Straight Lines
Geometrical figures formed by straight lines
Grade 11
Question:
<p><strong>Match the Column</strong></p><p>Match the items in Column I with Column II:</p><table border='1'><tr><th>Column I</th><th>Column II</th></tr><tr><td>(A) The lines \(y = 0\); \(y = 1\); \(x - 6y + 4 = 0\) and \(x + 6y - 9 = 0\) constitute a figure which is</td><td>(p) A cyclic quadrilateral</td></tr><tr><td>(B) The points \(A(a, 0)\), \(B(0, b)\), \(C(c, 0)\) and \(D(0, d)\) are such that \(ac = bd\) and \(a, b, c, d\) are all non-zero. The points \(A\), \(B\), \(C\) and \(D\) always constitute</td><td>(q) A rhombus</td></tr><tr><td>(C) The figure formed by the four lines \(ax \pm by \pm c = 0\) \((a \neq b)\) is</td><td>(r) A square</td></tr><tr><td>(D) The line pairs \(x^2 - 8x + 12 = 0\) and \(y^2 - 14y + 45 = 0\) constitute a figure which is</td><td>(s) A trapezium</td></tr></table>
<p>(a) A→q, B→r, C→p,r, D→s</p>
<p>(b) A→p,s, B→p, C→q, D→p,q,r</p>
<p>(c) A→s, B→r, C→s, D→p,q</p>
<p>(d) A→p,q,r, B→s, C→r, D→p,q</p>
Step-by-Step Solution
Key Concept: For item (B), use the condition ac = bd to show that the diagonals AC and BD bisect each other at the same point, making ABCD a parallelogram. Since A and C lie on x-axis while B and D lie on y-axis, the diagonals are perpendicular, confirming it's always a rhombus.
<p><strong>Step 1 (Item A):</strong> Find intersection points of y = 0, y = 1, x - 6y + 4 = 0, and x + 6y - 9 = 0. The four vertices are (−4, 0), (9, 0), (2, 1), and (3, 1). Check if opposite angles sum to 180° for cyclic property. <strong>Answer: (p) Cyclic quadrilateral</strong></p><p><strong>Step 2 (Item B):</strong> Points A(a, 0), B(0, b), C(c, 0), D(0, d) with condition ac = bd. Midpoint of AC = ((a+c)/2, 0) and midpoint of BD = (0, (b+d)/2). For a parallelogram, diagonals must bisect each other. The condition ac = bd ensures the diagonals AC and BD meet at their midpoints. Since diagonals lie on perpendicular axes (x and y), they are perpendicular. A parallelogram with perpendicular diagonals is a rhombus. <strong>Answer: (q) A rhombus</strong></p><p><strong>Step 3 (Item C):</strong> Four lines ax + by + c = 0, ax − by + c = 0, −ax + by + c = 0, −ax − by + c = 0 form a rectangle with sides parallel to axes. Since a ≠ b, the distances between opposite pairs differ, so it's not a square. <strong>Answer: (s) A trapezium (or rectangle)</strong></p><p><strong>Step 4 (Item D):</strong> Lines x² − 8x + 12 = 0 gives (x−2)(x−6) = 0, so x = 2 and x = 6 (two parallel vertical lines). Lines y² − 14y + 45 = 0 gives (y−5)(y−9) = 0, so y = 5 and y = 9 (two parallel horizontal lines). These form a rectangle. <strong>Answer: (r) A square (or rectangle)</strong></p><p><strong>Match: A→(p), B→(q), C→(s), D→(r)</strong></p>
Correct Answer: B