Question

Consider the lines, $L_1: \frac{x}{3} + \frac{y}{4} = 1; L_2: \frac{x}{4} + \frac{y}{3} = 1; L_3: \frac{x}{3} + \frac{y}{4} = 2$ and $L_4: \frac{x}{4} + \frac{y}{3} = 2$

Statement-1: The quadrilateral formed by these four lines is a rhombus.

because

Statement-2: If diagonals of a quadrilateral formed by any four lines are unequal and intersect at right angle then it is a rhombus.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (A)

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Answer 2

The four lines are given by: $L_1: \frac{x}{3} + \frac{y}{4} = 1 \implies 4x + 3y = 12$ $L_2: \frac{x}{4} + \frac{y}{3} = 1 \implies 3x + 4y = 12$ $L_3: \frac{x}{3} + \frac{y}{4} = 2 \implies 4x + 3y = 24$ $L_4: \frac{x}{4} + \frac{y}{3} = 2 \implies 3x + 4y = 24$

Lines $L_1$ and $L_3$ are parallel since they have the same equation form $4x+3y=k$. Lines $L_2$ and $L_4$ are parallel since they have the same equation form $3x+4y=k$. Thus, the four lines form a parallelogram.

Statement-1: The quadrilateral formed by these four lines is a rhombus. A parallelogram is a rhombus if its diagonals are perpendicular. The parallelogram is formed by the lines $4x+3y=12$, $4x+3y=24$, $3x+4y=12$, and $3x+4y=24$. Consider the general form of lines forming a parallelogram: $ax+by=c_1$, $ax+by=c_2$, $bx+ay=d_1$, $bx+ay=d_2$. Here, $a=4$, $b=3$. So $c_1=12, c_2=24$ and $d_1=12, d_2=24$. The diagonals of such a parallelogram are perpendicular if and only if $(c_2-c_1)^2 = (d_2-d_1)^2$. Let $C = c_2-c_1 = 24-12 = 12$. Let $D = d_2-d_1 = 24-12 = 12$. Since $C=12$ and $D=12$, we have $C^2 = D^2$. This condition implies that the diagonals of the parallelogram are perpendicular. Therefore, the quadrilateral is a rhombus. Statement-1 is True.

Statement-2: If diagonals of a quadrilateral formed by any four lines are unequal and intersect at right angle then it is a rhombus. A parallelogram with perpendicular diagonals is always a rhombus. The condition "diagonals are unequal" means that the rhombus is not a square. This does not negate the fact that it is a rhombus. Thus, Statement-2 is True.

Relationship between Statement-1 and Statement-2: The quadrilateral in Statement-1 is a parallelogram. We've shown its diagonals are perpendicular. Let's check if its diagonals are unequal. The vertices are: A: $4x+3y=12, 3x+4y=12 \implies A(\frac{12}{7}, \frac{12}{7})$ B: $4x+3y=12, 3x+4y=24 \implies B(-\frac{24}{7}, \frac{60}{7})$ C: $4x+3y=24, 3x+4y=24 \implies C(\frac{24}{7}, \frac{24}{7})$ D: $4x+3y=24, 3x+4y=12 \implies D(\frac{60}{7}, -\frac{24}{7})$

Diagonal AC length: $AC = \sqrt{(\frac{24}{7}-\frac{12}{7})^2 + (\frac{24}{7}-\frac{12}{7})^2} = \sqrt{(\frac{12}{7})^2 + (\frac{12}{7})^2} = \sqrt{2 \cdot \frac{144}{49}} = \frac{12\sqrt{2}}{7}$. Diagonal BD length: $BD = \sqrt{(\frac{60}{7}-(-\frac{24}{7}))^2 + (-\frac{24}{7}-\frac{60}{7})^2} = \sqrt{(\frac{84}{7})^2 + (-\frac{84}{7})^2} = \sqrt{12^2 + (-12)^2} = \sqrt{144+144} = \sqrt{288} = 12\sqrt{2}$. The diagonals AC and BD are unequal ($12\sqrt{2} \neq \frac{12\sqrt{2}}{7}$). The diagonals also intersect at right angles (slopes are 1 and -1).

Statement-2 states that IF diagonals are unequal AND intersect at right angles, THEN it is a rhombus. The quadrilateral in Statement-1 has unequal diagonals that intersect at right angles. Therefore, according to Statement-2, it is a rhombus. This means Statement-2 provides the reasoning for Statement-1 being true.