JEE Mathematics Straight Lines MCQs Set 24

Answer: (d) (i) and (iv)

Explanation:
1. Points at distance 5 from P that are farthest from the line lie along the normal direction \( \frac{(4,3)}{5} \): \( (2,3)\pm5\cdot\frac{(4,3)}5 \).
2. This gives \( (6,6) \) and \( (-2,0) \), each at distance 5 from the line.

Teacher's Note:
1. The farthest points lie along the normal to the line through P.
2. Move 5 units from P in both normal directions.

Answer: (a) (ii) and (iii)

Explanation:
1. The distance between the parallel lines is \( \frac{2}{\sqrt5} \), so the required line makes an angle with them whose tangent is \( \frac12 \) (since \( \sin\phi=\frac{2/\sqrt5}{2}=\frac1{\sqrt5} \)).
2. Slope of the lines is −2, so \( \left|\frac{m+2}{1-2m}\right|=\frac12 \): \( m=-\frac34 \) or the vertical line.
3. Lines: \( y-3=-\frac34(x-2) \) i.e. \( 3x+4y=18 \), and \( x=2 \).

Teacher's Note:
1. Use the angle between the required line and the given parallel lines.
2. Do not forget the vertical line, for which the slope formula fails.

Answer: (d) (7, −2), (4, 3)

Explanation:
1. The centroid is the intersection of the medians: G = (4, 1).
2. B = (x1, y1) lies on x + y = 5 and C = (x2, y2) lies on x = 4, so x2 = 4.
3. From the centroid: \( \frac{1+x_1+4}3=4\Rightarrow x_1=7,\ y_1=-2 \) and \( \frac{2-2+y_2}3=1\Rightarrow y_2=3 \).
4. So B = (7, −2) and C = (4, 3).

Teacher's Note:
1. The centroid lies on both medians.
2. Use the centroid formula with the known coordinate relations.

Answer: (d) \( (0,0) \)

Explanation:
1. The rays meet at P(0, 2) and are symmetric about the y-axis, so the bisector is the y-axis.
2. A is at distance \( \frac4{\sqrt3} \) from P along a ray of slope \( \mp\sqrt3 \): \( A=\left(\pm\frac2{\sqrt3},0\right) \).
3. The foot of the perpendicular from A on the y-axis is (0, 0).

Teacher's Note:
1. Sketch the two rays: a symmetric V-shape with vertex (0, 2).
2. The bisector is the axis of symmetry.

Answer: (c) 4x + 3y + 17 = 0

Explanation:
1. The side is 2 units from the centre. Sides parallel to 3x − 4y = 0: \( 3x-4y+k=0 \) with \( \frac{|3-4+k|}5=2 \), so \( k=11,\ -9 \).
2. Sides perpendicular: \( 4x+3y+k=0 \) with \( \frac{|4+3+k|}5=2 \), so \( k=3,\ -17 \).
3. So 4x + 3y + 17 = 0 is not a side (the correct one is 4x + 3y − 17 = 0).

Teacher's Note:
1. Half the side is 2 units, the distance from the centre to each side.
2. Treat the parallel and perpendicular pairs separately.

Answer: (b) x + 2y − 6 = 0

Explanation:
1. Let the vertex be A(4, −1). Its reflection in the bisector x = 1 lies on the opposite side: \( (-2,-1) \); its reflection in x − y − 1 = 0 also lies on that side: \( (0,3) \). So the third side is \( 2x-y+3=0 \).
2. The bisectors meet this side at the other two vertices: \( Q(1,5) \) and \( R(-4,-5) \).
3. Side AQ: \( 2x+y-7=0 \); side AR: \( x-2y-6=0 \).
4. So x + 2y − 6 = 0 is not a side.

Teacher's Note:
1. Reflect the vertex in each bisector to get points on the opposite sides.
2. Find the other two vertices from the bisector intersections.

Answer: (d) (ii), (iii) and (iv)

Explanation:
1. Line: \( y-2=m(x-8),\ m \lt 0 \): P = (8 − 2/m, 0), Q = (0, 2 − 8m).
2. OP + OQ \( =10+\frac2{(-m)}+8(-m)\ge10+2\sqrt{16}=18 \) (equality at m = −1/2), so (iii) is true and (i) is false.
3. Area is least when (8, 2) is the mid-point of PQ: P(16, 0), Q(0, 4), area \( \frac12\cdot16\cdot4=32 \) and slope \( -\frac4{16}=-\frac14 \). So (ii) and (iv) are true.

Teacher's Note:
1. Write P and Q in terms of the slope and apply AM-GM.
2. Minimum area occurs when the fixed point is the mid-point of the intercept.

Answer: (c) x = 3

Explanation:
1. The diagonals meet at \( P(\tfrac32,5) \). The sides are parallel to the bisectors of the angle between the diagonals: \( x=\tfrac32 \) and \( y=5 \).
2. With \( \tan2\theta=\left|\frac{8+8}{1-64}\right|=\frac{16}{63} \) we get \( \sin2\theta=\frac{16}{65},\cos2\theta=\frac{63}{65} \). Half-diagonal \( \ell \) satisfies \( \ell^2\sin2\theta=4 \), so \( \ell=\frac{\sqrt{65}}2 \).
3. Distances of the sides from P: \( \ell\cos\theta=4 \) and \( \ell\sin\theta=\frac12 \).
4. Sides: y = 5 ± 4, i.e. y = 1 and y = 9; x = 3/2 ± 1/2, i.e. x = 1 and x = 2. So x = 3 is not a side.

Teacher's Note:
1. The sides are parallel to the angle bisectors of the diagonals.
2. Use area = 4 x (area of the triangle formed with half diagonals).

Answer: (c) −16 or 10/3

Explanation:
1. The diagonals are the angle bisectors of the sides, so the centre (1, 2) is equidistant from the two sides.
2. Distances: \( \frac{|2+6-a+5|}{\sqrt{13}}=\frac{|3+4-4+2a|}{\sqrt{13}} \), i.e. \( |13-a|=|3+2a| \).
3. So \( 13-a=3+2a \) gives \( a=\frac{10}3 \) and \( 13-a=-3-2a \) gives \( a=-16 \).

Teacher's Note:
1. In a rhombus the centre is at equal distances from all sides.
2. Solve the modulus equation in both ways.

Answer: (d) (1, 1)

Explanation:
1. L1 meets the y-axis at A(0, 2). Let L2 be y = mx + 2. L1 meets x = 5 at B(5, 16/3); L2 meets it at D(5, 5m + 2) and the x-axis at E(−2/m, 0).
2. Area = area(ADB) − area(ECD), which gives \( \frac{110m+12}{6m}=\frac{49}3 \), so m = −1.
3. L2: \( x+y=2 \). Of the options, only (1, 1) satisfies it.

Teacher's Note:
1. The pivot point is A(0, 2).
2. Express the area through the unknown slope.

Answer: (c) (i) and (iii) only

Explanation:
1. Take A(0, 0), B(a, 0), C(a, a), D(0, a), M(a, a/2), P(a/2, a).
2. Area AMP = \( \frac{3a^2}8 \), MCP = \( \frac{a^2}8 \), ABM = ADP = \( \frac{a^2}4 \), AMD = \( \frac{a^2}2 \).
3. (i) true, (ii) false (1 : 4), (iii) true, (iv) false (AMCP = a^2/2, ratio 1 : 2).

Teacher's Note:
1. Place the square on coordinate axes.
2. Compute each area with the determinant formula.

Answer: (b) (4, 0)

Explanation:
1. The perpendicular bisector of AC through (3/2, 5/2) with slope 2 is \( 4x-2y-1=0 \).
2. C is the mirror image of A in this line: \( \frac{x+2}4=\frac{y-3}{-2}=-2\cdot\frac{-8-6-1}{20} \) gives \( C=(4,0) \).
3. (Similarly B = (2, −1); the area of ABC is 6 and its centroid is (4/3, 2/3).)

Teacher's Note:
1. The perpendicular bisector of a side is the mirror between its end points.
2. Reflect A in the bisector of AC.

Answer: (a) p − q = 17

Explanation:
1. The mid-point of BC is (35/2, 39/2): slope of the median \( \frac{39/2-q}{35/2-p}=-5 \) gives \( 5p+q=107 \).
2. Area 70: \( |11q-p-197|=140 \), i.e. \( 11q-p=337\text{ or }11q-p=57 \).
3. Solutions: (p, q) = (15, 32) or (20, 7).
4. p + q = 47 or 27; p − q = −17 or 13. So p − q = 17 is impossible.

Teacher's Note:
1. Write two linear equations in p and q: slope and area.
2. Solve both sign cases.

Answer: (b) 2x + 3y − 12 ≥ 0

Explanation:
1. A linear inequality holds on the whole triangle if it holds at all three vertices.
2. 3x + 2y: 9, 27, 1 (all positive). 2x + y + 1: 6, 17, 1. 2x + 11: 13, 21, 9.
3. 2x + 3y − 12: −1 at (1, 3) is negative, so this one fails.

Teacher's Note:
1. Test the vertices only.
2. If the signs differ at vertices, the inequality fails somewhere in the triangle.

Answer: (c) \( \left(-\frac{14}5,-\frac45\right) \)

Explanation:
1. A and B lie on the same side of the line, so reflect B: \( B'=\left(-\frac{118}{13},-\frac{44}{13}\right) \).
2. PA + PB is least when A, P, B' are collinear: line AB' is \( 7x-17y+6=0 \).
3. Intersecting with 3x + 2y + 10 = 0: \( P=\left(-\frac{14}5,-\frac45\right) \).

Teacher's Note:
1. Reflect one point in the line to remove the bend.
2. Intersect AB' with the line.

Answer: (a) \( (-22,28) \)

Explanation:
1. By the triangle inequality \( |PA-PB|\le AB \), with equality when P, A, B are collinear (P outside the segment).
2. Line AB: \( x+y-6=0 \).
3. Intersect with 3x + 2y + 10 = 0: \( P=(-22,28) \).

Teacher's Note:
1. Use |PA - PB| \lt = AB.
2. The maximum is reached when P lies on line AB.

Answer: (d) (ii) and (iii)

Explanation:
1. Points of the line: \( x=\sqrt3+\frac R2,\ y=\frac{\sqrt3}2R \). Substituting: \( 3R^2-2R-4(2+\sqrt3)=0 \).
2. Roots R1, R2 (signed distances): \( R_1+R_2=\frac23,\ R_1R_2=-\frac{4(2+\sqrt3)}3 \).
3. The roots have opposite signs (A, B on opposite sides of P), so \( |PA-PB|=|R_1+R_2|=\frac23 \) and \( PA\cdot PB=\frac{4(2+\sqrt3)}3 \).
4. PA + PB = |R1 − R2| is not 2/3, and (iv) is not equal to the stated value.

Teacher's Note:
1. Use the parametric form of the line and the quadratic in the distance R.
2. Opposite-sign roots mean the distances are the absolute values.

Answer: (b) u = 0 and v = 0

Explanation:
1. The two lines have the same normal-length factor, since u and v have equal norms \( \sqrt{a^2+b^2} \) (the normals (a, b) and (b, −a) are perpendicular of equal length).
2. Bisectors: \( \frac{k_1u-k_2v}{N}=\pm\frac{k_1u+k_2v}{N} \) with the same N, so the plus sign gives \( v=0 \) and the minus sign gives \( u=0 \).

Teacher's Note:
1. The normalising factors of the two lines are the same.
2. Take both signs in the bisector formula.

Answer: (a) centroid and incentre only

Explanation:
1. The vertices are the pairwise intersections of the three lines.
2. The angle at the intersection of 7y = x and sqrt3 y + x = 0 has \( \tan C=\frac{m_3-m_1}{1+m_3m_1} \) negative, so that angle is obtuse (about 127 degrees).
3. In an obtuse triangle the circumcentre and the orthocentre lie outside; the centroid and incentre are always inside.

Teacher's Note:
1. Find whether the triangle is acute or obtuse.
2. Remember where the centres lie for an obtuse triangle.

Answer: (a) y = x² and y = 2 − x² together (both curves)

Explanation:
1. For the intersection (h, k): \( m_1=\frac{k-1}{h-1},\ m_2=\frac{k-1}{h+1} \).
2. If \( m_1-m_2=2 \): \( \frac{2(k-1)}{h^2-1}=2\Rightarrow k=h^2 \).
3. If \( m_1-m_2=-2 \): \( k=2-h^2 \).
4. The condition 'difference is 2' is satisfied in either order, so both curves are included.

Teacher's Note:
1. Write the slopes using the intersection point.
2. The difference can be +2 or -2 depending on which line is which.

Answer: (d) The diagonals intersect at (−2, 2)

Explanation:
1. \( y^2-4y+3=(y-1)(y-3) \) and \( x^2+4xy+4y^2-5x-10y+4=(x+2y-1)(x+2y-4) \).
2. Vertices: A(−1, 1), B(2, 1), C(−2, 3), D(−5, 3); the parallelogram has base 3 and height 2, area 6.
3. Diagonals: \( \sqrt5 \) and \( \sqrt{53} \); their common mid-point is (−3/2, 2), not (−2, 2).

Teacher's Note:
1. Factorise both pairs and find the four vertices.
2. Parallel opposite sides give a parallelogram, whose diagonals bisect each other.