Question

If two sides of a triangle are represented by \( 3x^2 - 5xy + 2y^2 = 0 \) and its orthocentre is (2,1), then the equation of the third side is

• A. \( 2x+y-4=0 \)
• B. \( 6x+3y-13=0 \)
• C. \( 8x+4y-17=0 \)
• D. \( 10x+5y-21=0 \)

Hint:

1. Factorize the joint equation of lines to find individual side equations. The intersection is a vertex (often origin). 2. An altitude is perpendicular to a side and passes through the opposite vertex AND the orthocentre. 3. Find vertices B and C by intersecting side lines with corresponding altitudes (derived using the orthocentre). 4. Find the equation of the line BC (the third side).

Answer 1

The Correct Option is D

The joint equation \(3x^2 - 5xy + 2y^2 = 0\) can be factorized: \(3x(x-y) - 2y(x-y) = 0 \implies (3x-2y)(x-y) = 0\).
The two sides are \(L_1: 3x-2y=0\) (slope \(m_1=3/2\)) and \(L_2: x-y=0\) (slope \(m_2=1\)).
Vertex A is the intersection of \(L_1, L_2\), which is (0,0).
Orthocentre H = (2,1).
Altitude from vertex B (on \(L_1\)) to side AC (\(L_2\)): Slope of AC (\(L_2\)) is \(m_2=1\).
Slope of altitude BE is \(-1/1 = -1\).
Equation of altitude BE (through H(2,1), slope -1): \(y-1 = -1(x-2) \implies y-1=-x+2 \implies x+y-3=0\).
Vertex B is intersection of \(L_1: 3x-2y=0\) and BE: \(x+y-3=0 \implies y=3-x\).
\(3x-2(3-x)=0 \implies 3x-6+2x=0 \implies 5x=6 \implies x=6/5\).
So \(y=3-6/5=9/5\).
B = (6/5, 9/5).
Altitude from vertex C (on \(L_2\)) to side AB (\(L_1\)): Slope of AB (\(L_1\)) is \(m_1=3/2\).
Slope of altitude CF is \(-1/(3/2) = -2/3\).
Equation of altitude CF (through H(2,1), slope -2/3): \(y-1 = (-2/3)(x-2) \implies 3y-3=-2x+4 \implies 2x+3y-7=0\).
Vertex C is intersection of \(L_2: x-y=0 \implies x=y\) and CF: \(2x+3y-7=0\).
\(2y+3y-7=0 \implies 5y=7 \implies y=7/5\).
So \(x=7/5\).
C = (7/5, 7/5).
Equation of the third side BC, through B(6/5, 9/5) and C(7/5, 7/5): Slope \(m_{BC} = \frac{9/5 - 7/5}{6/5 - 7/5} = \frac{2/5}{-1/5} = -2\).
Equation: \(y - 7/5 = -2(x - 7/5)\) (using point C) \(y - 7/5 = -2x + 14/5\) Multiply by 5: \(5y - 7 = -10x + 14\) \[10x + 5y - 7 - 14 = 0 \implies 10x + 5y - 21 = 0\] This matches option (4).