Question:
The equations of the pairs of opposite sides of a parallelogram are
\[
x^2-5x+6=0
\]
and
\[
y^2-6y+5=0.
\]
Then the equations of its diagonals are
The equations of the pairs of opposite sides of a parallelogram are
\[
x^2-5x+6=0
\]
and
\[
y^2-6y+5=0.
\]
Then the equations of its diagonals are
Show Hint
When quadratic equations split into pairs of lines, first factorize them completely. The intersection points of those lines give the vertices of the required figure.
Updated On: Jun 17, 2026
- \(4x-y-7=0,\;4x+y=13\)
- \(4x+y-7=0,\;4x-y=13\)
- \(4x-y-7=0,\;x+4y=13\)
- \(4x+y+7=0,\;x-4y=13\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The equations
\[
x^2-5x+6=0
\]
and
\[
y^2-6y+5=0
\]
represent pairs of parallel lines. Their intersections form the vertices of a parallelogram.
The diagonals join opposite vertices.
Step 1: Factorize the equations.
First equation: \[ x^2-5x+6=0 \] gives \[ (x-2)(x-3)=0. \] Hence the pair of lines are \[ x=2,\qquad x=3. \] Second equation: \[ y^2-6y+5=0 \] gives \[ (y-1)(y-5)=0. \] Hence the pair of lines are \[ y=1,\qquad y=5. \]
Step 2: Find vertices of parallelogram.
The vertices are intersections: \[ (2,1),\;(2,5),\;(3,1),\;(3,5). \]
Step 3: Find equations of diagonals.
One diagonal joins \((2,1)\) and \((3,5)\). Slope: \[ m=\frac{5-1}{3-2}=4. \] Equation: \[ y-1=4(x-2). \] Thus, \[ y-1=4x-8 \] which gives \[ 4x-y-7=0. \] Second diagonal joins \((2,5)\) and \((3,1)\). Slope: \[ m=\frac{1-5}{3-2}=-4. \] Equation: \[ y-5=-4(x-2). \] Hence, \[ y-5=-4x+8 \] which simplifies to \[ 4x+y=13. \] Therefore the diagonals are \[ \boxed{4x-y-7=0,\;4x+y=13}. \]
Step 1: Factorize the equations.
First equation: \[ x^2-5x+6=0 \] gives \[ (x-2)(x-3)=0. \] Hence the pair of lines are \[ x=2,\qquad x=3. \] Second equation: \[ y^2-6y+5=0 \] gives \[ (y-1)(y-5)=0. \] Hence the pair of lines are \[ y=1,\qquad y=5. \]
Step 2: Find vertices of parallelogram.
The vertices are intersections: \[ (2,1),\;(2,5),\;(3,1),\;(3,5). \]
Step 3: Find equations of diagonals.
One diagonal joins \((2,1)\) and \((3,5)\). Slope: \[ m=\frac{5-1}{3-2}=4. \] Equation: \[ y-1=4(x-2). \] Thus, \[ y-1=4x-8 \] which gives \[ 4x-y-7=0. \] Second diagonal joins \((2,5)\) and \((3,1)\). Slope: \[ m=\frac{1-5}{3-2}=-4. \] Equation: \[ y-5=-4(x-2). \] Hence, \[ y-5=-4x+8 \] which simplifies to \[ 4x+y=13. \] Therefore the diagonals are \[ \boxed{4x-y-7=0,\;4x+y=13}. \]
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