Question

The line 3x + 2y = 6, will divide the quadrilateral formed by the line x + y = 5, y – 2x = 8, 3y + 2x = 0 and 4y – x = 0 in -

• A. two quadrilateral
• B. two triangle
• C. one pentagon and one triangle
• D. never cut the sides of quadrilateral

Answer 1

Answer: (A)

two quadrilateral

Answer 2

To determine how the line $L: 3x + 2y = 6$ divides the quadrilateral, we first need to find the vertices of the quadrilateral formed by the given lines:

$L_1: x + y = 5$ $L_2: y - 2x = 8 \implies 2x - y = -8$ $L_3: 3y + 2x = 0 \implies 2x + 3y = 0$ $L_4: 4y - x = 0 \implies x = 4y$

1. Find the vertices of the quadrilateral:

   • Vertex A ($L_1 \cap L_2$): $x + y = 5$ $y - 2x = 8$

     Substitute $y = 5 - x$ into the second equation: $(5 - x) - 2x = 8 \implies 5 - 3x = 8 \implies -3x = 3 \implies x = -1$. Then $y = 5 - (-1) = 6$. So, $A = (-1, 6)$.

   • Vertex B ($L_2 \cap L_3$): $y - 2x = 8$ $3y + 2x = 0$

     Add the two equations: $(y - 2x) + (3y + 2x) = 8 + 0 \implies 4y = 8 \implies y = 2$. Substitute $y = 2$ into $3y + 2x = 0$: $3(2) + 2x = 0 \implies 6 + 2x = 0 \implies 2x = -6 \implies x = -3$. So, $B = (-3, 2)$.

   • Vertex C ($L_3 \cap L_4$): $3y + 2x = 0$ $x = 4y$

     Substitute $x = 4y$ into the first equation: $3y + 2(4y) = 0 \implies 3y + 8y = 0 \implies 11y = 0 \implies y = 0$. Then $x = 4(0) = 0$. So, $C = (0, 0)$.

   • Vertex D ($L_4 \cap L_1$): $x = 4y$ $x + y = 5$

     Substitute $x = 4y$ into the second equation: $4y + y = 5 \implies 5y = 5 \implies y = 1$. Then $x = 4(1) = 4$. So, $D = (4, 1)$.

   The vertices of the quadrilateral are $A(-1, 6)$, $B(-3, 2)$, $C(0, 0)$, and $D(4, 1)$.

2. Determine which sides the line $L: 3x + 2y - 6 = 0$ intersects.

   Let $f(x, y) = 3x + 2y - 6$. A line segment connecting two points $(x_1, y_1)$ and $(x_2, y_2)$ is intersected by the line $f(x, y) = 0$ if $f(x_1, y_1)$ and $f(x_2, y_2)$ have opposite signs.

   Evaluate $f(x, y)$ at each vertex:

   • $f(A) = f(-1, 6) = 3(-1) + 2(6) - 6 = -3 + 12 - 6 = 3$ (positive)
   • $f(B) = f(-3, 2) = 3(-3) + 2(2) - 6 = -9 + 4 - 6 = -11$ (negative)
   • $f(C) = f(0, 0) = 3(0) + 2(0) - 6 = -6$ (negative)
   • $f(D) = f(4, 1) = 3(4) + 2(1) - 6 = 12 + 2 - 6 = 8$ (positive)

   Now, check the signs for each side:

   • Side AB (A to B): $f(A) = 3$ (positive), $f(B) = -11$ (negative). Since the signs are opposite, line $L$ intersects side AB.
   • Side BC (B to C): $f(B) = -11$ (negative), $f(C) = -6$ (negative). Since the signs are the same, line $L$ does not intersect side BC.
   • Side CD (C to D): $f(C) = -6$ (negative), $f(D) = 8$ (positive). Since the signs are opposite, line $L$ intersects side CD.
   • Side DA (D to A): $f(D) = 8$ (positive), $f(A) = 3$ (positive). Since the signs are the same, line $L$ does not intersect side DA.

3. Conclusion:

   The line $3x + 2y = 6$ intersects two opposite sides of the quadrilateral (AB and CD). When a line intersects two opposite sides of a convex quadrilateral, it divides the quadrilateral into two smaller quadrilaterals.