Question

The equations of the lines perpendicular to \(x^2 - 5xy + 4y^2 = 0\) and passing through (2, 1) is

• A. \( 4x^2+5xy+y^2-13x-1=0 \)
• B. \( 4x^2+5xy+y^2-5x-10y-7=0 \)
• C. \( 4x^2+5xy+y^2-4x-4y-15=0 \)
• D. \( 4x^2+5xy+y^2-21x-12y+27=0 \)

Hint:

Factorize the given pair of lines \(ax^2+2hxy+by^2=0\) into \( (m_1x-y)(m_2x-y)=0 \) or similar to find individual lines \(L_1, L_2\).
Find slopes \(m_1, m_2\). Slopes of perpendicular lines are \(-1/m_1, -1/m_2\).
Write equations of lines \(L_1', L_2'\) with these new slopes passing through the given point \((x_0, y_0)\).
The combined equation is \(L_1' \cdot L_2' = 0\).
A shortcut for pair of lines through origin perpendicular to \(ax^2+2hxy+by^2=0\) is \(bx^2-2hxy+ay^2=0\).

Answer 1

The Correct Option is D

The given equation \(x^2 - 5xy + 4y^2 = 0\) represents a pair of straight lines passing through the origin. We can factorize it: \(x^2 - 4xy - xy + 4y^2 = 0\) \(x(x-4y) - y(x-4y) = 0\) \((x-y)(x-4y) = 0\). So the two lines are \(L_1: x-y=0\) (or \(y=x\)) and \(L_2: x-4y=0\) (or \(y=x/4\)). Slope of \(L_1\) is \(m_1 = 1\). Slope of \(L_2\) is \(m_2 = 1/4\). The lines perpendicular to these and passing through (2,1) are: Line \(L_1'\) perpendicular to \(L_1\): Slope \(m_1' = -1/m_1 = -1/1 = -1\). Equation of \(L_1'\) through (2,1): \(y-1 = -1(x-2) \Rightarrow y-1 = -x+2 \Rightarrow x+y-3=0\). Line \(L_2'\) perpendicular to \(L_2\): Slope \(m_2' = -1/m_2 = -1/(1/4) = -4\). Equation of \(L_2'\) through (2,1): \(y-1 = -4(x-2) \Rightarrow y-1 = -4x+8 \Rightarrow 4x+y-9=0\). The combined equation of these two lines \(L_1'\) and \(L_2'\) is \((x+y-3)(4x+y-9)=0\). Expand this: \(x(4x+y-9) + y(4x+y-9) - 3(4x+y-9) = 0\) \(4x^2 + xy - 9x + 4xy + y^2 - 9y - 12x - 3y + 27 = 0\) Combine terms: \(4x^2 + (xy+4xy) + y^2 + (-9x-12x) + (-9y-3y) + 27 = 0\) \(4x^2 + 5xy + y^2 - 21x - 12y + 27 = 0\). This matches option (d). General method: The equation of pair of lines perpendicular to \(ax^2+2hxy+by^2=0\) and passing through origin is \(bx^2-2hxy+ay^2=0\). Here, \(a=1, 2h=-5 \Rightarrow h=-5/2, b=4\). Perpendicular pair through origin: \(4x^2 - 2(-5/2)xy + 1y^2 = 0 \Rightarrow 4x^2+5xy+y^2=0\). Let these lines be \(y-m_1'x=0\) and \(y-m_2'x=0\). The required lines are \(y-1 = m_1'(x-2)\) and \(y-1 = m_2'(x-2)\). This means the equation is of the form \(4(x-2)^2 + 5(x-2)(y-1) + (y-1)^2 = 0\), but this is for lines parallel to \(4x^2+5xy+y^2=0\) through (2,1). This is not right. The method of finding individual lines and then their combined equation is safer. \[ \boxed{4x^2+5xy+y^2-21x-12y+27=0} \]