CBSE Class 11 Mathematics Straight Lines Competitive Exam MCQs Set 04

Answer: (c) 9x − 20y + 96 = 0

Explanation:
1. Let the line meet the axes at A(a, 0) and B(0, b). The point dividing AB in the ratio 5 : 3 is \( \left( \frac{3a}{8}, \frac{5b}{8} \right) \).
2. \( \frac{3a}{8} = -4 \Rightarrow a = -\frac{32}{3} \); \( \frac{5b}{8} = 3 \Rightarrow b = \frac{24}{5} \).
3. \( -\frac{3x}{32} + \frac{5y}{24} = 1 \); multiplying by 96: −9x + 20y = 96 ⇒ 9x − 20y + 96 = 0.

Teacher's Note:
1. With A first, the ratio m : n gives the point (nA + mB)/(m + n).
2. Check: 9(−4) − 20(3) + 96 = 0.

Answer: (a) 83x − 35y + 92 = 0

Explanation:
1. Let the segment meet 5x − y − 4 = 0 at P(a, 5a − 4). Since (1, 5) is the mid-point, the other end is Q(2 − a, 14 − 5a).
2. Q lies on 3x + 4y − 4 = 0: 3(2 − a) + 4(14 − 5a) − 4 = 0 ⇒ 58 = 23a ⇒ \( a = \frac{58}{23} \), so \( P = \left( \frac{58}{23}, \frac{198}{23} \right) \).
3. Slope through (1, 5) and P = \( \frac{198/23 - 5}{58/23 - 1} = \frac{83}{35} \).
4. \( y - 5 = \frac{83}{35}(x - 1) \Rightarrow 83x - 35y + 92 = 0 \).

Teacher's Note:
1. If M is the mid-point of PQ, then Q = 2M − P; this links the two unknown end points.
2. Take a general point on one line using one parameter, then force the other end to lie on the second line.

Answer: (c) y + 3x − 8 = 0

Explanation:
1. The other diagonal is the perpendicular bisector of AB. Mid-point = (2, 2); slope of AB = \( \frac{2}{6} = \frac{1}{3} \).
2. Perpendicular slope = −3: y − 2 = −3(x − 2) ⇒ y + 3x − 8 = 0.

Teacher's Note:
1. The diagonals of a square bisect each other at right angles.
2. So the second diagonal is the perpendicular bisector of the first.

Answer: (b) x + y = 5

Explanation:
1. 3a + 2b = 13 and 4b − a = 5 ⇒ a = 4b − 5.
2. 3(4b − 5) + 2b = 13 ⇒ b = 2, a = 3. So P(3, 2) and Q(2, 3).
3. Line PQ: slope −1, y − 2 = −(x − 3) ⇒ x + y = 5.

Teacher's Note:
1. Note that the coordinates of Q are those of P interchanged.
2. P(a, b) and Q(b, a) always have the same sum of coordinates, so PQ is of the form x + y = a + b.

Answer: (b) x − 3y + 4 = 0

Explanation:
1. B = AB ∩ BC: y = x + 2 and x + 2(x + 2) = 1 ⇒ x = −1, y = 1. So B(−1, 1).
2. The altitude through B is perpendicular to CA (slope −3), so its slope is \( \frac{1}{3} \).
3. \( y - 1 = \frac{1}{3}(x + 1) \Rightarrow x - 3y + 4 = 0 \).

Teacher's Note:
1. The altitude from a vertex is perpendicular to the opposite side.
2. Vertex B is the common point of sides AB and BC.

Answer: (a) y(cos α − sin α) = x(cos α + sin α); (c) y(sin α + cos α) − x(sin α − cos α) = a

Explanation:
1. Take O(0, 0), A(a cos α, a sin α), C(−a sin α, a cos α); then B = A + C = (a(cos α − sin α), a(sin α + cos α)).
2. Diagonal OB: y(cos α − sin α) = x(sin α + cos α), which is option (a).
3. Diagonal AC: slope = \( \frac{a\cos\alpha - a\sin\alpha}{-a\sin\alpha - a\cos\alpha} = -\frac{\cos\alpha - \sin\alpha}{\sin\alpha + \cos\alpha} \); through A it gives y(sin α + cos α) + x(cos α − sin α) = a.
4. This is the same as y(sin α + cos α) − x(sin α − cos α) = a, option (c). Hence (a) and (c).

Teacher's Note:
1. Rotating the side OA through 90° about O gives the adjacent vertex C = (−a sin α, a cos α).
2. One diagonal passes through the origin (no constant term) and the other does not, so expect one of each among the answers.

Answer: (a) x − 7y + 13 = 0 and 7x + y = 9

Explanation:
1. The given lines are perpendicular (slopes \( -\frac{3}{4} \) and \( \frac{4}{3} \)), so △ABC is right-angled and isosceles at A; BC makes 45° with each line.
2. \( \left| \frac{m + \frac{3}{4}}{1 - \frac{3m}{4}} \right| = 1 \Rightarrow m = \frac{1}{7} \) or m = −7.
3. Through (1, 2): \( y - 2 = \frac{1}{7}(x - 1) \Rightarrow x - 7y + 13 = 0 \) and y − 2 = −7(x − 1) ⇒ 7x + y = 9.

Teacher's Note:
1. AB = AC with a right angle at A makes the base angles 45°.
2. Always check the angle between the given lines first; perpendicular lines simplify the problem.

Answer: (a) (2pq − 1)(px + qy − 1) = (p² + q² − 1)(qx + py − 1)

Explanation:
1. A is the common point of px + qy = 1 and qx + py = 1; by symmetry \( A = \left( \frac{1}{p + q}, \frac{1}{p + q} \right) \).
2. The median through A passes through A and the mid-point (p, q) of BC.
3. Every equation of the form λ(px + qy − 1) = μ(qx + py − 1) passes through A. Making (p, q) satisfy it: λ(p² + q² − 1) = μ(2pq − 1).
4. Taking λ = 2pq − 1, μ = p² + q² − 1 gives (2pq − 1)(px + qy − 1) = (p² + q² − 1)(qx + py − 1).

Teacher's Note:
1. Any equation built from the two side equations in this way is automatically satisfied at their common point A.
2. You can also check the answer by substituting (p, q): both sides become (2pq − 1)(p² + q² − 1).

Answer: (c) \( x + \sqrt{3}y = 0 \)

Explanation:
1. Take the centre of the clock as the origin, 12 along the positive y-axis and 3 along the positive x-axis.
2. At 4 o'clock the hour hand points at 4, which is 30° below the 3 o'clock direction, i.e., at −30° to the x-axis.
3. Slope = tan(−30°) = \( -\frac{1}{\sqrt{3}} \), so \( y = -\frac{x}{\sqrt{3}} \Rightarrow x + \sqrt{3}y = 0 \).

Teacher's Note:
1. Each hour mark on a clock is 360°/12 = 30° apart.
2. The hand passes through the centre, so the line has no constant term.

Answer: (b) (4, 4), (2, 0) and c = −4

Explanation:
1. The mid-point (3, 2) of the given diagonal lies on the other diagonal y = 2x + c ⇒ 2 = 6 + c ⇒ c = −4.
2. The other vertices are at a distance half the diagonal \( = \sqrt{5} \) from (3, 2): (x − 3)² + (2x − 4 − 2)² = 5 ⇒ 5(x − 3)² = 5 ⇒ x = 4 or 2.
3. So the vertices are (4, 4) and (2, 0), with c = −4.

Teacher's Note:
1. The diagonals of a rectangle are equal and bisect each other.
2. So all four vertices are at the same distance from the centre.

Answer: (b) 60°

Explanation:
1. \( \tan\theta = \left| \frac{(2 + \sqrt{3}) - (2 - \sqrt{3})}{1 + (2 + \sqrt{3})(2 - \sqrt{3})} \right| = \frac{2\sqrt{3}}{1 + 1} = \sqrt{3} \).
2. Hence θ = 60°.

Teacher's Note:
1. (2 + √3)(2 − √3) = 4 − 3 = 1 simplifies the denominator.
2. Also tan 75° = 2 + √3 and tan 15° = 2 − √3, so the angle is 75° − 15° = 60°.

Answer: (a) \( 2\tan^{-1}\frac{b}{a} \)

Explanation:
1. Slopes: \( m_1 = -\frac{b}{a} \), \( m_2 = \frac{b}{a} \).
2. \( \tan\theta = \left| \frac{-\frac{b}{a} - \frac{b}{a}}{1 - \frac{b^2}{a^2}} \right| = \frac{2ab}{|a^2 - b^2|} \).
3. If \( \tan\varphi = \frac{b}{a} \), then \( \tan 2\varphi = \frac{2b/a}{1 - b^2/a^2} = \frac{2ab}{a^2 - b^2} \), so θ = \( 2\tan^{-1}\frac{b}{a} \).

Teacher's Note:
1. The two lines are mirror images in the x-axis, so the angle between them is twice the angle each makes with the x-axis.
2. tan 2φ = 2 tan φ/(1 − tan² φ) converts the answer into the form of option (a).

Answer: (c) 90°

Explanation:
1. Slopes: m₁ = 2 and \( m_2 = -\frac{1}{2} \).
2. m₁m₂ = −1, so the lines are perpendicular: θ = 90°.

Teacher's Note:
1. When the product of slopes is −1, the angle is 90° and no tan formula is needed.
2. In coefficient form, a₁a₂ + b₁b₂ = (−2)(1) + (1)(2) = 0 confirms it.

Answer: (b) 135°

Explanation:
1. y = −2 is parallel to the x-axis; y = x + 2 makes 45° with the x-axis.
2. The acute angle between them is 45°, so the obtuse angle is 180° − 45° = 135°.

Teacher's Note:
1. The angles between two intersecting lines are θ and 180° − θ.
2. A horizontal line makes an angle with any other line equal to that line's inclination (or its supplement).

Answer: (b) 60°

Explanation:
1. y = 3 is horizontal.
2. y = √3x + 9 has slope √3, i.e., inclination 60°. Hence the acute angle is 60°.

Teacher's Note:
1. For a horizontal line, the angle with another line is that line's inclination.
2. tan 60° = √3.

Answer: (b) \( \tan^{-1}(3) \)

Explanation:
1. x − 3y = 6 has slope \( \frac{1}{3} \), so its inclination is \( \tan^{-1}\frac{1}{3} \).
2. x = 2 is vertical (inclination 90°). Angle = \( 90^\circ - \tan^{-1}\frac{1}{3} = \cot^{-1}\frac{1}{3} = \tan^{-1}3 \).

Teacher's Note:
1. The tan formula fails for a vertical line, so work with inclinations instead.
2. 90° − tan⁻¹ x = cot⁻¹ x = tan⁻¹(1/x) for x > 0.

Answer: (b) \( \cot^{-1}\frac{a_1a_2 + b_1b_2}{a_1b_2 - a_2b_1} \)

Explanation:
1. With \( m_1 = -\frac{a_1}{b_1} \), \( m_2 = -\frac{a_2}{b_2} \): \( \tan\theta = \frac{m_1 - m_2}{1 + m_1m_2} = \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2} \).
2. So \( \cot\theta = \frac{a_1a_2 + b_1b_2}{a_2b_1 - a_1b_2} \), i.e., θ = \( \cot^{-1}\frac{a_1a_2 + b_1b_2}{a_1b_2 - a_2b_1} \) for one of the two angles between the lines.

Teacher's Note:
1. Changing the sign of the fraction only changes θ to its supplement, the other angle between the lines.
2. Remember the numerator a₂b₁ − a₁b₂ and denominator a₁a₂ + b₁b₂ of tan θ.

Answer: (c) \( -\frac{1}{2} \)

Explanation:
1. Perpendicular ⇒ a₁a₂ + b₁b₂ = 0: (2)(3) + (3a)(4) = 0.
2. 6 + 12a = 0 ⇒ \( a = -\frac{1}{2} \).

Teacher's Note:
1. The condition a₁a₂ + b₁b₂ = 0 avoids dividing by coefficients that might be zero.
2. Check: slopes −2/(3a) = 4/3 and −3/4 multiply to −1.

Answer: (b) a₁a₂ + b₁b₂ = 0

Explanation:
1. \( m_1m_2 = -1 \Rightarrow \left(-\frac{a_1}{b_1}\right)\left(-\frac{a_2}{b_2}\right) = -1 \).
2. ⇒ a₁a₂ = −b₁b₂ ⇒ a₁a₂ + b₁b₂ = 0.

Teacher's Note:
1. a₁b₂ − a₂b₁ = 0 is the condition for parallel lines, not perpendicular ones.
2. Multiply the x-coefficients, multiply the y-coefficients, and add: zero means perpendicular.

Answer: (a) 90°

Explanation:
1. Slopes: \( m_1 = \frac{1}{\sqrt{3}} \), \( m_2 = -\sqrt{3} \).
2. m₁m₂ = −1, so the angle is 90°.

Teacher's Note:
1. Inclinations 30° and 120° differ by 90°.
2. Check with coefficients: (1)(√3) + (−√3)(1) = 0.

Answer: (a) 90°

Explanation:
1. a₁a₂ + b₁b₂ = (2)(1) + (−1)(2) = 0.
2. So the lines are perpendicular: 90°.

Teacher's Note:
1. Use the coefficient test a₁a₂ + b₁b₂ = 0 for a quick check of perpendicularity.
2. Slopes 2 and −1/2 confirm it.

Answer: (b) Perpendicular

Explanation:
1. Slopes: 2 and \( -\frac{1}{2} \).
2. Product = −1, so the lines are perpendicular.

Teacher's Note:
1. Rewrite x = −2y as y = −x/2 to read the slope.
2. Both lines pass through the origin, so they cannot be parallel.

Answer: (c) \( \tan^{-1}\frac{b^2 - a^2}{2ab} \)

Explanation:
1. First line \( \frac{x}{a} - \frac{y}{b} = 1 \): slope \( \frac{b}{a} \). Second line \( \frac{x}{b} - \frac{y}{a} = 1 \): slope \( \frac{a}{b} \).
2. \( \tan\theta = \frac{\frac{b}{a} - \frac{a}{b}}{1 + 1} = \frac{b^2 - a^2}{2ab} \).

Teacher's Note:
1. A line with intercepts p and q has slope −q/p.
2. The product of the two slopes is 1, so the denominator 1 + m₁m₂ = 2.

Answer: (d) \( \tan^{-1}(3/2) \)

Explanation:
1. A = (3, 0) and \( B = \left(0, \frac{9}{2}\right) \).
2. In right triangle OAB (right angle at O), \( \tan\angle OAB = \frac{OB}{OA} = \frac{9/2}{3} = \frac{3}{2} \).
3. So ∠OAB = \( \tan^{-1}\frac{3}{2} \).

Teacher's Note:
1. ∠OAB is the angle at A, so the opposite side is OB and the adjacent side is OA.
2. This equals the acute angle between the line and the x-axis, |slope| = 3/2.

Answer: (c) \( -\frac{1}{3}, 3 \)

Explanation:
1. \( \tan\frac{\pi}{4} = \left| \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} \right| \Rightarrow \frac{2m - 1}{2 + m} = \pm 1 \).
2. +1: 2m − 1 = 2 + m ⇒ m = 3; −1: 2m − 1 = −2 − m ⇒ \( m = -\frac{1}{3} \).

Teacher's Note:
1. The modulus in the angle formula always produces two answers.
2. The two slopes found, 3 and −1/3, are perpendicular to each other, as expected for ±45° turns.

Answer: (b) \( y - 3 = (2 - \sqrt{3})(x - 2) \)

Explanation:
1. The sides through (2, 3) make 60° with x + y = 2 (slope −1): \( \frac{m + 1}{1 - m} = \pm\sqrt{3} \).
2. \( m + 1 = \sqrt{3}(1 - m) \Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \); the other sign gives \( m = 2 + \sqrt{3} \).
3. So one side is \( y - 3 = (2 - \sqrt{3})(x - 2) \).

Teacher's Note:
1. Each angle of an equilateral triangle is 60°, so the two sides through a vertex make 60° with the opposite side.
2. Rationalise (√3 − 1)/(√3 + 1) by multiplying with (√3 − 1) to get 2 − √3.

Answer: (c) \( \tan^{-1}(2) \)

Explanation:
1. Diagonal through (2, −1) and (2, 3) is the vertical line x = 2.
2. Diagonal through (0, 2) and (4, 0) has slope \( -\frac{1}{2} \); its acute angle with the x-axis is \( \tan^{-1}\frac{1}{2} \).
3. Angle with the vertical diagonal = \( 90^\circ - \tan^{-1}\frac{1}{2} = \tan^{-1}2 \).

Teacher's Note:
1. Opposite vertices are the 1st and 3rd, and the 2nd and 4th, in the order given.
2. When one line is vertical, find the other line's angle with the vertical directly.

Answer: (b) Parallel

Explanation:
1. Slope of the first line = \( \frac{6 + 4}{-2 - 3} = -2 \).
2. Slope of the second line = \( \frac{-18 - 6}{9 + 3} = -2 \).
3. Equal slopes ⇒ the lines are parallel.

Teacher's Note:
1. Equal slopes mean parallel (or coincident) lines.
2. Here (−3, 6) is not on the first line (the line through (3, −4) with slope −2 is y = −2x + 2), so they are distinct.

Answer: (b) 3x − y − 7 = 0 and x + 3y − 9 = 0

Explanation:
1. Slope of x − 2y = 3 is \( \frac{1}{2} \). \( \left| \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} \right| = 1 \Rightarrow m = 3 \) or \( m = -\frac{1}{3} \).
2. m = 3: y − 2 = 3(x − 3) ⇒ 3x − y − 7 = 0.
3. \( m = -\frac{1}{3} \): 3(y − 2) = −(x − 3) ⇒ x + 3y − 9 = 0.

Teacher's Note:
1. Use the line-through-a-point-at-a-given-angle method: put the unknown slope in the tan formula.
2. The two answers are perpendicular to each other, since both make 45° with the same line.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. Distance between the first pair of sides = \( \frac{|n - n'|}{\sqrt{l^2 + m^2}} \); between the second pair = \( \frac{|n - n'|}{\sqrt{m^2 + l^2}} \).
2. The distances (heights) are equal, so all four sides of the parallelogram are equal: it is a rhombus.
3. The diagonals of a rhombus are perpendicular, so the angle is \( \frac{\pi}{2} \).

Teacher's Note:
1. A parallelogram with equal heights between both pairs of opposite sides has equal sides, i.e., it is a rhombus.
2. Equivalently, the lines are symmetric about y = x, and both diagonals turn out to be lines of slope 1 and −1.

Answer: (c) 90°

Explanation:
1. A = AB ∩ DA = (−3, 3); B = AB ∩ BC = (1, 1); C = BC ∩ CD = (1, −1); D = CD ∩ DA = (−2, −2).
2. Slope of AC = \( \frac{-1 - 3}{1 + 3} = -1 \); slope of BD = \( \frac{-2 - 1}{-2 - 1} = 1 \).
3. Product of slopes = −1, so the angle is 90°.

Teacher's Note:
1. Each vertex is the intersection of the two sides that share its letter.
2. Once the vertices are known, the diagonals are just two-point lines.

Answer: (c) 23x − 7y − 9 = 0, 7x + 23y − 53 = 0

Explanation:
1. The sides make 45° with the diagonal (slope \( \frac{8}{15} \)): \( \frac{15m - 8}{15 + 8m} = \pm 1 \).
2. +1: 7m = 23 ⇒ \( m = \frac{23}{7} \); −1: 23m = −7 ⇒ \( m = -\frac{7}{23} \).
3. \( y - 2 = \frac{23}{7}(x - 1) \Rightarrow 23x - 7y - 9 = 0 \); \( y - 2 = -\frac{7}{23}(x - 1) \Rightarrow 7x + 23y - 53 = 0 \).

Teacher's Note:
1. A diagonal of a square bisects its right angles, so each side makes 45° with it.
2. The vertex (1, 2) is not on the diagonal 8x − 15y = 0, but the sides through it still make 45° with that direction.

Answer: (c) aα + bγ = 0

Explanation:
1. Eliminating t: \( \frac{x - \beta}{\alpha} = \frac{y - \delta}{\gamma} \), a line with slope \( \frac{\gamma}{\alpha} \).
2. Parallel to ax + by + c = 0 (slope \( -\frac{a}{b} \)): \( \frac{\gamma}{\alpha} = -\frac{a}{b} \Rightarrow b\gamma = -a\alpha \).
3. So aα + bγ = 0.

Teacher's Note:
1. As t changes by 1, x changes by α and y by γ, so the slope is γ/α.
2. β and δ only fix a point of the line and do not affect parallelism.

Answer: (d) 3x − 4y + 4a = 0

Explanation:
1. The apex lies on the perpendicular bisector of the base. Mid-point = \( \left(a, \frac{a}{2}\right) \); base slope \( -\frac{1}{2} \), so bisector slope 2: \( y - \frac{a}{2} = 2(x - a) \).
2. The side x = 2a passes through the apex: at x = 2a, \( y = \frac{5a}{2} \). Apex = \( \left(2a, \frac{5a}{2}\right) \).
3. Other side through (0, a) and \( \left(2a, \frac{5a}{2}\right) \): slope \( \frac{3}{4} \), so \( y - a = \frac{3}{4}x \Rightarrow 3x - 4y + 4a = 0 \).

Teacher's Note:
1. The apex of an isosceles triangle lies on the perpendicular bisector of its base.
2. The side x = 2a passes through the base end (2a, 0), so the other side must pass through (0, a).

Answer: (c) \( \frac{\pi}{12} \) and \( \frac{5\pi}{12} \)

Explanation:
1. Perpendicular distance of (1, 2) from x + y = 4 is \( \frac{|1 + 2 - 4|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \).
2. If a required line meets x + y = 4 at angle φ, then \( \sin\varphi = \frac{1/\sqrt{2}}{\sqrt{6}/3} = \frac{3}{\sqrt{12}} = \frac{\sqrt{3}}{2} \Rightarrow \varphi = \frac{\pi}{3} \).
3. x + y = 4 has inclination \( \frac{3\pi}{4} \); so the lines have inclinations \( \frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12} \) and \( \frac{3\pi}{4} + \frac{\pi}{3} - \pi = \frac{\pi}{12} \).

Teacher's Note:
1. The perpendicular, the slanted segment and the given line form a right triangle, which gives the angle with the line.
2. Add and subtract that angle from the given line's inclination, keeping answers between 0 and π.

Answer: (c) 3x + y + 7 = 0 or x − 3y − 31 = 0

Explanation:
1. The third side (slope m) must be equally inclined to the lines with slopes 7 and −1: \( \left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{m + 1}{1 - m} \right| \).
2. For 3x + y + 7 = 0 (m = −3): \( \left| \frac{-10}{-20} \right| = \frac{1}{2} \) and \( \left| \frac{-2}{4} \right| = \frac{1}{2} \) — equal.
3. For x − 3y − 31 = 0 (\( m = \frac{1}{3} \)): \( \left| \frac{-20/3}{10/3} \right| = 2 \) and \( \left| \frac{4/3}{2/3} \right| = 2 \) — equal.
4. Both lines pass through (1, −10), so either can be the third side.

Teacher's Note:
1. The base of an isosceles triangle makes equal angles with the two equal sides.
2. Two perpendicular directions (slopes −3 and 1/3) both satisfy this, giving two possible triangles.

Answer: (b) 3/10

Explanation:
1. Write the second line as 3x + 4y = \( \frac{15}{2} \).
2. Distance = \( \frac{|9 - 15/2|}{\sqrt{3^2 + 4^2}} = \frac{3/2}{5} = \frac{3}{10} \).

Teacher's Note:
1. Make the x and y coefficients identical before using |c₁ − c₂|/√(a² + b²).
2. Forgetting to divide 6x + 8y = 15 by 2 is the most common mistake here.

Answer: (a) \( \frac{7}{13} \)

Explanation:
1. Write the line as 12x + 5y − 7 = 0.
2. Distance from the origin = \( \frac{|-7|}{\sqrt{12^2 + 5^2}} = \frac{7}{13} \).

Teacher's Note:
1. For the origin, the distance from ax + by + c = 0 is |c|/√(a² + b²).
2. 12, 5, 13 is a Pythagorean triple, so the root is exact.

Answer: (b) 7

Explanation:
1. Use p = |ax₁ + by₁ + c|/√(a² + b²) with (x₁, y₁) = (3, 1).
2. \( p = \frac{|4(3) + 3(1) + 20|}{\sqrt{16 + 9}} = \frac{35}{5} = 7 \).

Teacher's Note:
1. Substitute the point into the left side of the line and divide by √(a² + b²).
2. Take the absolute value since a distance is never negative.

Answer: (d) 1

Explanation:
1. Here a = 3, b = 4, c₁ = −8, c₂ = −3.
2. Distance = \( \frac{|-8 - (-3)|}{\sqrt{9 + 16}} = \frac{5}{5} = 1 \).

Teacher's Note:
1. The coefficients of x and y are already equal, so use |c₁ − c₂|/√(a² + b²) directly.
2. Both constants are negative, so both lines lie on the same side of the origin.

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

Explanation:
1. A line through (0, a) is y = mx + a, i.e., mx − y + a = 0.
2. Distance from (2a, 2a): \( \frac{|2am - 2a + a|}{\sqrt{m^2 + 1}} = a \Rightarrow (2m - 1)^2 = m^2 + 1 \Rightarrow 3m^2 - 4m = 0 \).
3. m = 0 gives y − a = 0; \( m = \frac{4}{3} \) gives 3y − 3a = 4x, i.e., 4x − 3y + 3a = 0.

Teacher's Note:
1. Set up the line with one unknown (its slope) and use the distance condition to find it.
2. The common factor a cancels, leaving a simple quadratic in m.

Answer: (d) \( \frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}} \)

Explanation:
1. Area = \( \frac{1}{2}|2(2 - 4) + 5(4 - 1) + 4(1 - 2)| = \frac{7}{2} \).
2. Opposite sides: from (2, 1) the side joining (5, 2), (4, 4) has length √5; from (5, 2) the side joining (2, 1), (4, 4) has length √13; from (4, 4) the side joining (2, 1), (5, 2) has length √10.
3. Altitude = \( \frac{2 \times \text{Area}}{\text{side}} \): \( \frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}} \).

Teacher's Note:
1. Using altitude = 2 × area ÷ base avoids writing three line equations.
2. You can still check one value with the point-line distance formula.

Answer: (c) \( \frac{2}{\sqrt{5}} \)

Explanation:
1. Lines: 2x − y + 7 = 0 and 2x − y + 5 = 0.
2. Distance = \( \frac{|7 - 5|}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}} \).

Teacher's Note:
1. For y = mx + c₁ and y = mx + c₂, distance = |c₁ − c₂|/√(1 + m²).
2. The difference of y-intercepts (2) is not the distance; it is the vertical gap.

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

Explanation:
1. The line is 4x − 3y − 12 = 0.
2. Distance from the origin = \( \frac{12}{5} = 2\frac{2}{5} \).

Teacher's Note:
1. Clear fractions in the intercept form before applying the distance formula.
2. Alternatively 1/p² = 1/a² + 1/b² = 1/9 + 1/16 = 25/144 gives p = 12/5.