JEE Mathematics Straight Lines MCQs Set 23

Answer: (b) (3, 3)

Explanation:
1. The line meets the axes at A(0, 2) and B(4, 0); the mid-point of AB is (2, 1) and \( AB=2\sqrt5 \).
2. The centre lies at distance \( \frac{AB}2=\sqrt5 \) from AB along the normal pointing away from the origin, direction \( \frac{(1,2)}{\sqrt5} \).
3. Centre: \( (2,1)+(1,2)=(3,3) \).

Teacher's Note:
1. The centre of a square is at half the side from each side.
2. Move from the mid-point of AB along the unit normal.

Answer: (d) 2

Explanation:
1. The line \( x\sin\alpha-y\cos\alpha=0 \) passes through the origin and is perpendicular to \( x\cos\alpha+y\sin\alpha=p \), so the common point F is the foot of the perpendicular from O, with \( OF=p \).
2. The line \( ax+by+p=0 \) passes through F and makes an angle \( \pi/4 \) with OF's perpendicular, so it makes \( \pi/4 \) with OF as well. Its distance from O is \( p\sin\frac\pi4=\frac p{\sqrt2} \).
3. Also this distance is \( \frac{p}{\sqrt{a^2+b^2}} \), so \( a^2+b^2=2 \).

Teacher's Note:
1. Identify the point common to the three lines.
2. Compute the distance from the origin to the line in two ways.

Answer: (d) \( \alpha\in[-1,3],\ \beta\in[-2,3] \)

Explanation:
1. The vertices are \( (-\tfrac{15}7,-\tfrac{24}7),\ (\tfrac{60}{17},\tfrac6{17}),\ (0,3) \).
2. On the x-axis the triangle spans from the line 3x − y + 3 = 0 (x = −1) to the line 2x − 3y − 6 = 0 (x = 3), so \( \alpha\in[-1,3] \).
3. On the y-axis it spans from 2x − 3y − 6 = 0 (y = −2) up to the vertex (0, 3), so \( \beta\in[-2,3] \).

Teacher's Note:
1. Find the three vertices of the triangle.
2. Intersect the triangle with each axis.

Answer: (c) \( -\frac{4\sqrt2}{3} \lt t \lt \frac{5\sqrt2}{6} \)

Explanation:
1. P lies between the lines when \( x+2y-1 \) and \( 2x+4y-15 \) have opposite signs.
2. On the first line: \( 5+\frac{3t}{\sqrt2}-1=0\Rightarrow t=-\frac{4\sqrt2}3 \). On the second: \( 10+\frac{6t}{\sqrt2}-15=0\Rightarrow t=\frac{5\sqrt2}6 \).
3. Since the first expression is increasing in t and the second is increasing in t, the product is negative exactly for \( -\frac{4\sqrt2}3 \lt t \lt \frac{5\sqrt2}6 \).

Teacher's Note:
1. Substitute P in both line expressions.
2. Between the lines the two expressions have opposite signs.

Answer: (c) \( (-2\sqrt6,\ 2\sqrt2) \)

Explanation:
1. After (i): (1, 4). After (ii): (4, 4).
2. The point (4, 4) is at distance \( 4\sqrt2 \) from O at angle 45 degrees; after rotating by 105 degrees the angle is 150 degrees.
3. Final: \( (4\sqrt2\cos150^\circ,\ 4\sqrt2\sin150^\circ)=(-2\sqrt6,\ 2\sqrt2) \).

Teacher's Note:
1. Apply the transformations one at a time.
2. Use polar form for the rotation.

Answer: (b) \( \left(\frac{21}{17},-\frac{37}{17}\right) \)

Explanation:
1. A and B are on the same side of the line (2x − 3y − 9 is negative for both), so reflect B in the line: \( B'=\left(\frac{84}{13},-\frac{74}{13}\right) \).
2. The perimeter is least when A, P, B' are collinear, so P is where AB' meets the line.
3. AB' has slope \( -\frac{74}{110}=-\frac{37}{55} \); intersecting with 2x − 3y = 9 gives \( P=\left(\frac{21}{17},-\frac{37}{17}\right) \).

Teacher's Note:
1. AB is fixed, so minimise AP + PB.
2. Use the reflection of one point in the line.

Answer: (d) \( \left(0,-\frac12\right) \)

Explanation:
1. The reflected ray from (3, 0) comes from the image of (1, 4) in the x-axis, (1, −4): its line is \( y=2x-6 \).
2. It meets x + y = 5 at \( C=\left(\frac{11}3,\frac43\right) \).
3. The image of (3, 0) in x + y = 5 is (5, 2), and the second reflected ray passes through C and (5, 2): \( x-2y=1 \).
4. It meets the y-axis at \( P=\left(0,-\frac12\right) \).

Teacher's Note:
1. Use images of the source point in each mirror.
2. Find the second reflected line through the meeting point and an image point.

Answer: (a) \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \)

Explanation:
1. Let A = (α, 0), B = (0, β) with \( \alpha^2+\beta^2=(a+b)^2 \).
2. P divides AB in the ratio b : a, so \( h=\frac{a\alpha}{a+b},\ k=\frac{b\beta}{a+b} \).
3. Hence \( \alpha=\frac{h(a+b)}a,\ \beta=\frac{k(a+b)}b \) and \( \frac{h^2}{a^2}+\frac{k^2}{b^2}=1 \).

Teacher's Note:
1. Use the section formula for P.
2. Eliminate alpha and beta using the length of AB.

Answer: (a) a square

Explanation:
1. The condition says the larger of |x| and |y| equals |a|.
2. If |x| ≥ |y| it gives x = ±a with |y| ≤ |a|; if |y| ≥ |x| it gives y = ±a with |x| ≤ |a|.
3. These four segments form the boundary of a square with vertices (±a, ±a).

Teacher's Note:
1. Split into the cases |x| \gt = |y| and |y| \gt = |x|.
2. Each case gives two segments.

Answer: (b) \( x^{2/3}+y^{2/3}=c^{2/3} \)

Explanation:
1. Let A = (a, 0), B = (0, b) with \( a^2+b^2=c^2 \) and P = (a, b). AB: \( bx+ay-ab=0 \).
2. The foot L = (x1, y1) of the perpendicular from P satisfies \( x_1=\frac{a^3}{c^2},\ y_1=\frac{b^3}{c^2} \), i.e. \( a^3=c^2x_1,\ b^3=c^2y_1 \).
3. Then \( (c^2x_1)^{2/3}+(c^2y_1)^{2/3}=c^2 \), which gives \( x_1^{2/3}+y_1^{2/3}=c^{2/3} \).

Teacher's Note:
1. Compute the foot of the perpendicular from (a, b) on bx + ay = ab.
2. Use a^2 + b^2 = c^2 to eliminate a and b.

Answer: (c) \( \frac xa-\frac yb=0 \)

Explanation:
1. Let PQ be \( \frac xa+\frac y b=\lambda \) so P = (aλ, 0), Q = (0, bλ), A = (a, 0), B = (0, b).
2. PB: \( bx+a\lambda y=a b\lambda \); AQ: \( b\lambda x+ay=ab\lambda \).
3. Solving gives \( h=\frac{a\lambda}{\lambda+1},\ k=\frac{b\lambda}{\lambda+1} \), so \( \frac ha=\frac kb \).

Teacher's Note:
1. Write P and Q using a parameter.
2. Eliminate the parameter from the coordinates of the intersection.

Answer: (b) x + y = 0

Explanation:
1. Let AC be \( y=-2x+\lambda \): A = (λ/2, 0), C = (0, λ), mid-point \( M=(\frac\lambda4,\frac\lambda2) \).
2. BD is the perpendicular through M, slope 1/2; it meets y = x at \( B=(\frac{3\lambda}4,\frac{3\lambda}4) \).
3. M is the mid-point of BD: \( D=2M-B=\left(-\frac\lambda4,\frac\lambda4\right) \), so \( x+y=0 \).

Teacher's Note:
1. Diagonals of a rhombus are perpendicular and bisect each other.
2. Find B first, then reflect it through the centre.

Answer: (c) \( (y-x)^2+(3x-y)^2=\frac{a^2}4 \)

Explanation:
1. Let the AB make angle θ: A = (0, a sinθ), B = (a cosθ, 0), C = (a cosθ + a sinθ, a cosθ).
2. In the right isosceles triangle AEF, \( AF=\frac{a}{2\sqrt2} \) and \( AC=a\sqrt2 \), so AF : FC = 1 : 3.
3. Foot F: \( x=\frac{a(\sin\theta+\cos\theta)}4,\ y=\frac{a(3\sin\theta+\cos\theta)}4 \), giving \( \sin\theta=\frac{2(y-x)}a,\ \cos\theta=\frac{6x-2y}a \).
4. Using \( \sin^2\theta+\cos^2\theta=1 \): \( (y-x)^2+(3x-y)^2=\frac{a^2}4 \).

Teacher's Note:
1. Note that AF = AE cos 45 and AC = a sqrt2.
2. Use sin^2 + cos^2 = 1 to eliminate theta.

Answer: (c) (5, 4)

Explanation:
1. AH is vertical (x = 4), so BC is horizontal: BC is y = 4, and C = (c, 4).
2. BH has slope \( \frac{6-4}{4+2}=\frac13 \), so AC has slope −3: \( y-7=-3(x-4) \).
3. Put y = 4: \( x=5 \), so C = (5, 4).

Teacher's Note:
1. The altitude from A is perpendicular to BC.
2. The altitude from B is perpendicular to AC.

Answer: (c) (10, 1)

Explanation:
1. For C = (10, 1): \( BC=14,\ CA=15,\ AB=13 \).
2. Incentre \( =\frac{14(1,13)+15(-4,1)+13(10,1)}{42}=\frac{(14-60+130,\ 182+15+13)}{42}=(2,5) \).
3. So C = (10, 1) (it is the only option giving the incentre (2, 5)).

Teacher's Note:
1. Use the incentre formula with side lengths.
2. Test the options.

Answer: (b) y + x = 0

Explanation:
1. The mid-point of AB is (−q, 0) and of AC is (−p, 0); the slopes are \( \frac{1+p}{p} \) and \( \frac{1+q}{q} \).
2. Perpendicular bisectors: \( (1+p)y+p(x+q)=0 \) and \( (1+q)y+q(x+p)=0 \).
3. Subtracting: \( y(p-q)+(p-q)x=0 \), so \( x+y=0 \) (indeed the circumcentre is (pq, −pq)).

Teacher's Note:
1. Find the perpendicular bisectors of AB and AC.
2. Subtract to eliminate the parameters.

Answer: (a) (x−1)² − (y−1)² = 10 or (y−1)² − (x−1)² = 10

Explanation:
1. For the centre (h, k), the distances to the two sides are \( \frac{|h-k|}{\sqrt2} \) and \( \frac{|h+k-2|}{\sqrt2} \).
2. Area = 4 times the product of these distances = 20, so \( |h-k|\,|h+k-2|=10 \).
3. Since \( (h-k)(h+k-2)=(h-1)^2-(k-1)^2 \), the locus is \( (x-1)^2-(y-1)^2=\pm10 \).

Teacher's Note:
1. The centre is at half the side lengths from the sides.
2. Write the product of the two distances as a difference of squares.

Answer: (c) 2 sq. units

Explanation:
1. \( x^2-y^2+2y-1=x^2-(y-1)^2=0 \) gives the lines \( x=\pm(y-1) \), which meet at (0, 1) and have slopes ±1.
2. Their bisectors are \( x=0 \) and \( y=1 \).
3. With x + y = 3 the vertices are (0, 1), (0, 3), (2, 1): area \( \frac12\cdot2\cdot2=2 \).

Teacher's Note:
1. Factorise the pair of lines.
2. The bisectors of lines with slopes +1 and -1 are the vertical and horizontal lines through their meeting point.

Answer: (b) xy + 3x² = 0

Explanation:
1. Factorise: \( xy-3y^2+y-2x+10=(y-2)(x-3y-5) \); check: \( xy-3y^2-5y-2x+6y+10 \).
2. The lines are y = 2 and x − 3y = 5. Perpendiculars through the origin: x = 0 and \( y=-3x \).
3. Joint equation: \( x(3x+y)=0 \), i.e. \( 3x^2+xy=0 \).

Teacher's Note:
1. Factorise the given second degree equation.
2. A line perpendicular to a given one has its slope as the negative reciprocal.

Answer: (c) 6x² − xy − y² − x − 12y − 35 = 0

Explanation:
1. Factorise: \( 6x^2-xy-y^2+x+12y-35=(2x-y+5)(3x+y-7) \).
2. The centre of the parallelogram is the origin, so the other two sides are the reflections of these lines in the origin: \( 2x-y-5=0 \) and \( 3x+y+7=0 \).
3. Combined: \( (2x-y-5)(3x+y+7)=6x^2-xy-y^2-x-12y-35 \).

Teacher's Note:
1. Find the two given lines by factorising.
2. Reflect each line in the origin (change the sign of the constant term).

Answer: (d) a rectangle which is not a square

Explanation:
1. \( AB=CD=\sqrt{13} \) and \( BC=AD=2\sqrt{13} \), so ABCD is a parallelogram.
2. The diagonals \( AC=BD=\sqrt{65} \) are equal, so it is a rectangle.
3. Adjacent sides are unequal, so it is not a square.

Teacher's Note:
1. Compare opposite and adjacent sides.
2. Compare the two diagonals.

Answer: (d) a rhombus which is not a square

Explanation:
1. The sides are two pairs of parallel lines, so the figure is a parallelogram.
2. The distance between the pair 2x + 5y + 4 = 0 and 2x + 5y + 3 = 0 is \( \frac1{\sqrt{29}} \); between 5x + 2y + 7 = 0 and 5x + 2y + 6 = 0 it is also \( \frac1{\sqrt{29}} \), so all sides are equal (heights are equal, hence sides equal) and it is a rhombus.
3. Adjacent sides are not perpendicular since \( 2\cdot5+5\cdot2\ne0 \), hence not a square.

Teacher's Note:
1. Check the two perpendicular distances between opposite sides.
2. Check whether adjacent sides are perpendicular.

Answer: (a) 36

Explanation:
1. Let P = (p, p), Q = (q, 2q) and the mid-point (h, k): \( p+q=2h,\ p+2q=2k \), so \( q=2k-2h,\ p=4h-2k \).
2. Then \( PQ^2=(q-p)^2+(2q-p)^2=(4k-6h)^2+(6k-8h)^2=16 \).
3. This simplifies to \( 25h^2-36hk+13k^2=4 \), so λ = 36.

Teacher's Note:
1. Express p and q through the mid-point.
2. Use PQ^2 = 16.

Answer: (c) 6

Explanation:
1. Arrange in powers of K: \( K^2(x-2)+K(2x+y)+(x-2)=0 \).
2. For this to hold for all K: \( x-2=0 \) and \( 2x+y=0 \), so (α, β) = (2, −4).
3. Hence \( \alpha-\beta=6 \).

Teacher's Note:
1. Collect powers of K.
2. All the coefficients must vanish.

Answer: (a) −4/5

Explanation:
1. Unit distance from C1: \( 3\alpha-4\beta+1=\pm5 \); from C2: \( 8\alpha+6\beta+1=\pm10 \).
2. Solving the four systems gives \( \alpha=\frac65,\ -\frac25,\ 0,\ -\frac85 \).
3. Sum: \( \frac{6-2+0-8}{5}=-\frac45 \).

Teacher's Note:
1. There are four sign combinations.
2. Add the four values of alpha.