Question:
The triangle formed by
\[
x^2-4xy+y^2=0
\]
and
\[
x+y+4\sqrt6=0
\]
is:
The triangle formed by
\[
x^2-4xy+y^2=0
\]
and
\[
x+y+4\sqrt6=0
\]
is:
Show Hint
For a homogeneous second-degree equation representing a pair of lines, first find the slopes of the two lines. Geometric symmetry often reveals the nature of the triangle formed with another line.
Updated On: Jun 26, 2026
- an equilateral triangle
- a right-angled triangle
- an isosceles triangle
- a scalene triangle
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The Correct Option is A
Solution and Explanation
Step 1: Factorize the homogeneous equation.
Given \[ x^2-4xy+y^2=0. \] Let \[ y=mx. \] Then \[ 1-4m+m^2=0. \] Solving, \[ m=2\pm\sqrt3. \] Since \[ 2+\sqrt3=\tan75^\circ, \qquad 2-\sqrt3=\tan15^\circ, \] the two lines make angles \[ 75^\circ \quad\text{and}\quad 15^\circ. \] Hence the angle between them is \[ 60^\circ. \]
Step 2: Observe symmetry.
The line \[ x+y+4\sqrt6=0 \] has slope \[ -1, \] and is symmetrically placed with respect to the two lines obtained above. Thus the intersections form a triangle whose two sides are equally inclined.
Step 3: Determine the nature of the triangle.
The angle between the two lines is \[ 60^\circ. \] The third line cuts the two arms symmetrically, making the intercepted sides equal. Therefore all three sides of the triangle are equal.
Step 4: Final conclusion.
Hence the triangle formed is an \[ \boxed{\text{equilateral triangle}}. \]
Given \[ x^2-4xy+y^2=0. \] Let \[ y=mx. \] Then \[ 1-4m+m^2=0. \] Solving, \[ m=2\pm\sqrt3. \] Since \[ 2+\sqrt3=\tan75^\circ, \qquad 2-\sqrt3=\tan15^\circ, \] the two lines make angles \[ 75^\circ \quad\text{and}\quad 15^\circ. \] Hence the angle between them is \[ 60^\circ. \]
Step 2: Observe symmetry.
The line \[ x+y+4\sqrt6=0 \] has slope \[ -1, \] and is symmetrically placed with respect to the two lines obtained above. Thus the intersections form a triangle whose two sides are equally inclined.
Step 3: Determine the nature of the triangle.
The angle between the two lines is \[ 60^\circ. \] The third line cuts the two arms symmetrically, making the intercepted sides equal. Therefore all three sides of the triangle are equal.
Step 4: Final conclusion.
Hence the triangle formed is an \[ \boxed{\text{equilateral triangle}}. \]
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