CBSE Class 11 Mathematics Straight Lines Competitive Exam MCQs Set 02

Answer: (b) 8

Explanation:
1. The lines are parallel with the same coefficients 5 and −12.
2. Distance = \( \frac{|65 - (-39)|}{\sqrt{5^2 + 12^2}} = \frac{104}{13} \).
3. = 8 units.

Teacher's Note:
1. Make sure the x and y coefficients are identical in both equations before using the formula.
2. A distance can never be negative, so −2 is ruled out at once.

Answer: (b) 2

Explanation:
1. Parallel lines: x + 3y = k, with intercepts k and k/3.
2. Length between the axes = \( \sqrt{k^2 + \frac{k^2}{9}} = \frac{|k|\sqrt{10}}{3} = 10 \), so |k| = 3√10.
3. k = ±3√10 gives two lines.

Teacher's Note:
1. The intercept between the axes is the length of the segment joining the two axis points.
2. Both signs of k give genuine lines, one on each side of the origin.

Answer: (a) rhombus

Explanation:
1. Two pairs of parallel lines form a parallelogram.
2. Distance between 8x + 4y = 1 and 8x + 4y = 5 is \( \frac{4}{\sqrt{80}} \); between 4x + 8y = 3 and 4x + 8y = 7 it is also \( \frac{4}{\sqrt{80}} \).
3. Equal distances between opposite sides mean equal sides; slopes −2 and −1/2 have product 1 ≠ −1, so it is not a square. Hence a rhombus.

Teacher's Note:
1. A parallelogram whose two pairs of opposite sides are equally far apart is a rhombus.
2. It becomes a square only if the two directions are also perpendicular.

Answer: (a) 4

Explanation:
1. Vertices: y = x and y = 2x meet at (0, 0); y = x and y = 3x + 4 meet at (−2, −2); y = 2x and y = 3x + 4 meet at (−4, −8).
2. Area = \( \frac{1}{2} |0 + (-2)(-8 - 0) + (-4)(0 + 2)| = \frac{1}{2} |16 - 8| \).
3. = 4 sq units.

Teacher's Note:
1. Find the three vertices by solving the lines in pairs.
2. When one vertex is the origin, area = ½|x₂y₃ − x₃y₂|.

Answer: (b) (7, −2)

Explanation:
1. B lies on its median x + y = 5, so B = (b, 5 − b).
2. The mid-point of AB, \( \left( \frac{1 + b}{2}, \frac{7 - b}{2} \right) \), lies on the median through C, x = 4.
3. (1 + b)/2 = 4 gives b = 7, so B = (7, −2).

Teacher's Note:
1. The median from C passes through the mid-point of AB; use this to link B and C.
2. Also, each vertex lies on its own median.

Answer: (c) \( x + 5y = \pm 5\sqrt{2} \)

Explanation:
1. Perpendicular to 5x − y = 1: x + 5y = c, with intercepts c and c/5.
2. Area = \( \frac{1}{2} \cdot |c| \cdot \frac{|c|}{5} = \frac{c^2}{10} = 5 \), so c² = 50.
3. c = ±5√2, giving x + 5y = ±5√2.

Teacher's Note:
1. Write the perpendicular line with an unknown constant, then use the area condition.
2. Both signs of c are allowed because area uses absolute values.

Answer: (b) y = x

Explanation:
1. The required line is the perpendicular bisector of the segment joining the two points.
2. Mid-point = (0, 0); slope of the segment = \( \frac{-1 - 1}{1 + 1} = -1 \), so the bisector has slope 1.
3. Line through (0, 0) with slope 1: y = x.

Teacher's Note:
1. Two points are symmetric about a line exactly when that line is the perpendicular bisector of their join.
2. Check: (−1, 1) and (1, −1) swap into each other when x and y are interchanged, which is reflection in y = x.

Answer: (a) \( 2x + 2y + \sqrt{2} = 0 \)

Explanation:
1. Slope of BC = \( \frac{-3 + 1}{-1 + 3} = -1 \), so the line is x + y + k = 0.
2. Distance from O: \( \frac{|k|}{\sqrt{2}} = \frac{1}{2} \), so \( k = \pm\frac{1}{\sqrt{2}} \).
3. To cut OB and OC the line must lie between O and BC, where x + y < 0, so \( k = +\frac{1}{\sqrt{2}} \): \( x + y + \frac{1}{\sqrt{2}} = 0 \), i.e. \( 2x + 2y + \sqrt{2} = 0 \).

Teacher's Note:
1. A line parallel to BC has the same x and y coefficients as BC; only the constant changes.
2. Of the two lines at the given distance, choose the one on the same side of O as the triangle.

Answer: (c) \( \frac{a^2}{5} \)

Explanation:
1. The triangle is right-angled at O and isosceles, so the altitude from O to the hypotenuse 2x + y = a equals half the hypotenuse.
2. Altitude p = \( \frac{|a|}{\sqrt{5}} \); hypotenuse = 2p.
3. Area = \( \frac{1}{2} \times 2p \times p = p^2 = \frac{a^2}{5} \).

Teacher's Note:
1. In an isosceles right triangle the altitude to the hypotenuse is half the hypotenuse.
2. So the area is simply the square of the perpendicular distance from the right-angle vertex.

Answer: (a) 8

Explanation:
1. |x| + |y| = 2 gives the four lines ±x ± y = 2, forming a square with vertices (±2, 0), (0, ±2).
2. Its diagonals are both of length 4.
3. Area = \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4 \times 4 = 8 \).

Teacher's Note:
1. Open the modulus in each quadrant to get four straight lines.
2. For a square or rhombus, area = half the product of the diagonals.

Answer: (c) (3, 3)

Explanation:
1. The line meets the axes at A(4, 0) and B(0, 2), so AB = (−4, 2).
2. A perpendicular of the same length away from the origin is (2, 4); the other vertices are A + (2, 4) = (6, 4) and B + (2, 4) = (2, 6).
3. The diagonals meet at the mid-point of A(4, 0) and (2, 6), i.e. (3, 3).

Teacher's Note:
1. Turning the side AB = (−4, 2) through 90° gives (2, 4) or (−2, −4); choose the one pointing away from O.
2. The diagonals of a square bisect each other, so their meeting point is the mid-point of any diagonal.

Answer: (b) π/4

Explanation:
1. The line meets the axes at A(−5, 0) and B(0, −10). Points dividing AB as 1 : 2 : 2 are at 1/5 and 3/5 of AB: P(−4, −2) and Q(−2, −6).
2. Slope of MP = \( \frac{-2 + 8}{-4 + 6} = 3 \); slope of MQ = \( \frac{-6 + 8}{-2 + 6} = \frac{1}{2} \).
3. \( \tan \theta = \left| \frac{3 - \frac{1}{2}}{1 + \frac{3}{2}} \right| = 1 \), so θ = π/4.

Teacher's Note:
1. The ratio 1 : 2 : 2 splits the segment into 5 equal parts; the division points are at 1/5 and 3/5 of the way.
2. Then the angle formula with the two slopes gives the answer.

Answer: (a) 3x − 4y + 11 = 0, 3x − 4y − 9 = 0, 4x + 3y + 3 = 0, 4x + 3y − 17 = 0

Explanation:
1. Each side is at distance 2 from the centre (1, 1).
2. Sides 3x − 4y + k = 0: \( \frac{|3 - 4 + k|}{5} = 2 \) gives k − 1 = ±10, so k = 11 or −9.
3. Sides 4x + 3y + k = 0: \( \frac{|4 + 3 + k|}{5} = 2 \) gives k = 3 or −17.

Teacher's Note:
1. The centre of a square is at distance half the side from every side.
2. The other pair of sides is perpendicular, so it has the form 4x + 3y + k = 0.

Answer: (c) \( \frac{4\sqrt{7}}{\sqrt{3}} \)

Explanation:
1. Take A at the origin and the lines as y = 4 (through B) and y = −2 (through C). Let AB make angle θ with the x-axis; AC makes θ − 60°.
2. s sin θ = 4 and s sin(θ − 60°) = −2, which give \( \frac{2}{s} - \frac{\sqrt{3}}{2}\cos\theta = -\frac{2}{s} \), so \( \cos\theta = \frac{8}{\sqrt{3}s} \).
3. sin²θ + cos²θ = 1: \( \frac{16}{s^2} + \frac{64}{3s^2} = 1 \), so s² = 112/3 and \( s = \frac{4\sqrt{7}}{\sqrt{3}} \).

Teacher's Note:
1. Place A at the origin with the parallel lines horizontal; the distances 4 and 2 become y-coordinates of B and C.
2. Writing B and C as s(cos θ, sin θ) and s(cos(θ − 60°), sin(θ − 60°)) uses the 60° angle of the equilateral triangle.

Answer: (a) 3x + 2y ≥ 0; (b) 2x + y + 1 ≥ 0; (d) −2x + 11 ≥ 0

Explanation:
1. A linear expression is largest and smallest over a triangle at its vertices, so test the vertices (1, 3), (5, 6), (−1, 2).
2. 3x + 2y: 9, 27, 1 ✓; 2x + y + 1: 6, 17, 1 ✓; −2x + 11: 9, 1, 13 ✓.
3. 2x + 3y − 12 at (1, 3) is 2 + 9 − 12 = −1 < 0, so (c) fails.

Teacher's Note:
1. If all three vertices satisfy a linear inequality, the whole triangle (inside and on the sides) satisfies it.
2. A single vertex that fails is enough to reject an option.

Answer: (a) x − y − 1 = 0; (b) x − y − 3 = 0

Explanation:
1. A parallel line is x − y + c = 0; its constant divides the constants −7 and 3 in the ratio 3 : 2 (or 2 : 3).
2. From x − y − 7 = 0 side: c = \( \frac{3(3) + 2(-7)}{5} = -1 \), giving x − y − 1 = 0.
3. From the other side: c = \( \frac{2(3) + 3(-7)}{5} = -3 \), giving x − y − 3 = 0.

Teacher's Note:
1. For parallel lines with identical x, y coefficients, distances are proportional to differences in the constants.
2. The ratio can be measured from either line, so two answers are possible.

Answer: (a) \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} + \frac{1}{d^2} \); (c) \( \frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d} \)

Explanation:
1. By symmetry the two lines meet where x = y: \( x\left( \frac{1}{a} + \frac{1}{b} \right) = 1 \), so \( x = y = \frac{ab}{a + b} \).
2. This point lies on \( \frac{x}{c} + \frac{y}{d} = 1 \): \( \frac{ab}{a + b}\left( \frac{1}{c} + \frac{1}{d} \right) = 1 \), i.e. \( \frac{1}{c} + \frac{1}{d} = \frac{1}{a} + \frac{1}{b} \).
3. Perpendicular from O on \( \frac{x}{a} + \frac{y}{b} = 1 \) is \( \frac{1}{\sqrt{1/a^2 + 1/b^2}} \); equal perpendiculars give \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} + \frac{1}{d^2} \).

Teacher's Note:
1. Interchanging x and y swaps the two given lines, so their intersection lies on y = x.
2. For the intercept form x/a + y/b = 1 the distance from the origin is 1/√(1/a² + 1/b²).

Answer: (a) 9x − 7y = 1; (c) 7x + 9y = 73

Explanation:
1. Slopes of the given lines are 3/4 and 12/5. For slope m equally inclined: \( \frac{\frac{3}{4} - m}{1 + \frac{3}{4}m} = -\frac{\frac{12}{5} - m}{1 + \frac{12}{5}m} \).
2. This gives 63m² − 32m − 63 = 0, i.e. (7m − 9)(9m + 7) = 0, so m = 9/7 or −7/9.
3. Through (4, 5): 7y − 35 = 9x − 36, i.e. 9x − 7y = 1; and 9y − 45 = −7x + 28, i.e. 7x + 9y = 73.

Teacher's Note:
1. A line equally inclined to two lines makes angles of equal size with them, so the two tan formulas are equal in magnitude.
2. The two answers are perpendicular to each other (slopes 9/7 and −7/9).

Answer: (d) A is false, but R is true.

Explanation:
1. aa′ + bb′ = 0 only makes adjacent sides perpendicular, so the parallelogram is a rectangle.
2. The diagonals of a rectangle are perpendicular only when it is a square, i.e. when the distances between opposite sides are equal, which is not implied. So the Assertion is false.
3. The diagonals of a rhombus are always perpendicular, so the Reason is true.

Teacher's Note:
1. Perpendicular diagonals need equal sides (a rhombus), not perpendicular sides.
2. Here the two gaps are |c − c′|/√(a² + b²) and |c − c′|/√(a′² + b′²); they are equal only if a² + b² = a′² + b′².

Answer: (c) \( \left( -1, \frac{2}{3} \right) \cup \left( \frac{2}{3}, 2 \right) \)

Explanation:
1. Angle A is obtuse when BC² > AB² + AC²: 10 > (4 + β²) + (1 + (1 − β)²).
2. This simplifies to β² − β − 2 < 0, i.e. (β − 2)(β + 1) < 0, so −1 < β < 2.
3. A must not lie on BC (y = (x + 2)/3 gives β = 2/3 at x = 0), so β ∈ (−1, 2/3) ∪ (2/3, 2).

Teacher's Note:
1. By the converse of Pythagoras, the angle opposite the longest side is obtuse when its square exceeds the sum of the other two squares.
2. Always remove the value of β that makes the three points collinear.

Answer: (b) I₁ is a subset of I₂

Explanation:
1. I₁ = (−1, 2/3) ∪ (2/3, 2) (angle A obtuse).
2. A is the largest angle when BC² = 10 exceeds AB² = 4 + β² and AC² = 1 + (1 − β)²: β² < 6 and −2 < β < 4, so I₂ = (−2, 2/3) ∪ (2/3, √6).
3. Every β in I₁ is in I₂, so I₁ is a subset of I₂.

Teacher's Note:
1. An obtuse angle is always the largest angle of a triangle, so I₁ must lie inside I₂.
2. The largest angle is opposite the longest side, which gives the conditions for I₂.

Answer: (c) \( \left( -2, \frac{2}{3} \right) \cup \left( \frac{2}{3}, \sqrt{6} \right) \)

Explanation:
1. Angle A is largest when BC is the longest side: BC² = 10.
2. 10 > AB² = 4 + β² gives −√6 < β < √6; 10 > AC² = 1 + (1 − β)² gives −2 < β < 4.
3. Together −2 < β < √6, excluding β = 2/3 (collinear): (−2, 2/3) ∪ (2/3, √6).

Teacher's Note:
1. In a triangle the largest angle is opposite the largest side.
2. Use the distance formula for all three sides and compare squares to avoid roots.

Answer: 7

Explanation:
1. Line BC: slope \( \frac{-2 - 5}{4 + 3} = -1 \), so BC is x + y − 2 = 0 and BC = 7√2.
2. Area PBC = \( \frac{1}{2} \times 7\sqrt{2} \times \frac{|x + y - 2|}{\sqrt{2}} = \frac{7}{2}|x + y - 2| \); area ABC = \( \frac{7}{2}|6 + 3 - 2| = \frac{49}{2} \).
3. Ratio = \( \frac{|x + y - 2|}{7} \), so λ = 7.

Teacher's Note:
1. Two triangles on the same base BC have areas in the ratio of their heights, i.e. their distances from line BC.
2. So the ratio is |value of BC's equation at P| ÷ |value at A|.

Answer: 130

Explanation:
1. Let the side be s; the vertices on the given lines are A = s(4/5, 3/5) and C = s(5/13, 12/13), and the fourth vertex is D = A + C = \( s\left( \frac{77}{65}, \frac{99}{65} \right) \).
2. OD is the longer diagonal: \( s \cdot \frac{11\sqrt{130}}{65} = 12 \), so \( D = \frac{12(7, 9)}{\sqrt{130}} \).
3. The side through D parallel to 3x − 4y = 0: \( 3x - 4y = \frac{12(21 - 36)}{\sqrt{130}} = -\frac{180}{\sqrt{130}} \); similarly \( 12x - 5y = \frac{468}{\sqrt{130}} \). So k = 130.

Teacher's Note:
1. In a rhombus OADC with O at the origin, the fourth vertex is the vector sum of the two adjacent vertices.
2. Unit vectors along the sides, (4/5, 3/5) and (5/13, 12/13), make the side length easy to use.

Answer: (c) x + y + 1 = 0

Explanation:
1. A line with equal intercepts a on both axes is \( \frac{x}{a} + \frac{y}{a} = 1 \), i.e., x + y = a.
2. It passes through (1, −2), so a = 1 + (−2) = −1.
3. Hence the line is x + y = −1, i.e., x + y + 1 = 0.

Teacher's Note:
1. Equal intercepts always give a line of the form x + y = a, so only one constant has to be found.
2. Substitute the given point directly into x + y = a; there is no need to find slopes.

Answer: (c) x + y − 6 = 0

Explanation:
1. Equal intercepts ⇒ the line is x + y = a.
2. Through (2, 4): a = 2 + 4 = 6.
3. Hence x + y − 6 = 0.

Teacher's Note:
1. The intercept form x/a + y/b = 1 with a = b reduces to x + y = a.
2. Check the answer by putting (2, 4) in the chosen equation: 2 + 4 − 6 = 0.

Answer: (b) There will be a set of parallel lines

Explanation:
1. The slope of every such line is m, which is fixed.
2. Changing y₁ only changes the point (x₁, y₁) on the vertical line x = x₁, so the lines are different.
3. Different lines with the same slope are parallel, so we get a set of parallel lines.

Teacher's Note:
1. In point-slope form the slope decides the direction of the line; the point only fixes its position.
2. Lines with equal slopes are parallel, so fixing m always produces a family of parallel lines.

Answer: (b) x + y = 5

Explanation:
1. Slope of y = x is 1, so the slope of the perpendicular line is −1.
2. Equation: y − 2 = −1(x − 3) ⇒ y − 2 = −x + 3.
3. Hence x + y = 5.

Teacher's Note:
1. For perpendicular lines m₁m₂ = −1, so the perpendicular to a line of slope 1 has slope −1.
2. Always verify by substituting the given point: 3 + 2 = 5.

Answer: (d) \( \frac{x}{b} + \frac{y}{a} = \frac{a}{b} \)

Explanation:
1. The line cuts the x-axis at (a, 0). Its slope is \( \frac{b}{a} \), so the perpendicular slope is \( -\frac{a}{b} \).
2. Equation: \( y = -\frac{a}{b}(x - a) \Rightarrow by = -ax + a^2 \Rightarrow ax + by = a^2 \).
3. Dividing by ab: \( \frac{x}{b} + \frac{y}{a} = \frac{a}{b} \).

Teacher's Note:
1. Put y = 0 in the intercept form to get the x-intercept point directly.
2. Writing the line as bx − ay = ab shows its slope b/a quickly.

Answer: (b) y − x − 1 = 0

Explanation:
1. Slope of x + y + 1 = 0 is −1, so the perpendicular slope is 1.
2. Equation: y − 2 = 1(x − 1) ⇒ y − x − 1 = 0.

Teacher's Note:
1. A line perpendicular to ax + by + c = 0 can be written directly as bx − ay + λ = 0.
2. Here that gives x − y + λ = 0 and (1, 2) gives λ = 1, the same line.

Answer: (a) 3x + 4y + 3 = 0

Explanation:
1. From 2x + y = 5, y = 5 − 2x. Then x + 3(5 − 2x) + 8 = 0 ⇒ x = \( \frac{23}{5} \), y = \( -\frac{21}{5} \).
2. A line parallel to 3x + 4y = 7 is 3x + 4y = λ.
3. λ = \( 3 \cdot \frac{23}{5} + 4 \cdot \left(-\frac{21}{5}\right) = \frac{69 - 84}{5} = -3 \), so 3x + 4y + 3 = 0.

Teacher's Note:
1. A parallel line keeps the same x and y coefficients; only the constant changes.
2. Substituting the intersection point straight into 3x + 4y saves finding the slope.

Answer: (c) 3x + 4y − 5 = 0

Explanation:
1. Subtracting the equations: 5y = 1 ⇒ y = \( \frac{1}{5} \), x = \( \frac{7}{5} \).
2. The parallel line is 3x + 4y = λ.
3. λ = \( \frac{21}{5} + \frac{4}{5} = 5 \), so 3x + 4y − 5 = 0.

Teacher's Note:
1. Eliminate one variable by subtracting when the x-coefficients are equal.
2. Lines parallel to 3x + 4y = 0 all have the form 3x + 4y = λ.

Answer: (a) 5x + 4y = 0

Explanation:
1. Slope = \( \frac{5 - 0}{-4 - 0} = -\frac{5}{4} \).
2. Equation: \( y = -\frac{5}{4}x \Rightarrow 5x + 4y = 0 \).

Teacher's Note:
1. A line through the origin has no constant term, i.e., it is of the form y = mx.
2. Check: 5(−4) + 4(5) = 0, so (−4, 5) lies on the line.

Answer: (a) \( \frac{x}{3} - \frac{y}{2} = 1 \)

Explanation:
1. Intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \) with a = 3, b = −2.
2. So \( \frac{x}{3} + \frac{y}{-2} = 1 \), i.e., \( \frac{x}{3} - \frac{y}{2} = 1 \).

Teacher's Note:
1. A negative intercept simply appears with a minus sign in the intercept form.
2. Check: putting y = 0 gives x = 3 and putting x = 0 gives y = −2.

Answer: (d) \( y + 4 = \frac{4}{3}(x - 3) \)

Explanation:
1. Slope of 3x + 4y = 5 is \( -\frac{3}{4} \), so the perpendicular slope is \( \frac{4}{3} \).
2. Equation: \( y + 4 = \frac{4}{3}(x - 3) \).

Teacher's Note:
1. The slopes of perpendicular lines are negative reciprocals of each other.
2. Keep the answer in point-slope form when the options are written that way.

Answer: (c) y − 2 = 3(x − 1)

Explanation:
1. The parallel line has the same slope 3.
2. Point-slope form through (1, 2): y − 2 = 3(x − 1).

Teacher's Note:
1. Parallel lines have equal slopes.
2. In y − y₁ = m(x − x₁) the coordinates enter with minus signs.

Answer: (a) 2(y − 1) = 3(x + 1)

Explanation:
1. Slope of 2x + 3y + 4 = 0 is \( -\frac{2}{3} \), so the perpendicular slope is \( \frac{3}{2} \).
2. Equation: \( y - 1 = \frac{3}{2}(x + 1) \Rightarrow 2(y - 1) = 3(x + 1) \).

Teacher's Note:
1. Slope of ax + by + c = 0 is −a/b.
2. Clear the fraction before comparing with the options.

Answer: (c) a(x − c) + b(y − d) = 0

Explanation:
1. A line parallel to ax + by + c = 0 is ax + by + λ = 0.
2. Through (c, d): λ = −ac − bd.
3. So ax + by − ac − bd = 0, i.e., a(x − c) + b(y − d) = 0.

Teacher's Note:
1. Parallel lines differ only in the constant term.
2. The form a(x − x₁) + b(y − y₁) = 0 is a quick way to write a line through (x₁, y₁) parallel to ax + by + c = 0.

Answer: (c) y = x

Explanation:
1. The lines x = 0 and y = 0 meet at the origin (0, 0).
2. The line through (0, 0) and (2, 2) has slope 1, so y = x.

Teacher's Note:
1. x = 0 is the y-axis and y = 0 is the x-axis; they meet at the origin.
2. A line through the origin and (a, a) is always y = x.

Answer: (b) x = 0

Explanation:
1. The points (a, 0) and (−a, 0) both lie on the x-axis, so the line joining them is the x-axis.
2. The line through the origin perpendicular to the x-axis is the y-axis, x = 0.

Teacher's Note:
1. The line perpendicular to the x-axis is vertical and has equation x = constant.
2. Through the origin that constant must be 0.

Answer: (d) \( a = -\frac{8}{3}, b = 4 \)

Explanation:
1. Intercepts of 2x − 3y + 6 = 0: x-intercept −3, y-intercept 2.
2. Intercepts of ax + by + 8 = 0: \( -\frac{8}{a} \) and \( -\frac{8}{b} \). These must be 3 and −2.
3. \( -\frac{8}{a} = 3 \Rightarrow a = -\frac{8}{3} \); \( -\frac{8}{b} = -2 \Rightarrow b = 4 \).

Teacher's Note:
1. The x-intercept of ax + by + c = 0 is −c/a and the y-intercept is −c/b.
2. Write down both sets of intercepts first, then compare them one at a time.

Answer: (b) 2

Explanation:
1. A line is fixed by two conditions, e.g., two points, or a point and a slope, or two intercepts.
2. Hence two geometrical parameters are needed.

Teacher's Note:
1. Every standard form of a line (y = mx + c, x/a + y/b = 1) contains exactly two constants.
2. One condition alone gives a whole family of lines, not a single line.

Answer: (c) x − y = π(1 − y)

Explanation:
1. From x − 4y + 3 = 0, x = 4y − 3. Then 3(4y − 3) − 2y − 1 = 0 ⇒ y = 1, x = 1. The lines meet at (1, 1).
2. Line through (1, 1) and (π, 0): \( y - 1 = \frac{0 - 1}{\pi - 1}(x - 1) \Rightarrow (\pi - 1)(y - 1) = -(x - 1) \).
3. Expanding: πy − π − y + 1 = −x + 1 ⇒ x − y = π − πy = π(1 − y).

Teacher's Note:
1. π here is only a number, so treat it like any other constant in the two-point form.
2. A quick check: (1, 1) gives 0 = 0 and (π, 0) gives π = π in x − y = π(1 − y).

Answer: (c) bx − ay = 0

Explanation:
1. A line perpendicular to ax + by + c = 0 is bx − ay + λ = 0.
2. Through (a, b): ba − ab + λ = 0 ⇒ λ = 0.
3. Hence bx − ay = 0.

Teacher's Note:
1. Swap the coefficients of x and y and change one sign to get a perpendicular line.
2. The constant is then fixed by the given point.