JEE Mathematics Straight Lines MCQs Set 19

Answer: (a) (ii) and (iv) only

Explanation:
1. By the intersecting chords theorem \( PS\cdot ST=QS\cdot SR \).
2. AM-HM: \( \frac1{PS}+\frac1{ST}\ge\frac{2}{\sqrt{PS\cdot ST}}=\frac2{\sqrt{QS\cdot SR}} \) with equality only if \( PS=ST \) (excluded), so (ii) holds.
3. Also \( QS\cdot SR\le\left(\frac{QR}{2}\right)^2 \), so \( \frac2{\sqrt{QS\cdot SR}}\ge\frac4{QR} \) and (iv) holds.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 1

Explanation:
1. Solving: \( x=\frac{|a|+1}{a+1},\ y=|a|-x \). For \( a<0 \) the point has \( y<0 \): no intersection in the first quadrant.
2. For \( a>0 \): \( x=1 \), \( y=a-1 \). Need \( y>0 \), so \( a>1 \). Hence \( a_0=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( k=-9 \) or \( k=-\frac65 \)

Explanation:
1. Slopes: \( L_1:-\frac13,\ L_2:\frac3k,\ L_3:-\frac52 \).
2. [\( L_1\parallel L_2 \)]: \( \frac3k=-\frac13 \), \( k=-9 \). \( L_2\parallel L_3 \): \( \frac3k=-\frac52 \), \( k=-\frac65 \). \( L_1 \) and \( L_3 \) are never parallel.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( 2x+9y+7=0 \)

Explanation:
1. \( S=\left(\frac{13}2,1\right) \), the midpoint of \( QR \).
2. Slope of \( PS \) = \( \frac{1-2}{13/2-2}=-\frac29 \).
3. Line: \( y+1=-\frac29(x-1) \), i.e. \( 2x+9y+7=0 \).

Teacher's Note:
1. Find the midpoint of QR, then the slope of PS.
2. Use the point-slope form.

Answer: (b) \( k=\frac56 \)

Explanation:
1. They fail to form a triangle only when concurrent (\( k=5 \)) or when two are parallel (\( k=-9,-\frac65 \)).
2. For \( k=\frac56 \) none of these happens, so a triangle is formed.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) a straight line

Explanation:
1. The vertices are \( A(-p,0),\ B(-q,0) \) and \( C(pq,(1+p)(1+q)) \).
2. The altitude from \( C \) is \( x=pq \). The altitude from \( A \) has slope \( -\frac{q}{1+q} \), giving \( y=-pq \) at \( x=pq \).
3. The orthocentre \( (pq,-pq) \) lies on \( x+y=0 \), a straight line.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \sqrt3x+y=0 \)

Explanation:
1. \( \vec{QP} \) points along \( 180^\circ \) and \( \vec{QR} \) has direction \( \tan^{-1}\sqrt3=60^\circ \).
2. The bisector is at \( \frac{180^\circ+60^\circ}{2}=120^\circ \), slope \( -\sqrt3 \), through the origin: \( \sqrt3x+y=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac1{|m-n|} \)

Explanation:
1. With vertices \( O(0,0) \), \( A\left(\frac{1}{m-n},\frac{m}{m-n}\right) \), \( B\left(\frac1{n-m},\frac n{n-m}\right) \) the area is \( |x_Ay_B-x_By_A|=\frac1{|m-n|} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \left(-3,-\frac1{\sqrt2}\right)\cup\left(\frac1{\sqrt2},3\right) \)

Explanation:
1. Inside the square: \( |a|<3 \).
2. Outside the circle: \( a^2+a^2>1 \), i.e. \( |a|>\frac1{\sqrt2} \).
3. Both hold for \( -3<a<-\frac1{\sqrt2} \) or \( \frac1{\sqrt2}<a<3 \).

Teacher's Note:
1. Translate each geometric condition into an inequality in a.
2. Combine them; the point lies on the line y = x.

Answer: (d) 2

Explanation:
1. Substituting: \( 3x+4mx+4=9 \), so \( x=\frac5{3+4m} \).
2. For an integer \( x \), \( 3+4m\in\{\pm1,\pm5\} \) gives \( m=-\frac12,-1,\frac12,-2 \).
3. Integer values of \( m \): \( -1 \) and \( -2 \) only, so 2 values.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( 3:4 \)

Explanation:
1. Write the lines as \( 2x+y=\frac92 \) and \( 2x+y=-6 \). Along any line through \( O \), the distances to the two lines are proportional to \( \frac92 \) and \( 6 \).
2. So \( OP:OQ=\frac92:6=3:4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) Statement 1 is true and Statement 2 is false

Explanation:
1. \( P=(-2,-2) \) and \( Q=(1,-2) \), with \( OP=2\sqrt2 \) and \( OQ=\sqrt5 \). The acute angle between the lines is angle \( POQ \) (about \( 71.6^\circ \)).
2. By the angle bisector theorem \( PR:RQ=OP:OQ=2\sqrt2:\sqrt5 \): Statement 1 is true.
3. Statement 2 is false: the two parts are generally not similar.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \left(1,\frac1{\sqrt3}\right) \)

Explanation:
1. All sides equal 2 (e.g. distance from \( (0,0) \) to \( (1,\sqrt3) \) is 2), so the triangle is equilateral.
2. The incentre equals the centroid: \( \left(\frac{1+0+2}{3},\frac{\sqrt3}{3}\right)=\left(1,\frac1{\sqrt3}\right) \).

Teacher's Note:
1. In an equilateral triangle all the centres coincide.
2. Average the vertex coordinates.

Answer: (c) rational numbers

Explanation:
1. The centroid G divides the line joining circumcentre O and orthocentre H in the ratio 1 : 2, so \( G=\frac{2O+H}{3} \).
2. If O and H have integer coordinates, G has coordinates of the form integer divided by 3, i.e. rational.
3. Example: O = (0,0), H = (1,0) gives G = (1/3, 0), which is not an integer, so integers are not guaranteed.

Teacher's Note:
1. O, G, H are collinear on the Euler line with OG : GH = 1 : 2.
2. The centroid is rational but need not be an integer.

Answer: (b) \( P,Q,R \) are non-collinear

Explanation:
1. The area determinant of \( P,Q,R \) is non-zero for admissible values: e.g. \( \alpha=\beta=\frac\pi6 \), \( \theta=\frac\pi{12} \) gives a non-zero value, and it does not vanish identically for the stated ranges.
2. So the three points are not collinear.

Teacher's Note:
1. Collinearity requires the 3 x 3 determinant to vanish.
2. A numerical check with admissible angles shows it does not.

Answer: (d) two straight lines and a circle, when \( a=b \) and \( c \) is of sign opposite to that of \( a \)

Explanation:
1. \( x^2-5xy+6y^2=(x-2y)(x-3y) \) gives two straight lines.
2. The factor \( ax^2+by^2+c=0 \) is a circle if \( a=b \) and \( c \) has sign opposite to \( a \).
3. If \( c=0 \) with \( a,b \) of the same sign it is just the origin; with \( a,b \) of the same sign and \( c \) opposite it is an ellipse, not a hyperbola; and the other factor is a pair of lines, not an ellipse.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( k=5 \)

Explanation:
1. Intersection of \( L_1 \) and \( L_3 \): \( x=2,\ y=1 \).
2. This lies on \( L_2 \) when \( 6-k=1 \), i.e. \( k=5 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. A point dividing the triangle into three equal areas is the centroid.
2. Centroid \( =\left(\frac{0+3+6}{3},\frac{0+4+0}{3}\right)=\left(3,\frac43\right) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.