JEE Mathematics Straight Lines MCQs Set 15

Answer: (d) x − 3y − 31 = 0

Explanation:
1. Slopes of the equal sides are 7 and −1. The third side makes equal angles with them, so it is perpendicular to a bisector of the angle between them.
2. Bisectors: (7x − y + 3)/√50 = ±(x + y − 3)/√2, that is 7x − y + 3 = ±5(x + y − 3). This gives 2x − 6y + 18 = 0 (slope 1/3) and 12x + 4y − 12 = 0 (slope −3).
3. The third side has slope −3 or 1/3. Through (1, −10) the slope −3 gives 3x + y + 7 = 0 and the slope 1/3 gives y + 10 = (1/3)(x − 1), that is x − 3y − 31 = 0.

Teacher's Note:
1. In an isosceles triangle the base is perpendicular to the bisector of the angle between the equal sides.
2. Two bisectors exist, so two answers are possible; only one is listed.

Answer: (b) 3

Explanation:
1. The triangle has base 8 (from (1, 1) to (9, 1)) and height 1, so area 4. Each region must have area 2.
2. For 1 < c < 9 the part to the right of x = c is a triangle with vertex (9, 1), bounded by y = 1, y = x/9 and x = c. Its height is 1 − c/9 and its base is 9 − c, so its area is (9 − c)²/18.
3. (9 − c)²/18 = 2 gives 9 − c = 6, so c = 3 (c = 15 is outside the triangle).

Teacher's Note:
1. Find which side of x = c has the simple shape: the right-hand piece is a triangle.
2. Reject any root that lies outside the triangle.

Answer: (b) 5x + 4y = 13

Explanation:
1. Reflect A in the x-axis: A′ = (1, −2). The reflected ray lies on the line through A′ and (5, 3).
2. Slope = 5/4. At y = 0: 0 + 2 = (5/4)(x − 1), so B = (13/5, 0).
3. AB: slope = (2 − 0)/(1 − 13/5) = −5/4, so y − 2 = (−5/4)(x − 1), i.e. 5x + 4y = 13.

Teacher's Note:
1. The image method turns a reflection into a straight-line problem.
2. Find B first, then the line through A and B.

Answer: (b) 83x − 35y + 92 = 0

Explanation:
1. Let the end on the first line be (p, 5p − 4). The other end is (2 − p, 14 − 5p) since (1, 5) is the midpoint.
2. It lies on 3x + 4y − 4 = 0: 3(2 − p) + 4(14 − 5p) − 4 = 0, so p = 58/23.
3. The line joins (1, 5) and (58/23, 198/23): slope = (83/23)/(35/23) = 83/35, so y − 5 = (83/35)(x − 1), i.e. 83x − 35y + 92 = 0.

Teacher's Note:
1. Use a parameter for one end and the midpoint formula to get the other end.
2. Always verify that (1, 5) satisfies the final equation: 83 − 175 + 92 = 0.

Answer: (c) \( 2\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}+\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_3&y_3&1\end{vmatrix}=0 \)

Explanation:
1. BC = 3BD means D = B + (C − B)/3 = (2B + C)/3, so D divides BC in the ratio 1 : 2 from B.
2. The line AD is the set of points P with det[P, A, D] = 0. Since the determinant is linear in the last row, det[P, A, D] = (2 det[P, A, B] + det[P, A, C])/3.
3. So the line is 2·det[x, y; A; B] + det[x, y; A; C] = 0, which is the last option.

Teacher's Note:
1. A determinant with a row equal to (2B + C)/3 splits linearly.
2. D is closer to B, so B carries the larger weight 2.

Answer: (b) 288

Explanation:
1. Area = ab/2 = 12, so ab = 24. The line passes through (2, 3): 2/a + 3/b = 1, so 2b + 3a = ab = 24.
2. 4b² + 9a² = (2b + 3a)² − 12ab = 576 − 288 = 288.
3. (Check that the line exists: 3a + 2b = 24 and ab = 24 give (a − 4)² = 0, so a = 4, b = 6.)

Teacher's Note:
1. Use (2b + 3a)² = 4b² + 9a² + 12ab.
2. Always check that the values of a and b are real.

Answer: (b) \( \left(1,-\frac{11}{3}\right) \)

Explanation:
1. Let O be the circumcentre and H the orthocentre. Then H − O = (A − O) + (B − O) + (C − O), so B + C − 2O = H − A.
2. The mid-point M of BC = O + (H − A)/2.
3. M = (−1/3 + (11/3 − 1)/2, 2/3 + (4/3 − 10)/2) = (−1/3 + 4/3, 2/3 − 13/3) = (1, −11/3).

Teacher's Note:
1. Taking the circumcentre as origin, the orthocentre is the sum of the three vertices.
2. Therefore AH = 2 × OM (vector), which gives the midpoint directly.

Answer: (c) x + y = a + b

Explanation:
1. Let the intersection be (x, y). From AB′: x/a + y/b′ = 1, so b′ = ay/(a − x). From A′B: x/a′ + y/b = 1, so a′ = bx/(b − y).
2. Impose a′ + b′ = a + b: bx/(b − y) + ay/(a − x) = a + b. Clearing denominators and factorising gives (bx + ay − ab)(x + y − a − b) = 0.
3. The factor bx + ay = ab is the line AB itself (the trivial case A′ = A), so the locus is x + y = a + b.

Teacher's Note:
1. Express a′ and b′ in terms of the intersection point.
2. Discard the factor that gives the fixed line AB.

Answer: (b) (cosα − sinα)x + (cosα + sinα)y = 4

Explanation:
1. Take O(0, 0), A(4cosα, 4sinα). The square lies above the x-axis, so the next vertices are B = A + 4(−sinα, cosα) and C = 4(−sinα, cosα).
2. The diagonal not through O is AC, joining (4cosα, 4sinα) and (−4sinα, 4cosα).
3. The line (cosα − sinα)x + (cosα + sinα)y = 4 satisfies both points: at A: 4cos²α − 4sinαcosα + 4sinαcosα + 4sin²α = 4, and at C: −4sinαcosα + 4sin²α + 4cos²α + 4sinαcosα = 4.

Teacher's Note:
1. Rotate the side vector OA by 90° to get the adjacent side direction.
2. Verify the final line by substituting both end points.

Answer: (b) with slope 2/3

Explanation:
1. Let P = (α, β) with 2α − 3β + 4 = 0 and let the centroid be (h, k).
2. h = (α + 4)/3 and k = (β + 2)/3, so α = 3h − 4 and β = 3k − 2.
3. 2(3h − 4) − 3(3k − 2) + 4 = 0 gives 6h − 9k + 2 = 0, a line of slope 2/3 (parallel to the original line).

Teacher's Note:
1. Express the moving point in terms of the centroid and substitute in its line.
2. The locus is a line parallel to the original line.

Answer: (a) \( \sqrt{20} \)

Explanation:
1. Line AB: x/8 + y/16 = 1, i.e. 2x + y − 16 = 0.
2. Distance from (2, 2) = |4 + 2 − 16|/√5 = 10/√5 = 2√5 = √20.

Teacher's Note:
1. Write the line in the form ax + by + c = 0 first.
2. Use d = |ax1 + by1 + c|/√(a² + b²).

Answer: (a) 11

Explanation:
1. P = (12, 0) and Q = (0, 9). Area of POQ = (1/2)(12)(9) = 54.
2. Area of TOP = 54/3 = 18 = (1/2)(OP)(s) = 6s, so s = 3.
3. T lies on the line: 3 = −3r/4 + 9, so r = 8. Hence r + s = 11.

Teacher's Note:
1. The area of triangle TOP depends only on the height s of T above the x-axis.
2. Use the line equation to get r.

Answer: (c) 18

Explanation:
1. Let the intercepts be a and b: 8/a + 2/b = 1.
2. OP + OQ = a + b = (a + b)(8/a + 2/b) = 10 + 8b/a + 2a/b ≥ 10 + 2√16 = 18 by AM-GM.
3. Equality holds when 8b/a = 2a/b, i.e. a = 2b, giving a = 12, b = 6. The minimum is 18.

Teacher's Note:
1. Use the intercept form through the given point and apply AM-GM to the product (a + b)(8/a + 2/b).
2. The same result follows from calculus.

Answer: (d) \( (-3,3) \)

Explanation:
1. Slope of BC = (−5 − 1)/(5 + 7) = −1/2, so the altitude from A has slope 2: y − 7 = 2(x + 1), i.e. y = 2x + 9.
2. Slope of AC = (−5 − 7)/(5 + 1) = −2, so the altitude from B has slope 1/2: y − 1 = (1/2)(x + 7), i.e. x − 2y + 9 = 0.
3. Intersection: x − 2(2x + 9) + 9 = 0 gives x = −3, y = 3.

Teacher's Note:
1. Altitudes are perpendicular to the opposite sides.
2. Two altitudes are enough to locate the orthocentre.

Answer: (d) (x² + y²)³ = 4R²x²y²

Explanation:
1. Angle AOB is a right angle, so AB is a diameter of length 2R: if A = (a, 0) and B = (0, b), then a² + b² = 4R².
2. The foot of the perpendicular from O on x/a + y/b = 1 is (h, k) = (ab²/(a² + b²), a²b/(a² + b²)).
3. Then h² + k² = a²b²/(a² + b²) and h²k² = a⁶b⁶/(a² + b²)⁴, so (h² + k²)³/(h²k²) = a² + b² = 4R², that is (h² + k²)³ = 4R²h²k².

Teacher's Note:
1. The chord AB is a diameter because the angle at O is a right angle.
2. Eliminate a and b using a² + b² = 4R².

Answer: (c) square of side 2√2 units

Explanation:
1. The boundary lines are x − y = ±2 and x + y = ±2. Two pairs of these are perpendicular (slopes 1 and −1).
2. The vertices are (2, 0), (0, 2), (−2, 0) and (0, −2). Each side has length √(4 + 4) = 2√2.
3. All sides are equal and the angles are right angles, so it is a square of side 2√2 and area 8.

Teacher's Note:
1. Find the vertices by intersecting adjacent boundary lines.
2. Perpendicular boundaries with equal sides give a square.

Answer: (d) \( \left(3,\frac43\right) \)

Explanation:
1. Equal areas of the three sub-triangles mean R is the centroid.
2. R = ((0 + 3 + 6)/3, (0 + 4 + 0)/3) = (3, 4/3).

Teacher's Note:
1. The centroid is the only interior point giving three equal areas with the vertices.
2. Add the coordinates and divide by 3.

Answer: (d) 5

Explanation:
1. Equal areas of the three triangles mean P is the centroid: P = ((1 + 6 + 3/2)/3, (0 + 2 + 6)/3) = (17/6, 8/3).
2. PQ² = (17/6 + 7/6)² + (8/3 + 1/3)² = 4² + 3² = 25.
3. PQ = 5.

Teacher's Note:
1. P is the centroid, which is the average of the vertices.
2. Use the distance formula.

Answer: (c) (4/3, 2/5)

Explanation:
1. Horizontal moves: 1, 1/4, 1/16, ... a geometric series with ratio 1/4, sum = 1/(1 − 1/4) = 4/3.
2. Vertical moves: +1/2, −1/8, +1/32, ... with ratio −1/4, sum = (1/2)/(1 + 1/4) = 2/5.
3. The limit point is (4/3, 2/5).

Teacher's Note:
1. Separate the horizontal and vertical displacements.
2. Use S = a/(1 − r) for each infinite geometric series.

Answer: (d) \( \left(\frac1k,\frac1k\right) \)

Explanation:
1. The line x/a + y/b = 1 with 1/a + 1/b = k.
2. Then x/a + y/b = 1 can be written as 1/a · x + (k − 1/a) y = 1, i.e. (1/a)(x − y) + (ky − 1) = 0.
3. This holds for every a if x − y = 0 and ky − 1 = 0, so the fixed point is (1/k, 1/k).

Teacher's Note:
1. Replace 1/b by k − 1/a to reduce the family to one parameter.
2. The family (1/a)L1 + L2 = 0 passes through the point L1 = L2 = 0.

Answer: (b) 5x − 7y = 0

Explanation:
1. A point on x = 2y is (2s, s). Its foot on x = y is (3s/2, 3s/2).
2. The midpoint of the perpendicular is ((2s + 3s/2)/2, (s + 3s/2)/2) = (7s/4, 5s/4).
3. With x = 7s/4 and y = 5s/4: x/y = 7/5, so 5x − 7y = 0.

Teacher's Note:
1. The foot of the perpendicular from (p, q) on y = x is ((p + q)/2, (p + q)/2).
2. Eliminate the parameter s.

Answer: (b) (2, 8/3)

Explanation:
1. M = (6, 0) and N = (0, 8).
2. Three equal areas means P is the centroid of triangle OMN.
3. P = ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3).

Teacher's Note:
1. A point that splits a triangle into three triangles of equal area is the centroid.
2. The centroid is the average of the three vertices.

Answer: (b) 5

Explanation:
1. Slope of 2x − 3y + 17 = 0 is 2/3, so the other line has slope −3/2.
2. (β − 17)/(15 − 7) = −3/2, so β − 17 = −12 and β = 5.

Teacher's Note:
1. Perpendicular slopes multiply to −1.
2. Use the two-point slope formula.

Answer: (b) 3

Explanation:
1. Incentre I = (6, 6). The circumcentre is the midpoint of the hypotenuse AB: (9, 12). The centroid is (6, 8).
2. Area = (1/2)|6(12 − 8) + 9(8 − 6) + 6(6 − 12)| = (1/2)|24 + 18 − 36| = 3.

Teacher's Note:
1. For a right triangle the circumcentre is the midpoint of the hypotenuse.
2. Apply the determinant formula for area.

Answer: (a) \( \left(-\frac14,\frac23\right) \)

Explanation:
1. Parallel lines: 4x − 3y + c = 0 with |c|/5 = 3/5, so c = ±3.
2. Test the points: for (−1/4, 2/3): 4x − 3y = −1 − 2 = −3, so 4x − 3y + 3 = 0 holds.
3. Other points give 4x − 3y = 0, 1, and 2, none equal to ±3 (negated).

Teacher's Note:
1. The distance of ax + by + c = 0 from the origin is |c|/√(a² + b²).
2. Compute 4x − 3y for each option and compare with −c = ∓3.