CBSE Class 11 Mathematics Straight Lines Competitive Exam MCQs Set 05

Answer: (b) \( \frac{12}{5} \)

Explanation:
1. Here a = 3, b = 4, c₁ = 7, c₂ = −5.
2. Distance = \( \frac{|7 - (-5)|}{\sqrt{9 + 16}} = \frac{12}{5} \).

Teacher's Note:
1. The constants have opposite signs, so the origin lies between the lines and the distances from O add up.
2. 7/5 + 5/5 = 12/5 confirms the answer.

Answer: (a) True

Explanation:
1. 12x + 3y = 3 and 7x − 3y = 35; adding, 19x = 38 ⇒ x = 2, y = −7.
2. Line through (3, 5) and (2, −7): slope 12, y − 5 = 12(x − 3) ⇒ 12x − y − 31 = 0.
3. Distance of (0, 0) = \( \frac{31}{\sqrt{145}} \); distance of (8, 34) = \( \frac{|96 - 34 - 31|}{\sqrt{145}} = \frac{31}{\sqrt{145}} \). Equal, so the statement is true.

Teacher's Note:
1. Find the line first, then compare the two distances using the same denominator.
2. Only the numerators |ax₁ + by₁ + c| need comparing, since √(a² + b²) is common.

Answer: (c) \( \frac{35}{3\sqrt{34}} \)

Explanation:
1. Divide the second equation by 3: \( 5x + 3y + \frac{14}{3} = 0 \).
2. Distance = \( \frac{|-7 - \frac{14}{3}|}{\sqrt{25 + 9}} = \frac{35/3}{\sqrt{34}} = \frac{35}{3\sqrt{34}} \).

Teacher's Note:
1. Make the x and y coefficients equal before subtracting constants.
2. −7 − 14/3 = −35/3; the modulus gives 35/3.

Answer: (b) \( \frac{11}{\sqrt{52}} \)

Explanation:
1. Second line: 6x − 4y + 9 = 0, i.e., \( 3x - 2y + \frac{9}{2} = 0 \). First line: 3x − 2y − 1 = 0.
2. Distance = \( \frac{|-1 - \frac{9}{2}|}{\sqrt{9 + 4}} = \frac{11/2}{\sqrt{13}} = \frac{11}{2\sqrt{13}} = \frac{11}{\sqrt{52}} \).

Teacher's Note:
1. Bring both equations to the form 3x − 2y + c = 0 first.
2. 2√13 = √52, which matches the option form.

Answer: (a) \( \sqrt{2} \)

Explanation:
1. The line through (1, 1) parallel to x + y = 1 is x + y = 2.
2. It meets 2x − 3y = 4 where 2x − 3(2 − x) = 4 ⇒ x = 2, y = 0.
3. Required distance = distance between (1, 1) and (2, 0) = \( \sqrt{1 + 1} = \sqrt{2} \).

Teacher's Note:
1. "Measured parallel to a line" means travel along a line in that direction, not along the perpendicular.
2. So find where the parallel line meets the given line and use the distance formula.

Answer: (d) \( 17\sqrt{5}/15 \)

Explanation:
1. Lines: 2x − y + 4 = 0 and 6x − 3y − 5 = 0, i.e., \( 2x - y - \frac{5}{3} = 0 \).
2. Distance = \( \frac{|4 + \frac{5}{3}|}{\sqrt{4 + 1}} = \frac{17/3}{\sqrt{5}} = \frac{17}{3\sqrt{5}} = \frac{17\sqrt{5}}{15} \).

Teacher's Note:
1. Divide 6x − 3y − 5 = 0 by 3 so that both equations have the same x and y coefficients.
2. Rationalise 17/(3√5) by multiplying numerator and denominator by √5.

Answer: (d) None of these

Explanation:
1. Distance between the lines = \( \frac{|12 - 7|}{\sqrt{13}} = \frac{5}{\sqrt{13}} \), so (c) is false.
2. Distances of A: \( \frac{|6 - 15 - 7|}{\sqrt{13}} = \frac{16}{\sqrt{13}} \) and \( \frac{|6 - 15 - 12|}{\sqrt{13}} = \frac{21}{\sqrt{13}} \); their sum is \( \frac{37}{\sqrt{13}} \), so (b) is false.
3. If A were between the lines, the sum of its distances would equal \( \frac{5}{\sqrt{13}} \); it does not, so (a) is false. Hence (d).

Teacher's Note:
1. A point lies between two parallel lines exactly when its two distances add up to the distance between the lines.
2. Check each option separately; 'None of these' is right only after all others fail.

Answer: (a) (3, 1), (−7, 11)

Explanation:
1. A point on x + y = 4 is (t, 4 − t).
2. Distance from 4x + 3y − 10 = 0: \( \frac{|4t + 12 - 3t - 10|}{5} = \frac{|t + 2|}{5} = 1 \Rightarrow t + 2 = \pm 5 \).
3. t = 3 or t = −7, giving (3, 1) and (−7, 11).

Teacher's Note:
1. Use one parameter to describe points of the first line.
2. The modulus gives two answers, one on each side of the second line.

Answer: (d) \( \frac{25}{16} \)

Explanation:
1. L: slope −1, x + y = 2. L′: slope 1 through \( \left(\frac{1}{2}, 0\right) \), \( y = x - \frac{1}{2} \).
2. L meets the y-axis at (0, 2) and L′ at \( \left(0, -\frac{1}{2}\right) \); base on the y-axis = \( \frac{5}{2} \).
3. L ∩ L′: \( x + x - \frac{1}{2} = 2 \Rightarrow x = \frac{5}{4} \) (height from the y-axis).
4. Area = \( \frac{1}{2} \times \frac{5}{2} \times \frac{5}{4} = \frac{25}{16} \).

Teacher's Note:
1. Take the side on the y-axis as the base; the height is then the x-coordinate of the third vertex.
2. Draw a quick sketch to see which vertices lie on the y-axis.

Answer: (a) \( x\sqrt{3} + y + 8 = 0 \)

Explanation:
1. Slope = tan 120° = \( -\sqrt{3} \), so the line is \( y = -\sqrt{3}x + c \), i.e., \( \sqrt{3}x + y - c = 0 \).
2. Distance from the origin = \( \frac{|c|}{2} = 4 \Rightarrow c = \pm 8 \).
3. c = −8 gives \( x\sqrt{3} + y + 8 = 0 \), which is in the options.

Teacher's Note:
1. Fix the slope from the angle, then use the perpendicular distance to find the intercept.
2. There are two such lines, on opposite sides of the origin; pick the one listed.

Answer: (c) \( \sqrt{2/3} \)

Explanation:
1. Altitude = distance of (2, −1) from x + y − 2 = 0 = \( \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \).
2. For an equilateral triangle of side a, altitude = \( \frac{\sqrt{3}}{2}a \). So \( a = \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}} = \sqrt{\frac{2}{3}} \).

Teacher's Note:
1. The altitude of an equilateral triangle is the distance of the vertex from the base.
2. Side = 2 × altitude/√3.

Answer: (c) 3x − 4y + 6 = 0 and 4x − 3y + 1 = 0

Explanation:
1. x = y − 1; 2(y − 1) − 3y + 5 = 0 ⇒ y = 3, x = 2. The lines meet at (2, 3).
2. Line through (2, 3): mx − y + 3 − 2m = 0. Distance from (3, 2): \( \frac{|m + 1|}{\sqrt{m^2 + 1}} = \frac{7}{5} \Rightarrow 12m^2 - 25m + 12 = 0 \Rightarrow m = \frac{4}{3}, \frac{3}{4} \).
3. \( m = \frac{4}{3} \): 4x − 3y + 1 = 0; \( m = \frac{3}{4} \): 3x − 4y + 6 = 0.

Teacher's Note:
1. Squaring the distance condition gives a quadratic in m, hence two lines.
2. Check one answer: |12 − 6 + 1|/5 = 7/5 for 4x − 3y + 1 = 0.

Answer: (c) (0, 0)

Explanation:
1. Distances of (0, 0): \( \frac{10}{5} = 2 \), \( \frac{26}{13} = 2 \), \( \frac{50}{25} = 2 \).
2. All three are equal, so (0, 0) is the required point.

Teacher's Note:
1. With options given, test each point with the distance formula.
2. The origin is the easiest to test since the distance is just |c|/√(a² + b²).

Answer: (a) \( \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) \)

Explanation:
1. (x − a₁)² + (y − b₁)² = (x − a₂)² + (y − b₂)².
2. ⇒ −2a₁x + a₁² − 2b₁y + b₁² = −2a₂x + a₂² − 2b₂y + b₂².
3. ⇒ 2(a₁ − a₂)x + 2(b₁ − b₂)y + (a₂² + b₂² − a₁² − b₁²) = 0; dividing by 2, \( c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) \).

Teacher's Note:
1. The set of points equidistant from two points is the perpendicular bisector of the segment joining them.
2. The x² and y² terms cancel, which is why the result is a straight line.

Answer: (a) a²

Explanation:
1. \( p = \frac{|a|}{\sqrt{\sec^2\theta + \csc^2\theta}} = \frac{|a|}{\sqrt{\frac{1}{\sin^2\theta\cos^2\theta}}} = |a\sin\theta\cos\theta| = \frac{|a\sin 2\theta|}{2} \).
2. \( p' = \frac{|a\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} = |a\cos 2\theta| \).
3. 4p² + p′² = a² sin² 2θ + a² cos² 2θ = a².

Teacher's Note:
1. sec² θ + cosec² θ = 1/(sin² θ cos² θ) because sin² θ + cos² θ = 1.
2. The factor 4 in the expression is chosen to turn sin θ cos θ into sin 2θ.

Answer: (c) |a| < 1

Explanation:
1. |x + y| = 2 represents the parallel lines x + y = 2 and x + y = −2.
2. Points between them satisfy −2 < x + y < 2. For (a, a): −2 < 2a < 2.
3. So −1 < a < 1, i.e., |a| < 1.

Teacher's Note:
1. The region between two parallel lines x + y = ±k is |x + y| < k.
2. Points on the lines themselves give |a| = 1, which is the boundary, not 'between'.

Answer: (b) (5, 6)

Explanation:
1. Foot (h, k): \( \frac{h - 2}{1} = \frac{k - 3}{1} = -\frac{2 + 3 - 11}{1 + 1} = 3 \).
2. h = 5, k = 6. So the foot is (5, 6).

Teacher's Note:
1. The foot lies on the line and the joining segment is perpendicular to the line; check: 5 + 6 − 11 = 0.
2. Slope of the segment from (2, 3) to (5, 6) is 1, perpendicular to the slope −1 of the line.

Answer: (b) \( \left(\frac{-1}{10}, \frac{37}{10}\right) \)

Explanation:
1. Write the line as 3x − y + 4 = 0.
2. \( \frac{h - 2}{3} = \frac{k - 3}{-1} = -\frac{6 - 3 + 4}{9 + 1} = -\frac{7}{10} \).
3. \( h = 2 - \frac{21}{10} = -\frac{1}{10} \), \( k = 3 + \frac{7}{10} = \frac{37}{10} \).

Teacher's Note:
1. The foot of the perpendicular can also be found by intersecting the given line with the perpendicular line through the point.
2. Check: 3(−1/10) + 4 = 37/10.

Answer: (a) 2x + 3y = 12

Explanation:
1. Intercepts are 6 and 4: \( \frac{x}{6} + \frac{y}{4} = 1 \).
2. Multiplying by 12: 2x + 3y = 12.

Teacher's Note:
1. If (h, k) is the mid-point of the intercept, the line is x/h + y/k = 2.
2. Here x/3 + y/2 = 2 gives the same answer.

Answer: (b) \( \left[\frac{a}{2}(\cos\alpha + \cos\beta), \frac{a}{2}(\sin\alpha + \sin\beta)\right] \)

Explanation:
1. Both points are at distance a from the origin, so the origin O is equidistant from them.
2. Hence the perpendicular from O to the line joining them meets it at the mid-point of the segment.
3. Mid-point = \( \left[\frac{a}{2}(\cos\alpha + \cos\beta), \frac{a}{2}(\sin\alpha + \sin\beta)\right] \).

Teacher's Note:
1. In an isosceles triangle the perpendicular from the apex bisects the base.
2. (a cos α, a sin α) always lies on the circle x² + y² = a², so OA = OB = a.

Answer: (b) \( \left(\frac{3}{5}, \frac{4}{5}\right) \)

Explanation:
1. \( \frac{h - 0}{3} = \frac{k - 0}{4} = -\frac{0 + 0 - 5}{9 + 16} = \frac{1}{5} \).
2. \( h = \frac{3}{5} \), \( k = \frac{4}{5} \).

Teacher's Note:
1. The perpendicular from the origin to ax + by + c = 0 is along the direction (a, b).
2. Check: 3(3/5) + 4(4/5) = 25/5 = 5.

Answer: (a) \( \left(\frac{b^2x_1 - aby_1 - ac}{a^2 + b^2}, \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2}\right) \)

Explanation:
1. Foot (h, k): \( \frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2} \).
2. \( h = x_1 - \frac{a(ax_1 + by_1 + c)}{a^2 + b^2} = \frac{b^2x_1 - aby_1 - ac}{a^2 + b^2} \).
3. \( k = y_1 - \frac{b(ax_1 + by_1 + c)}{a^2 + b^2} = \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2} \).

Teacher's Note:
1. The vector from (x₁, y₁) to the foot is parallel to the normal direction (a, b).
2. A quick test: if (x₁, y₁) is on the line, the formula must return the same point.

Answer: (b) \( \frac{1}{2}\frac{c^2}{|ab|} \)

Explanation:
1. Intercepts: \( -\frac{c}{a} \) and \( -\frac{c}{b} \).
2. Area = \( \frac{1}{2}\left|\frac{c}{a}\right|\left|\frac{c}{b}\right| = \frac{c^2}{2|ab|} \).

Teacher's Note:
1. The triangle is right-angled at the origin with legs equal to the absolute values of the intercepts.
2. If c = 0 the line passes through the origin and there is no triangle, which is why c ≠ 0 is given.

Answer: (a) Isosceles

Explanation:
1. Vertices: x + y = 0 and 3x + y = 4 give (2, −2); x + y = 0 and x + 3y = 4 give (−2, 2); 3x + y = 4 and x + 3y = 4 give (1, 1).
2. Sides: (2, −2) to (1, 1) = √10; (−2, 2) to (1, 1) = √10; (2, −2) to (−2, 2) = √32.
3. Two sides are equal (and 10 + 10 ≠ 32), so the triangle is isosceles.

Teacher's Note:
1. The lines 3x + y = 4 and x + 3y = 4 are mirror images in y = x, which hints at an isosceles triangle.
2. Always check the right-angle option with Pythagoras before choosing.

Answer: (d) Rhombus

Explanation:
1. Slopes of the diagonals: \( -\frac{1}{3} \) and 3. Their product is −1, so the diagonals are perpendicular.
2. A parallelogram with perpendicular diagonals is a rhombus.

Teacher's Note:
1. Perpendicular diagonals alone make a parallelogram a rhombus; it is a square only if the diagonals are also equal.
2. Nothing here tells us the diagonals are equal, so 'square' cannot be concluded.

Answer: (b) 91 sq. units

Explanation:
1. A(0, 12), B(8, 0). Mid-point (4, 6); slope of AB = \( -\frac{3}{2} \), so the bisector has slope \( \frac{2}{3} \): \( y - 6 = \frac{2}{3}(x - 4) \).
2. It meets y = −1: \( -7 = \frac{2}{3}(x - 4) \Rightarrow x = -\frac{13}{2} \). So \( C\left(-\frac{13}{2}, -1\right) \).
3. Area = \( \frac{1}{2}\left|0(0 + 1) + 8(-1 - 12) - \frac{13}{2}(12 - 0)\right| = \frac{1}{2}|-104 - 78| = 91 \).

Teacher's Note:
1. The line through (0, −1) parallel to the x-axis is simply y = −1.
2. Use the coordinate area formula once all three vertices are known.

Answer: (b) \( \frac{2c^2}{ab} \)

Explanation:
1. The four lines meet the axes at \( \left(\pm\frac{c}{a}, 0\right) \) and \( \left(0, \pm\frac{c}{b}\right) \), which are the vertices.
2. The figure is a rhombus with diagonals \( \frac{2c}{a} \) and \( \frac{2c}{b} \).
3. Area = \( \frac{1}{2} \times \frac{2c}{a} \times \frac{2c}{b} = \frac{2c^2}{ab} \) (taking a, b, c positive).

Teacher's Note:
1. Area of a rhombus = half the product of its diagonals.
2. The four lines are symmetric about both axes, so the vertices lie on the axes.

Answer: (b) ab/2

Explanation:
1. The triangle has vertices (0, 0), (a, 0) and (0, b).
2. Area = \( \frac{1}{2} \times a \times b = \frac{ab}{2} \) (for positive a, b).

Teacher's Note:
1. A line in intercept form makes a right triangle with the axes, with legs a and b.
2. For signed intercepts use |ab|/2.

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

Explanation:
1. A line perpendicular to 5x − y = 1 is x + 5y = c.
2. Intercepts: c and \( \frac{c}{5} \). Area = \( \frac{1}{2} \cdot \frac{c^2}{5} = 5 \Rightarrow c^2 = 50 \Rightarrow c = \pm 5\sqrt{2} \).
3. So L: \( x + 5y = \pm 5\sqrt{2} \).

Teacher's Note:
1. The area condition fixes only c², so there are two lines, one on each side of the origin.
2. Area with axes = c²/(2|ab|) = c²/10 here.

Answer: (a) A fixed point

Explanation:
1. Take the perpendicular lines as the axes and the moving line as \( \frac{x}{a} + \frac{y}{b} = 1 \) with \( \frac{1}{a} + \frac{1}{b} = k \) (constant).
2. Putting x = y = \( \frac{1}{k} \): \( \frac{1}{k}\left(\frac{1}{a} + \frac{1}{b}\right) = \frac{1}{k} \cdot k = 1 \).
3. So every such line passes through the fixed point \( \left(\frac{1}{k}, \frac{1}{k}\right) \).

Teacher's Note:
1. Look for a point whose coordinates make the given condition appear in the equation of the line.
2. Choosing the two perpendicular lines as coordinate axes keeps the algebra simple.

Answer: (c) \( \frac{23}{3} \) sq. units

Explanation:
1. A(6, 0), B(0, 4). The perpendicular to AB through (5, 5) has slope \( \frac{3}{2} \): 3x − 2y = 5.
2. C = \( \left(\frac{5}{3}, 0\right) \); E: 2x + 3y = 12 and 3x − 2y = 5 give (3, 2).
3. Area of OCEB with O(0, 0), \( C\left(\frac{5}{3}, 0\right) \), E(3, 2), B(0, 4): \( \frac{1}{2}\left|\frac{5}{3} \cdot 2 - 0 + 3 \cdot 4 - 0\right| = \frac{1}{2} \cdot \frac{46}{3} = \frac{23}{3} \).

Teacher's Note:
1. For a quadrilateral, list the vertices in order around the figure before using the shoelace formula.
2. Point D is not needed for the area of OCEB.

Answer: (a) 7x − 4y + 25 = 0, 4x + 7y = 11 and 7x − 4y − 3 = 0

Explanation:
1. (−3, 1) satisfies 4x + 7y + 5 = 0, but (1, 1) does not; so (1, 1) is a vertex on the opposite side.
2. Side through (1, 1) parallel to the given side: 4x + 7y = 11.
3. Sides perpendicular to it are 7x − 4y + λ = 0: through (−3, 1), λ = 25; through (1, 1), λ = −3.
4. So the other sides are 7x − 4y + 25 = 0, 4x + 7y = 11 and 7x − 4y − 3 = 0.

Teacher's Note:
1. First test which given vertices lie on the given side.
2. Adjacent sides of a rectangle are perpendicular and opposite sides are parallel.

Answer: (c) x − y = 0

Explanation:
1. The sides meet at O(0, 0); the diagonal 11x + 7y = 9 joins the other ends A and C of these sides.
2. A: 4x + 5y = 0 and 11x + 7y = 9 ⇒ \( A\left(\frac{5}{3}, -\frac{4}{3}\right) \). C: 7x + 2y = 0 and 11x + 7y = 9 ⇒ \( C\left(-\frac{2}{3}, \frac{7}{3}\right) \).
3. Mid-point of AC = \( \left(\frac{1}{2}, \frac{1}{2}\right) \). The other diagonal passes through O and this mid-point: y = x, i.e., x − y = 0.

Teacher's Note:
1. The diagonals of a parallelogram bisect each other.
2. The given diagonal does not pass through the origin, so the other one does.

Answer: (a) x − 5y = 0, 5x + y = 0, \( \Delta = \frac{36}{13} \)

Explanation:
1. The lines through O make 45° with 2x + 3y = 6 (slope \( -\frac{2}{3} \)): \( \frac{3m + 2}{3 - 2m} = \pm 1 \Rightarrow m = \frac{1}{5} \) or m = −5.
2. Lines: x − 5y = 0 and 5x + y = 0.
3. Altitude from O to the hypotenuse = \( \frac{6}{\sqrt{13}} \); in an isosceles right triangle the area = (altitude)² = \( \frac{36}{13} \).

Teacher's Note:
1. In an isosceles right triangle the altitude to the hypotenuse equals half the hypotenuse, so area = h × h.
2. The two lines found are perpendicular (slopes 1/5 and −5), confirming the right angle at O.

Answer: (d) A straight line passing through the point \( \left(\frac{\pi}{2}, -\sin^2 1\right) \) and parallel to the x-axis

Explanation:
1. \( \cos x\cos(x + 2) = \frac{1}{2}[\cos(2x + 2) + \cos 2] \) and \( \cos^2(x + 1) = \frac{1}{2}[1 + \cos(2x + 2)] \).
2. Difference = \( \frac{1}{2}(\cos 2 - 1) = -\sin^2 1 \), a constant.
3. So the graph is the horizontal line y = −sin² 1, which passes through \( \left(\frac{\pi}{2}, -\sin^2 1\right) \).

Teacher's Note:
1. Product-to-sum and cos 2A = 1 − 2 sin² A make the x-terms cancel.
2. A function with a constant value has a horizontal straight-line graph (slope 0).

Answer: (b) 9/2

Explanation:
1. x − y = 2 has inclination 45°; lines making 45° with it have inclination 0° or 90°.
2. So the lines through (3, 4) are y = 4 and x = 3.
3. They meet x − y = 2 at (6, 4) and (3, 1). Triangle (3, 4), (6, 4), (3, 1) is right-angled with legs 3 and 3: area = \( \frac{9}{2} \).

Teacher's Note:
1. Adding and subtracting 45° from the inclination of the given line gives the inclinations directly.
2. The triangle meant is the one formed by the two new lines and the given line.

Answer: (d) \( \frac{1}{|m - n|} \)

Explanation:
1. Vertices: O(0, 0) (y = mx, y = nx); y = mx and y = nx + 1 meet at \( \left(\frac{1}{m - n}, \frac{m}{m - n}\right) \); y = mx + 1 and y = nx meet at \( \left(\frac{1}{n - m}, \frac{n}{n - m}\right) \).
2. Area of parallelogram = 2 × area of triangle formed by O and these two adjacent vertices = \( \left| x_1y_2 - x_2y_1 \right| = \left| \frac{1}{m - n} \cdot \frac{n}{n - m} - \frac{1}{n - m} \cdot \frac{m}{m - n} \right| \).
3. = \( \left| \frac{-n + m}{(m - n)^2} \right| = \frac{1}{|m - n|} \).

Teacher's Note:
1. The area of a parallelogram with one vertex at O and adjacent vertices (x₁, y₁), (x₂, y₂) is |x₁y₂ − x₂y₁|.
2. The vertex opposite O is not needed.

Answer: (c) 4x − 3y = 0

Explanation:
1. Zero intercepts ⇒ AB passes through the origin.
2. Perpendicular to 3x + 4y + 6 = 0: 4x − 3y + λ = 0; through the origin λ = 0.
3. So AB: 4x − 3y = 0.

Teacher's Note:
1. A line making zero intercepts on both axes must pass through the origin.
2. Perpendicular lines: swap the coefficients and change one sign.

Answer: (b) \( \frac{5a^2}{2} \) sq. units

Explanation:
1. The apex lies on the perpendicular bisector of the base, \( y - \frac{a}{2} = 2(x - a) \), and on x = 2a, so apex = \( \left(2a, \frac{5a}{2}\right) \).
2. Area = \( \frac{1}{2}\left|2a\left(a - \frac{5a}{2}\right) + 0\left(\frac{5a}{2} - 0\right) + 2a(0 - a)\right| = \frac{1}{2}|-3a^2 - 2a^2| = \frac{5a^2}{2} \).

Teacher's Note:
1. The apex of an isosceles triangle is on the perpendicular bisector of the base.
2. Alternatively, base x = 2a from (2a, 0) to (2a, 5a/2) has length 5a/2 and the third vertex is 2a away: area = ½ · (5a/2) · 2a.

Answer: (c) x − y + 3 = 0

Explanation:
1. Total area by the shoelace formula: ½|(−4 + 4) + (−2 − 2) + (4 − 4) + (8 + 8)| = ½ × 12 = 6.
2. x − y + 3 = 0 passes through (−1, 2) and meets the top side y = 4 at (1, 4).
3. Region (−1, 2), (1, 2), (2, 4), (1, 4): area = ½|(−2 − 2) + (4 − 4) + (8 − 4) + (2 + 4)| = ½ × 6 = 3, exactly half.
4. Hence x − y + 3 = 0 divides the quadrilateral into two equal parts.

Teacher's Note:
1. The quadrilateral is a trapezium with parallel sides on y = 2 and y = 4.
2. With options given, it is quickest to test each line by computing the area it cuts off.

Answer: (b) b, 2a, c are in G.P.; (d) b, −2a, c are in G.P.

Explanation:
1. Intercepts: \( \frac{a}{b} \) and \( \frac{a}{c} \). Area = \( \frac{1}{2} \cdot \frac{a^2}{bc} = \frac{1}{8} \Rightarrow 4a^2 = bc \).
2. (2a)² = bc ⇒ b, 2a, c are in G.P.
3. Also (−2a)² = 4a² = bc ⇒ b, −2a, c are in G.P. Hence (b) and (d).

Teacher's Note:
1. Three numbers x, y, z are in G.P. when y² = xz, so the sign of the middle term does not matter.
2. Since a, b, c have the same sign, both intercepts are positive and bc > 0.

Answer: (a) \( \frac{a^2(a + 2)}{2(a + 1)} \), if a > −1; (c) \( \frac{-a^2(a + 2)}{2(a + 1)} \), if −2 < a < −1; (d) \( \frac{a^2(a + 2)}{2(a + 1)} \), if a < −2

Explanation:
1. Vertices: O(0, 0), \( \left(\frac{a}{m_1}, a\right) \), \( \left(\frac{a}{m_2}, a\right) \). Area = \( \frac{1}{2}|a| \cdot \left|\frac{a}{m_1} - \frac{a}{m_2}\right| = \frac{a^2}{2} \cdot \frac{|m_1 - m_2|}{|m_1m_2|} \).
2. m₁ + m₂ = a, m₁m₂ = −(a + 1), so (m₁ − m₂)² = a² + 4a + 4 = (a + 2)².
3. Area = \( \frac{a^2|a + 2|}{2|a + 1|} \).
4. a > −1: both factors positive ⇒ (a). −2 < a < −1: a + 2 > 0, a + 1 < 0 ⇒ (c). a < −2: both negative ⇒ (d). Option (b) fails for a < −2, where it gives a negative value.

Teacher's Note:
1. Use the sum and product of roots instead of solving the quadratic.
2. Removing the modulus signs needs a separate check of the signs of a + 2 and a + 1 in each interval.

Answer: (a) PR = 3 sec θ; (b) PS = 4 cosec θ; (c) \( PR + PS = \frac{2(3\sin\theta + 4\cos\theta)}{\sin 2\theta} \); (d) \( \frac{9}{(PR)^2} + \frac{16}{(PS)^2} = 1 \)

Explanation:
1. From P to x = 6 the horizontal run is 3, so PR cos θ = 3 ⇒ PR = 3 sec θ.
2. From P to y = 8 the vertical rise is 4, so PS sin θ = 4 ⇒ PS = 4 cosec θ.
3. PR + PS = \( \frac{3}{\cos\theta} + \frac{4}{\sin\theta} = \frac{3\sin\theta + 4\cos\theta}{\sin\theta\cos\theta} = \frac{2(3\sin\theta + 4\cos\theta)}{\sin 2\theta} \).
4. \( \frac{9}{PR^2} + \frac{16}{PS^2} = \cos^2\theta + \sin^2\theta = 1 \). All four statements are true.

Teacher's Note:
1. Along a line of inclination θ, a length L has horizontal component L cos θ and vertical component L sin θ.
2. The angle is acute, so both sec θ and cosec θ are positive and no modulus is needed.