JEE Mathematics Straight Lines MCQs Set 22

Answer: (c) (1023, 2046)

Explanation:
1. Each step halves the distance to B, so \( A_n=B-\frac{1}{2^n}(B-A) \).
2. For n = 10: \( (1024-1,\ 2048-2)=(1023,2046) \).

Teacher's Note:
1. Each mid-point step halves the remaining distance from B.
2. After 10 steps the gap is 1/1024 of the original.

Answer: (d) 2

Explanation:
1. P is the perpendicular bisector of BC. Q is the pair of bisector lines (internal and external) of angle A.
2. The perpendicular bisector is parallel to neither bisector (AB is not equal to AC), so it meets each bisector in exactly one point.
3. Hence \( n(P\cap Q)=2 \).

Teacher's Note:
1. P is a line, Q is a pair of perpendicular lines.
2. A line meets each of them once unless it is parallel.

Answer: (a) Statement-1 is true, Statement-2 is true and Statement-2 is a correct explanation of Statement-1

Explanation:
1. Take A = 0, B = 2, M = 1, P = 2r/(r+1), Q = 2r/(r−1), r > 1.
2. AP, AB, AQ: \( \frac1{AP}+\frac1{AQ}=\frac{r+1}{2r}+\frac{r-1}{2r}=\frac1{1}=\frac{2}{AB} \), so they are in H.P. (Statement-2 true).
3. Subtracting AB/2 = 1 from each term gives AP − AM, AB − AM, AQ − AM, i.e. MP, MB, MQ, which are in G.P.: \( MP\cdot MQ=\frac{r-1}{r+1}\cdot\frac{r+1}{r-1}=1=MB^2 \).
4. The G.P. property follows from the H.P. property, so Statement-2 explains Statement-1.

Teacher's Note:
1. Put the points on a number line with M as the origin.
2. Check both statements with a general ratio r.

Answer: (c) 1.5 sq. units

Explanation:
1. Area of the original triangle: \( \frac12|2(3+3)+(-2)(-3-1)+4(1-3)|=\frac12|12+8-8|=6 \).
2. The medial triangle has one-quarter of the area: \( \frac64=1.5 \).

Teacher's Note:
1. Find the area of the big triangle by the determinant formula.
2. The medial triangle is similar with ratio 1/2.

Answer: (b) 2x + y − 4 = 0

Explanation:
1. Let the line be \( \frac xa+\frac yb=1 \). The mid-point of the intercepts is \( (\frac a2,\frac b2)=(1,2) \), so a = 2 and b = 4.
2. The line is \( \frac x2+\frac y4=1 \), i.e. \( 2x+y-4=0 \).

Teacher's Note:
1. The intercept points are (a, 0) and (0, b).
2. Set their mid-point equal to P.

Answer: (c) \( \sqrt3x-y-2\sqrt3=0 \)

Explanation:
1. Slope of AB is 1, so AB makes 45 degrees with the x-axis.
2. After rotating by 15 degrees the inclination is 60 degrees, slope \( \sqrt3 \).
3. Through A(2, 0): \( y=\sqrt3(x-2) \), i.e. \( \sqrt3x-y-2\sqrt3=0 \).

Teacher's Note:
1. Find the inclination of AB first.
2. Add 15 degrees to the inclination.

Answer: (a) x + y = 11

Explanation:
1. Let B = (x1, y1) on AB and C = (x2, y2) on AC with \( x_1+x_2=10,\ y_1+y_2=12 \).
2. Substitute C = (10 − x1, 12 − y1) in AC: \( (10-x_1)+2(12-y_1)=16\Rightarrow x_1+2y_1=18 \).
3. With \( 2x_1+3y_1=29 \): \( x_1=4,\ y_1=7 \), so B = (4, 7).
4. BC passes through (4, 7) and (5, 6): \( x+y=11 \).

Teacher's Note:
1. Use the mid-point to express C in terms of B.
2. Solve for B, then join B to the mid-point.

Answer: (d) \( \sqrt3x+y=8 \)

Explanation:
1. The normal makes an angle of 120 − 90 = 30 degrees with the x-axis.
2. Normal form: \( x\cos30^\circ+y\sin30^\circ=4 \), i.e. \( \sqrt3x+y=8 \).
3. Its slope is \( -\sqrt3=\tan120^\circ \).

Teacher's Note:
1. Use the normal form x cos a + y sin a = p.
2. The normal is perpendicular to the line.

Answer: (c) \( 4\sqrt2 \)

Explanation:
1. The line x − y + 1 = 0 has slope 1 (inclination 45 degrees). Points on it through (2, 3): \( \left(2+\frac r{\sqrt2},\ 3+\frac r{\sqrt2}\right) \).
2. Put in 2x − 3y + 9 = 0: \( 4-9+9-\frac{r}{\sqrt2}=0 \), so \( r=4\sqrt2 \).

Teacher's Note:
1. Use the parametric form of the line through the point in the given direction.
2. Substitute into the other line.

Answer: (a) \( (2-\sqrt2,\sqrt2-1),\ (2\sqrt5+2,\sqrt5-1) \)

Explanation:
1. Along x + y = 1 the slope is −1; moving towards increasing y the direction is 135 degrees: \( (2+2\cos135^\circ,\ -1+2\sin135^\circ)=(2-\sqrt2,\ \sqrt2-1) \).
2. Along x − 2y = 4 the slope is 1/2, so \( \sin\theta=\frac1{\sqrt5},\cos\theta=\frac2{\sqrt5} \): \( (2+5\cos\theta,\ -1+5\sin\theta)=(2+2\sqrt5,\ \sqrt5-1) \).

Teacher's Note:
1. Use the point plus r times direction cosines.
2. Choose the direction with positive y-component.

Answer: (c) y = (5/12)x

Explanation:
1. For y = 0.2x, \( \tan\theta=\frac15 \).
2. Then \( \tan2\theta=\frac{2/5}{1-1/25}=\frac{2/5}{24/25}=\frac5{12} \).
3. The line is \( y=\frac5{12}x \).

Teacher's Note:
1. Apply the double angle formula for tan.
2. The line passes through the origin.

Answer: (c) 7y + x = 31

Explanation:
1. Check: \( 7(-4)-5+8=-25\ne0 \), so (−4, 5) is not on the given diagonal; it lies on the other diagonal.
2. The diagonals of a square are perpendicular, so the slope of the other is \( -\frac17 \).
3. Equation: \( y-5=-\frac17(x+4) \), i.e. \( x+7y=31 \).

Teacher's Note:
1. First decide which diagonal the vertex lies on.
2. Diagonals of a square are perpendicular.

Answer: (c) 3x + 6y = 14

Explanation:
1. A line parallel to both is \( x+2y=c \). Along the normal the lines are at \( c=2 \) and \( c=6 \); a point dividing the gap 2 : 1 from the line c = 2 has \( c=2+\frac23(6-2)=\frac{14}3 \).
2. So the line is \( x+2y=\frac{14}3 \), i.e. \( 3x+6y=14 \).

Teacher's Note:
1. Parallel lines have the same x and y coefficients.
2. The constant term changes linearly across the gap.

Answer: (c) \( a\in(-3,0)\cup(\frac13,1) \)

Explanation:
1. The origin gives +1 for 3x − y + 1 and −5 for x + 2y − 5, so the point needs \( 3a^2-(a+1)+1>0 \) and \( a^2+2(a+1)-5<0 \).
2. First: \( a(3a-1)>0\Rightarrow a<0\text{ or }a>\frac13 \). Second: \( (a+3)(a-1)<0\Rightarrow -33. Intersection: \( (-3,0)\cup(\frac13,1) \).

Teacher's Note:
1. The point must give the same signs as the origin for both lines.
2. Intersect the two solution sets.