JEE Mathematics Straight Lines MCQs Set 13

Answer: (a) 5x - 7y = 0

Explanation:
1. Take P = (2t, t). Its foot on x = y is (3t/2, 3t/2).
2. Mid-point = (7t/4, 5t/4), so y/x = 5/7.
3. The locus is 5x - 7y = 0.

Teacher's Note:
1. Parametrise the points on the given line first.
2. The foot of the perpendicular to y = x from (p, q) is ((p + q)/2, (p + q)/2).

Answer: (b) 4 sq. units

Explanation:
1. Each pair of parallel lines is 10/5 = 2 units apart, and the two pairs are perpendicular.
2. So the figure is a square of side 2, with area 4.

Teacher's Note:
1. A pair of parallel lines against a perpendicular pair forms a rectangle.
2. Area = product of the two distances between parallel lines.

Answer: (d) \( \frac{49\sqrt3}{6} \)

Explanation:
1. The altitude is the distance from (7, 0) to y = x: h = 7/√2.
2. Area = \( \frac{h^2}{\sqrt3}=\frac{49/2}{\sqrt3}=\frac{49\sqrt3}{6} \).

Teacher's Note:
1. For an equilateral triangle, area = h²/√3 where h is the altitude.
2. The altitude here is the distance from the vertex to the opposite side.

Answer: (c) 15° or 75°

Explanation:
1. Take P = (1 + r cos θ, 2 + r sin θ) with r = √6/3.
2. P on x + y = 4: 3 + r(cos θ + sin θ) = 4, so cos θ + sin θ = 1/r = 3/√6 = √6/2.
3. √2 sin(θ + 45°) = √6/2 gives sin(θ + 45°) = √3/2, so θ + 45° = 60° or 120° and θ = 15° or 75°.

Teacher's Note:
1. Use the distance form x = x1 + r cos θ, y = y1 + r sin θ.
2. Convert cos θ + sin θ to √2 sin(θ + 45°).

Answer: (d) \( (5-\sqrt3,\ 3+2\sqrt3) \)

Explanation:
1. AB = √20. C lies on the perpendicular bisector of AB with CA = CB = √20.
2. Mid-point of AB is (5, 3); AB has direction (4, 2); a perpendicular direction is (-1, 2)/√5.
3. Distance from mid-point to C = (√3/2)√20 = √15, so C = (5, 3) ± √15(-1, 2)/√5 = (5 ∓ √3, 3 ± 2√3).
4. So (5 - √3, 3 + 2√3) and (5 + √3, 3 - 2√3) are the two possibilities; the first is in the options.

Teacher's Note:
1. C lies on the perpendicular bisector at height (√3/2) × side.
2. There are always two such points, one on each side of AB.

Answer: (b) 83x - 35y + 92 = 0

Explanation:
1. Let the points be (p, 5p - 4) on the first line and (2 - p, 10 - 5p + 4) on the second.
2. Substituting in 3x + 4y - 4 = 0: 3(2 - p) + 4(14 - 5p) - 4 = 0 gives p = 58/23.
3. The line joins (1, 5) to (58/23, 198/23), with slope (198/23 - 5)/(58/23 - 1) = 83/35.
4. y - 5 = (83/35)(x - 1), i.e. 83x - 35y + 92 = 0.

Teacher's Note:
1. If a segment is bisected at (1, 5), the end points are symmetric about it.
2. Only one parameter is needed after using that symmetry.

Answer: (c) \( 3x-4y+6\sqrt2=0 \)

Explanation:
1. The longer diagonal bisects the small angle between the sides, so it lies along y = x and C = (6√2, 6√2).
2. The other sides pass through C parallel to the given sides.
3. Parallel to y = 3x/4: 4(y - 6√2) = 3(x - 6√2), i.e. 3x - 4y + 6√2 = 0. (The other is 4x - 3y - 6√2 = 0.)

Teacher's Note:
1. In a rhombus the diagonals bisect the angles at the vertices.
2. The other sides are parallel to the given sides through C.

Answer: (d) \( \frac{a_1x+b_1y}{c_1}+\frac{a_2x+b_2y}{c_2}+2=0 \)

Explanation:
1. Along direction (cos θ, sin θ): 1/OP = -(a1 cos θ + b1 sin θ)/c1 and 1/OQ = -(a2 cos θ + b2 sin θ)/c2.
2. With R = (r cos θ, r sin θ) and 2/r = 1/OP + 1/OQ, multiply by r: 2 = -(a1x + b1y)/c1 - (a2x + b2y)/c2.
3. So the locus is a straight line: (a1x + b1y)/c1 + (a2x + b2y)/c2 + 2 = 0.

Teacher's Note:
1. Write each distance along the line using the direction cosines.
2. Multiply by r to convert to x and y.

Answer: (c) \( \left(-\tfrac32,-1\right)\cup\left(\tfrac12,1\right) \)

Explanation:
1. The vertices are (-7, 5), (5/4, 7/8) and (1/3, 1/9); the centroid shows that inside the triangle 2x + 3y - 1 > 0, x + 2y - 3 < 0 and 5x - 6y - 1 < 0.
2. Substituting (α, α²): 3α² + 2α - 1 > 0 gives α < -1 or α > 1/3; 2α² + α - 3 < 0 gives -3/2 < α < 1; 6α² - 5α + 1 > 0 gives α < 1/3 or α > 1/2.
3. The intersection of the three conditions is (-3/2, -1) ∪ (1/2, 1).

Teacher's Note:
1. Find the sign of each line's expression at an interior point, such as the centroid.
2. Then require the same signs for the moving point and solve the inequalities.

Answer: (a) (1, 0)

Explanation:
1. Group the terms: (x - 1) + λy = 0.
2. This holds for all λ if x - 1 = 0 and y = 0, giving (1, 0).

Teacher's Note:
1. Write the equation as L1 + λL2 = 0; the fixed point is where L1 = L2 = 0.
2. Here L1 = x - 1 and L2 = y.

Answer: (d) \( \left(-\tfrac15,-\tfrac35\right) \)

Explanation:
1. The condition is a + 3b - 5c = 0, so c = (a + 3b)/5.
2. Substituting: ax + by + (a + 3b)/5 = 0, i.e. a(x + 1/5) + b(y + 3/5) = 0.
3. This holds for all a and b at (-1/5, -3/5).

Teacher's Note:
1. Eliminate c using the given relation.
2. Then the coefficients of a and b must both vanish.

Answer: (a) \( \left(\tfrac12,-\tfrac32\right)\text{ and }\left(-\tfrac12,\tfrac32\right) \)

Explanation:
1. a² - 6ab + 9b² = 4c² gives (a - 3b)² = (2c)², so a - 3b = ±2c.
2. Case a - 3b = 2c: c = (a - 3b)/2 and ax + by + c = a(x + 1/2) + b(y - 3/2) = 0, so the point is (-1/2, 3/2).
3. Case a - 3b = -2c: c = (3b - a)/2, giving a(x - 1/2) + b(y + 3/2) = 0, so the point is (1/2, -3/2).

Teacher's Note:
1. Recognise the perfect square on the left.
2. Each sign choice gives one fixed point.

Answer: (b) centroid of triangle ABC

Explanation:
1. Sum of signed distances = (a(x1 + x2 + x3) + b(y1 + y2 + y3) + 3c)/sqrt(a² + b²) = 0.
2. This is 3 × (a x_G + b y_G + c) = 0 with G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).
3. So the line passes through the centroid.

Teacher's Note:
1. The signed distance is linear in the point's coordinates.
2. The sum is zero exactly when the average point, the centroid, lies on the line.

Answer: (c) (2, -1)

Explanation:
1. Collect a and b: a(x + y - 1) + b(2x + 3y - 1) = 0.
2. This holds for all a, b if x + y = 1 and 2x + 3y = 1, giving y = -1 and x = 2.

Teacher's Note:
1. Treat a and b as independent parameters.
2. Solve the two coefficient equations.

Answer: (a) 5x + y + 2 = 0 and x + 3y = 0

Explanation:
1. The vertices are the intersections of the lines of different families: (-9/7, 31/7), (15/7, -5/7), (3/7, -29/7) and (-3, 1) taken around the figure.
2. The diagonals join opposite vertices. Joining (-9/7, 31/7) and (3/7, -29/7) gives 5x + y + 2 = 0; joining (15/7, -5/7) and (-3, 1) gives x + 3y = 0.

Teacher's Note:
1. A diagonal joins two opposite vertices; list the vertices in order around the figure.
2. The diagonals meet at the centre of the parallelogram.

Answer: (b) (-1, 7) and (10, -3)

Explanation:
1. Reflection in the y-axis changes the sign of x: (1, 7) goes to (-1, 7).
2. Reflection in the x-axis changes the sign of y: (10, 3) goes to (10, -3).

Teacher's Note:
1. A reflection in an axis flips the sign of the other coordinate.
2. Check which coordinate stays fixed.

Answer: (c) \( \left(\tfrac{11}{7},0\right) \)

Explanation:
1. A and B are on the same side of the x-axis, so reflect A to A'(1, -2).
2. P is where A'B meets the x-axis: x = 1 + 2t, y = -2 + 7t, with y = 0 at t = 2/7.
3. x = 1 + 4/7 = 11/7, so P = (11/7, 0).

Teacher's Note:
1. Minimum path via a mirror: reflect one point and join to the other.
2. Reflect only when both points are on the same side of the mirror.

Answer: (d) (2, -6)

Explanation:
1. X = x - 1 = 2 and Y = y - 2 = -6.

Teacher's Note:
1. New coordinates = old coordinates minus the new origin.
2. Slopes and distances do not change on shifting the origin.

Answer: (a) 2X - 3Y + 14 = 0

Explanation:
1. x = X + 3 and y = Y - 1.
2. 2(X + 3) - 3(Y - 1) + 5 = 0, so 2X - 3Y + 14 = 0.

Teacher's Note:
1. Substitute x = X + h and y = Y + k for a shift of origin to (h, k).
2. The slope of the line does not change.

Answer: (b) \( \left(\tfrac52,-1\right) \)

Explanation:
1. Complete the squares: (x - 5/2)² + (y + 1)² = 5 + 25/4 + 1.
2. So the new origin is (5/2, -1).

Teacher's Note:
1. Complete the square in x and in y.
2. The shifted origin is the centre of the circle.

Answer: (d) 4 sq. units

Explanation:
1. Area is unchanged when the origin is shifted.
2. Original area = ½ × 2 × 4 = 4 sq. units.

Teacher's Note:
1. Shifting the origin is a translation, which preserves area, slopes and lengths.
2. Compute the area in the original coordinates.

Answer: (a) XY = 0

Explanation:
1. xy - x - y + 1 = (x - 1)(y - 1).
2. With X = x - 1 and Y = y - 1 this is XY = 0.

Teacher's Note:
1. Factorise the expression before substituting.
2. The equation represents a pair of lines through the new origin.

Answer: (d) 21x - 27y + 2 = 0 and 99x + 77y + 28 = 0

Explanation:
1. Distances to the two lines are equal: (3x + 4y + 1)/5 = ±(12x + 5y + 3)/13.
2. + sign: 13(3x + 4y + 1) = 5(12x + 5y + 3) gives -21x + 27y - 2 = 0, i.e. 21x - 27y + 2 = 0.
3. - sign: 13(3x + 4y + 1) = -5(12x + 5y + 3) gives 99x + 77y + 28 = 0.

Teacher's Note:
1. Equate the perpendicular distances, taking both signs.
2. Normalise each equation by sqrt(a² + b²) first.

Answer: (b) 4x + y - 7 = 0

Explanation:
1. AB = sqrt(64 + 361) = √425 and AC = sqrt(256 + 169) = √425, so the triangle is isosceles with AB = AC.
2. The bisector at A passes through the mid-point of BC, which is (3, -5).
3. Through (-1, 11) and (3, -5): slope = -16/4 = -4, so y - 11 = -4(x + 1), i.e. 4x + y - 7 = 0.

Teacher's Note:
1. In an isosceles triangle the bisector from the apex is also the median.
2. Check AB and AC first.