JEE Mathematics Straight Lines MCQs Set 14

Answer: (d) \( \left(-7,\frac18\right) \)

Explanation:
1. A lies on the perpendicular bisector of the base. The midpoint is (−1/2, 5) and the base has slope −4/3, so the bisector has slope 3/4: 6x − 8y + 43 = 0.
2. Test the options: (−1/2, 5): −3 − 40 + 43 = 0 satisfies it, but it is the midpoint of the base, so it gives a degenerate (collinear) triangle. (−1/8, 5): −3/4 − 40 + 43 ≠ 0. (5/6, −5): 5 + 40 + 43 ≠ 0.
3. (−7, 1/8): −42 − 1 + 43 = 0, so it lies on the bisector and is a genuine vertex.

Teacher's Note:
1. The apex of an isosceles triangle lies on the perpendicular bisector of the base but cannot be the midpoint of the base.
2. Substitute each option in the equation of the bisector.

Answer: (c) 14x + 23y = 40

Explanation:
1. B is the reflection of A in x − y + 5 = 0: d = (1 + 2 + 5)/2 = 4, so B = (1 − 2·4, −2 + 2·4) = (−7, 6).
2. C is the reflection of A in x + 2y = 0: d = (1 − 4)/5 = −3/5, so C = (1 + 6/5, −2 + 12/5) = (11/5, 2/5).
3. Slope of BC = (2/5 − 6)/(11/5 + 7) = (−28/5)/(46/5) = −14/23, so the line is 14x + 23y = 40 (check: −98 + 138 = 40).

Teacher's Note:
1. A point on a perpendicular bisector is equidistant from the two vertices, so B and C are reflections of A in the given bisectors.
2. Reflection of (p, q) in ax + by + c = 0 is (p − 2ad/(a²+b²), q − 2bd/(a²+b²)) with d = ap + bq + c.

Answer: (d) 26x − 122y − 1675 = 0

Explanation:
1. The vertex on both given sides is A = (10/23, 84/23). Let B be on the first side and C on the second.
2. The altitude from B passes through H(1, 1) and is perpendicular to the second side (slope −4/5), so BH has slope 5/4: B is on 3x − 2y + 6 = 0 and y − 1 = (5/4)(x − 1), giving B = (−13, −33/2). Similarly the altitude from C has slope −2/3 (perpendicular to the first side of slope 3/2), giving C = (35/2, −10).
3. Line BC: slope = (−10 + 33/2)/(35/2 + 13) = (13/2)/(61/2) = 13/61, so y + 10 = (13/61)(x − 35/2), which simplifies to 26x − 122y − 1675 = 0.

Teacher's Note:
1. Each altitude through the orthocentre is perpendicular to the side opposite the vertex.
2. Find the two unknown vertices and join them.

Answer: (d) (a, b)

Explanation:
1. Solving x + 2y = 9 and 3x − 5y = 5 gives x = 5, y = 2.
2. The third line passes through (5, 2), so 5a + 2b = 1.
3. The line 5x + 2y = 1 is therefore satisfied by (a, b).

Teacher's Note:
1. Find the common point of the first two lines and substitute in the third.
2. Compare the resulting relation with the line 5x + 2y = 1.

Answer: (b) (0, 2√3)

Explanation:
1. OA² = 9 + 3 = 12. Let P = (h, k). Then OP² = 12 and AP² = 12.
2. OP² = AP² gives h² + k² = (h − 3)² + (k − √3)², so 3h + √3k = 6, h = 2 − k/√3.
3. Substituting: (2 − k/√3)² + k² = 12 gives k² − √3k − 6 = 0, so k = 2√3 or k = −√3. Then P = (0, 2√3) or (3, −√3).

Teacher's Note:
1. Equal sides give two conditions: OP² = OA² and AP² = OA².
2. Both positions of P (on either side of OA) satisfy the equations.

Answer: (a) (0, a)

Explanation:
1. The line x − √3y = 0 has inclination 30° (or 210° in the opposite direction). The second vertex is at distance a along it, at angle 30° (or 210°).
2. The third vertex is at distance a from the origin at an angle 60° away: 30° + 60° = 90° gives (0, a); 30° − 60° = −30° gives (√3a/2, −a/2); and similarly on the other side (−√3a/2, a/2) and (0, −a).
3. Among the options, only (0, a) is one of these positions; (a, 0), (a/2, a) and (√3a, a) are not.

Teacher's Note:
1. The two other vertices of an equilateral triangle with a vertex at O are at the same distance a from O and 60° apart.
2. List all possible positions by adding ±60° to the direction of the line.

Answer: (b) \( \frac{1-\sqrt7}{1+\sqrt7} \)

Explanation:
1. Let the line have inclination θ. Points at distance 4: (2 + 4cosθ, 3 + 4sinθ). On x + y = 7: 5 + 4(cosθ + sinθ) = 7, so cosθ + sinθ = 1/2.
2. Squaring: 1 + sin2θ = 1/4, so sin2θ = −3/4. Using tanθ = m: 2m/(1 + m²) = −3/4 gives 3m² + 8m + 3 = 0.
3. m = (−4 ± √7)/3. Now (1 − √7)/(1 + √7) = (1 − √7)²/(1 − 7) = (8 − 2√7)/(−6) = (−4 + √7)/3, which matches the root m = (−4 + √7)/3.

Teacher's Note:
1. Use the distance (parametric) form of the line through P.
2. Rationalise the option to compare with the root of the quadratic.

Answer: (c) The centroid

Explanation:
1. The vertices are (0, 0), (7/8, 1/8) and (2.366, −1.366) (intersections of the lines in pairs).
2. The angle at (7/8, 1/8) is about 149°, so the triangle is obtuse. The circumcentre and the orthocentre of an obtuse triangle lie outside it.
3. The centroid always lies inside the triangle.

Teacher's Note:
1. For an obtuse triangle the circumcentre and orthocentre are outside the triangle.
2. The centroid is the average of the vertices, so it is always interior.

Answer: (d) \( \left(\frac{11}{5},\frac{28}{5}\right) \)

Explanation:
1. L: x/3 + y/1 = 1, that is x + 3y − 3 = 0.
2. For (−1, −4): d = (−1 − 12 − 3)/(1 + 9) = −16/10 = −8/5.
3. Image = (−1 − 2(1)(−8/5), −4 − 2(3)(−8/5)) = (−1 + 16/5, −4 + 48/5) = (11/5, 28/5).

Teacher's Note:
1. Image of (p, q) in ax + by + c = 0: subtract 2a·d and 2b·d with d = (ap + bq + c)/(a² + b²).
2. Check that the midpoint of the point and its image lies on the line.

Answer: (c) N

Explanation:
1. The centroid is (h, k) = ((6 + x)/3, (−6 + y)/3) where R = (x, y).
2. So x = 3h − 6 and y = 3k + 6. Substituting in 9x + 7y + 4 = 0: 27h − 54 + 21k + 42 + 4 = 0, i.e. 27h + 21k − 8 = 0.
3. This is parallel to 9x + 7y = 0, hence parallel to N.

Teacher's Note:
1. The centroid is an affine function of R, so its locus is parallel to the path of R.
2. The line 27x + 21y − 8 = 0 has the same slope as N.

Answer: (b) \( 3\sqrt{\frac52} \)

Explanation:
1. The centroid divides the line from orthocentre H to circumcentre C in the ratio 2 : 1, so G = (H + 2C)/3 and C = (3G − H)/2.
2. C = ((9 + 3)/2, (9 − 5)/2) = (6, 2).
3. AC = √(81 + 9) = √90 = 3√10, so the radius is 3√10/2 = 3√(5/2).

Teacher's Note:
1. Euler line: G divides HC in the ratio 2 : 1 measured from H.
2. The radius of a circle on AC as diameter is half of AC.

Answer: (d) −4

Explanation:
1. The point (0, −4) lies on the perpendicular bisector, so it is equidistant from P and Q.
2. 1² + (−4 − 4)² = k² + (−4 − 3)², so 65 = k² + 49 and k² = 16.
3. k = ±4. Among the options, k = −4 is the only value of k that satisfies this.

Teacher's Note:
1. Every point of a perpendicular bisector is equidistant from the end points.
2. Use the y-intercept point (0, −4).

Answer: (d) 50

Explanation:
1. A = (1, −2). PB is perpendicular to CA (direction (1, 1)): with B = (t, −2t), (2 − t) + (3 + 2t) = 0, so t = −5 and B = (−5, 10).
2. PC is perpendicular to AB (direction (1, −2)): with C = (3 + s, s), (−1 − s) − 2(3 − s) = 0, so s = 7 and C = (10, 7).
3. BC: −5 + 10p = q and 10 + 7p = q give p = 5, q = 45. Hence p + q = 50.

Teacher's Note:
1. The orthocentre lies on each altitude, so join it to a vertex and use perpendicularity with the opposite side.
2. Check by substituting B and C into x + py = q.

Answer: (a) 4x − 3y + 24 = 0

Explanation:
1. P is the midpoint of the intercepts, so the intercepts are 2(−3) = −6 on the x-axis and 2(4) = 8 on the y-axis.
2. The line is x/(−6) + y/8 = 1, i.e. 4x − 3y + 24 = 0 after multiplying by −24.

Teacher's Note:
1. The midpoint of (a, 0) and (0, b) is (a/2, b/2).
2. Use the intercept form.

Answer: (c) \( 6\sqrt2 \)

Explanation:
1. A = (18, 0) and B = (0, 24), so OA = 18, OB = 24 and AB = 30.
2. Inradius r = (18 + 24 − 30)/2 = 6, so the incentre is (6, 6). The orthocentre of a right triangle is the right-angle vertex O = (0, 0).
3. Distance = √(36 + 36) = 6√2.

Teacher's Note:
1. For a right triangle the orthocentre is at the right-angle vertex.
2. The incentre is (r, r) when the right angle is at the origin.

Answer: (c) (−9, −6)

Explanation:
1. C = ((3 + 1 + 2)/3, (−1 + 3 + 4)/3) = (2, 2).
2. Solving x + 3y = 1 and 3x − y = −1: x = −1/5, y = 2/5, so P = (−1/5, 2/5).
3. Slope of CP = (2 − 2/5)/(2 + 1/5) = (8/5)/(11/5) = 8/11. Check (−9, −6): slope from C = (−6 − 2)/(−9 − 2) = 8/11, so it lies on the line.

Teacher's Note:
1. Find C and P, then test each option by slope.
2. A point lies on line CP if its slope from C equals the slope of CP.

Answer: (c) 5x + y + 17 = 0

Explanation:
1. B is the intersection of the altitude and the bisector: B = (−3, −2).
2. The reflection of A in the bisector 3x − 2y + 5 = 0 lies on BC. For A(2, −1): d = (6 + 2 + 5)/13 = 1, so A′ = (2 − 6, −1 + 4) = (−4, 3).
3. BC passes through B(−3, −2) and A′(−4, 3): slope −5, so y + 2 = −5(x + 3), that is 5x + y + 17 = 0.

Teacher's Note:
1. The reflection of a vertex in an angle bisector lies on the other side through that vertex.
2. The altitude from B only locates B here.

Answer: (a) Second

Explanation:
1. Let the third vertex be (p, q). The altitude from (0, 2) passes through O and is perpendicular to the side joining (4, 3) and (p, q): (0, −2) · (p − 4, q − 3) = 0, so q = 3.
2. The altitude from (4, 3) passes through O: (−4, −3) · (p − 0, q − 2) = 0, so 4p + 3(q − 2) = 0, i.e. 4p + 3 = 0 and p = −3/4.
3. The vertex (−3/4, 3) lies in the second quadrant.

Teacher's Note:
1. Each altitude through the origin is perpendicular to the opposite side.
2. Use dot products of the altitude direction with the opposite side.

Answer: (a) k = −9 or k = −6/5

Explanation:
1. L1 has slope −1/3 and L3 has slope −5/2, so L1 and L3 are not parallel.
2. L2 has slope 3/k. L2 ∥ L1 when 3/k = −1/3, so k = −9.
3. L2 ∥ L3 when 3/k = −5/2, so k = −6/5.

Teacher's Note:
1. Compare slopes pairwise.
2. L1 and L3 are never parallel, so only L2 can be parallel to one of them.

Answer: (a) 6

Explanation:
1. Points of the line: (√3 + r cos 30°, 2 + r sin 30°) = (√3 + (√3/2)r, 2 + r/2).
2. Substitute in √3x − 4y + 8 = 0: 3 + (3/2)r − 8 − 2r + 8 = 0, so 3 − r/2 = 0 and r = 6.
3. PQ = 6.

Teacher's Note:
1. Use the parametric (distance) form of a line, where r is the distance from the fixed point.
2. Substitute the general point in the second line to find r.

Answer: (a) 780

Explanation:
1. Interior points satisfy x ≥ 1, y ≥ 1 and x + y ≤ 40.
2. For x = 1, 2, ..., 39 the number of y values is 40 − x, so the total is 39 + 38 + ... + 1.
3. Sum = (39)(40)/2 = 780.

Teacher's Note:
1. Strict interior means x > 0, y > 0 and x + y < 41, which for integers is x + y ≤ 40.
2. Count column by column and add the arithmetic series.

Answer: (d) \( \sqrt3y=x-\sqrt3 \)

Explanation:
1. The ray meets the x-axis (y = 0) at (√3, 0).
2. The reflected ray passes through the same point with the slope reversed. The incident slope is −1/√3, so the reflected slope is +1/√3.
3. y = (1/√3)(x − √3), that is √3y = x − √3.

Teacher's Note:
1. On reflection from the x-axis the slope changes sign and the point of incidence stays on the line.
2. Replace y by −y in the equation of the incident ray.

Answer: (d) 30

Explanation:
1. Write the lines as 2x − y + α/2 = 0 and 2x − y + β/3 = 0, all parallel to 2x − y + 3 = 0.
2. Distance conditions: |α/2 − 3|/√5 = 1/√5 gives α/2 = 4 or 2, so α = 8 or 4. |β/3 − 3|/√5 = 2/√5 gives β/3 = 5 or 1, so β = 15 or 3.
3. Sum of all possible values = 8 + 4 + 15 + 3 = 30.

Teacher's Note:
1. Make the coefficients of x and y equal before using the distance between parallel lines.
2. Each condition gives two values because of the modulus.

Answer: (d) 7 and 31/9

Explanation:
1. A = ((3K − 5)/(K + 1), (5K + 1)/(K + 1)).
2. The determinant of A, B, C (rows x, y, 1) simplifies to 2(7K − 33)/(K + 1), so Area = |7K − 33|/|K + 1|.
3. Area = 2 gives 7K − 33 = 2(K + 1) or 7K − 33 = −2(K + 1), that is K = 7 or K = 31/9.

Teacher's Note:
1. Write the area determinant with A in terms of K and clear the denominator (K + 1).
2. The two values come from the two signs of the determinant.

Answer: (b) \( \sqrt{\frac25} \)

Explanation:
1. Slopes: −1/(a − 1) and −2/a². Perpendicular means (1)(2) + (a − 1)(a²) = 0, so a³ − a² + 2 = 0, i.e. (a + 1)(a² − 2a + 2) = 0 and a = −1 is the only real root.
2. The lines are x − 2y = 1 and 2x + y = 1, meeting at (3/5, −1/5).
3. Distance from the origin = √(9/25 + 1/25) = √(10)/5 = √(2/5).

Teacher's Note:
1. Use a1a2 + b1b2 = 0 for perpendicular lines.
2. Factor the cubic by trying a = −1.