}{3}=\frac83\)

⇒ k = 8.

 

The angle θ between the pair of lines ax2 + 2hxy + by2 = 0, is given by: \(\rm \tan \theta=\frac{2\sqrt{h^2-ab}}{a+b}\).

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