Question

The lines x+y+4=0, x-2y-4=0 and 3x+4y-2=0

• A. are concurrent
• B. form an isosceles triangle
• C. form a right-angled triangle
• D. form a scalene triangle

Hint:

To classify a triangle formed by three lines, first find their slopes. Check products for -1 (right angle). If no right angle, calculate the tangents of the angles between the lines. If any two tangents are equal, the triangle is isosceles. If all are different, it's scalene.

Answer 1

The Correct Option is D

Let the three lines be L1: \(x+y+4=0\), L2: \(x-2y-4=0\), and L3: \(3x+4y-2=0\).
Step 1: Check for concurrence. Find the intersection of L1 and L2.
Subtracting L2 from L1: \((x+y+4) - (x-2y-4) = 0 \implies 3y+8=0 \implies y = -8/3\).
Substituting y back into L1: \(x - 8/3 + 4 = 0 \implies x = 8/3 - 12/3 = -4/3\).
The intersection point is \((-4/3, -8/3)\). Now check if this point lies on L3.
\(3(-4/3) + 4(-8/3) - 2 = -4 - 32/3 - 2 = -6 - 32/3 \neq 0\).
The lines are not concurrent; they form a triangle.
Step 2: Find the slopes to check for right angles or equal angles.
Slope of L1, \(m_1 = -1\).
Slope of L2, \(m_2 = 1/2\).
Slope of L3, \(m_3 = -3/4\).
Since \(m_1 m_2 \neq -1\), \(m_1 m_3 \neq -1\), and \(m_2 m_3 \neq -1\), the triangle is not right-angled.
Step 3: Check the angles between the lines to see if it's isosceles.
Let's find the tangent of the angles between pairs of lines using the formula \( \tan\theta = |\frac{m_a - m_b}{1+m_a m_b}| \).
Angle between L1 and L2: \( \tan\theta_{12} = |\frac{-1 - 1/2}{1+(-1)(1/2)}| = |\frac{-3/2}{1/2}| = 3 \).
Angle between L2 and L3: \( \tan\theta_{23} = |\frac{1/2 - (-3/4)}{1+(1/2)(-3/4)}| = |\frac{5/4}{5/8}| = 2 \).
Angle between L3 and L1: \( \tan\theta_{31} = |\frac{-3/4 - (-1)}{1+(-3/4)(-1)}| = |\frac{1/4}{7/4}| = 1/7 \).
Since all three tangents are different, all three angles of the triangle are different.
Therefore, the triangle is a scalene triangle.