JEE Mathematics Straight Lines MCQs Set 10

Answer: (c) \( (3,3) \)

Explanation:
1. The intercepts are \( (4,0) \) and \( (0,2) \); the mid-point of the side is \( (2,1) \) and its length is \( \sqrt{20} \).
2. The centre lies at distance half the side, \( \sqrt5 \), from the mid-point along the normal \( (1,2)/\sqrt5 \) away from the origin: \( (2,1)+(1,2)=(3,3) \).

Teacher's Note:
1. Find the mid-point of the intercepted segment.
2. Move half a side length along the unit normal directed away from the origin.

Answer: (a) \( \dfrac{400}{49} \)

Explanation:
1. Base \( b=10 \) and height \( h=4 \). If the side of the square is \( s \), similar triangles give \( \dfrac{s}{b}=\dfrac{h-s}{h} \).
2. So \( s=\dfrac{bh}{b+h}=\dfrac{40}{14}=\dfrac{20}7 \) and the area is \( \dfrac{400}{49} \).

Teacher's Note:
1. The small triangle above the square is similar to the whole triangle.
2. Equate the ratios of top side to base and remaining height to height.

Answer: (d) \( \dfrac{2\sqrt2}3 \)

Explanation:
1. The bisector of the right angle makes \( 45^\circ \) with both axes, so D lies on \( y=x \). Line AB: \( 2x+y=2 \).
2. Intersection: \( 3x=2 \), so \( D=\left(\tfrac23,\tfrac23\right) \) and \( CD=\tfrac23\sqrt2=\dfrac{2\sqrt2}{3} \).

Teacher's Note:
1. The bisector of the angle at the origin is the line y = x.
2. Find where it meets AB.

Answer: (a) 220

Explanation:
1. Area gives \( ab=22 \). The line \( \tfrac xa+\tfrac yb=1 \) passes through \( (2,3) \): \( \tfrac2a+\tfrac3b=1 \), so \( 2b+3a=ab=22 \).
2. Square: \( (3a+2b)^2=484=9a^2+4b^2+12ab \), hence \( 9a^2+4b^2=484-264=220 \).

Teacher's Note:
1. Use intercept form and the area to get ab and 3a + 2b.
2. Square 3a + 2b.

Answer: (a) \( 5x+4y=13 \)

Explanation:
1. Reflect \( (5,3) \) in the x-axis to get \( (5,-3) \); the incident ray AB points towards this image.
2. Slope of the line from \( (1,2) \) to \( (5,-3) \) is \( -\tfrac54 \): \( y-2=-\tfrac54(x-1) \), i.e. \( 5x+4y=13 \).

Teacher's Note:
1. The reflected ray appears to come from the mirror image of the final point.
2. Join A to that image.

Answer: (b) \( m(m+3n) \)

Explanation:
1. The vertices are \( A(m,n) \), \( B(n,m) \), \( C(-n,m) \), \( D(-n,-m) \), \( E(n,-m) \).
2. The shoelace formula gives area \( =m^2+3mn \). (Check with \( m=5,n=2 \): 55 and \( m(m+3n)=55 \).)

Teacher's Note:
1. Write all five vertices explicitly.
2. Use the shoelace formula, or split into a rectangle and a triangle.

Answer: (b) \( \left(\tfrac54,1\right) \)

Explanation:
1. The apex \( A=(2,t) \) lies on \( x=2 \) with \( AB=AC \) where \( B(2,0) \), \( C(0,1) \): \( t^2=4+(t-1)^2 \) gives \( t=\tfrac52 \).
2. The altitude from C is perpendicular to the vertical side AB, so \( y=1 \). The altitude from A is perpendicular to BC (slope \( -\tfrac12 \)), so \( y-\tfrac52=2(x-2) \).
3. Setting \( y=1 \): \( x=\tfrac54 \). Orthocentre \( \left(\tfrac54,1\right) \).

Teacher's Note:
1. The apex is on x = 2 and equidistant from the base ends.
2. Intersect two altitudes.

Answer: (a) 10

Explanation:
1. \( y_1=0 \) and \( y_2=\ln1000=3\ln10 \). C is one third of the way from A to B, so \( y_C=\ln10 \).
2. The horizontal line meets the curve where \( \ln x_3=\ln10 \), so \( x_3=10 \).

Teacher's Note:
1. The y-coordinate of C is the weighted average of y1 and y2.
2. Solve ln x = y_C.

Answer: (c) \( (a,b) \)

Explanation:
1. Solving the first two lines: \( x=5 \), \( y=2 \).
2. Concurrency means \( 5a+2b=1 \), i.e. the point \( (a,b) \) lies on \( 5x+2y=1 \).

Teacher's Note:
1. Find the common point of the first two lines.
2. Substitute it into the third.

Answer: (b) \( \dfrac{11}{8} \)

Explanation:
1. A line divides a parallelogram into congruent halves exactly when it passes through the centre (the intersection of the diagonals).
2. Centre \( =\left(\tfrac{3+13}2,\tfrac{1+21}2\right)=(8,11) \), so the slope from the origin is \( \tfrac{11}8 \).

Teacher's Note:
1. Congruent halves require the line to pass through the centre.
2. The centre is the mid-point of AC.

Answer: (b) 3

Explanation:
1. The triangle has area \( \tfrac12\cdot8\cdot1=4 \). For \( 1\le c\le9 \) the area to the left of \( x=c \) is \( \tfrac49+(c-1)-\dfrac{c^2-1}{18} \).
2. Setting this equal to 2: \( c^2-18c+45=0 \), so \( c=3 \) or \( 15 \); only \( c=3 \) lies inside the triangle.

Teacher's Note:
1. Compute the total area first.
2. Integrate between the lines y = x/9 and the top boundary.

Answer: (b) \( (2008,0) \)

Explanation:
1. If the line passed through \( (2008,0) \) then \( 2008m+b=0 \), so \( b=-2008m \) and \( mb=-2008m^2\le0 \), contradicting \( mb>0 \).
2. The other points are possible: for example \( (0,2008) \) with \( m,b>0 \), and \( (20,-100) \) with \( m=-6,\ b=20 \).

Teacher's Note:
1. Substitute the point into the equation and look at the sign of mb.
2. Check that the other points can be reached by some admissible m and b.

Answer: (d) \( \tfrac18 \)

Explanation:
1. With \( y=m_1x \) and \( C=(1,\,y+m_2(1-x)) \), the area is \( \tfrac12\left|x\cdot C_y-y\right|=\tfrac12x(1-x)(m_1-m_2) \).
2. So \( f(x)=\tfrac12x(1-x) \), whose maximum on \( (0,1) \) is \( \tfrac12\cdot\tfrac14=\tfrac18 \) at \( x=\tfrac12 \).

Teacher's Note:
1. Write the coordinates of C using the slope m2.
2. Maximise x(1-x).

Answer: (d) \( 21x-11y-15=0 \)

Explanation:
1. The vertex opposite to the origin is on \( 2x+3y=5 \) and \( 3x-4y=3 \); the other two vertices are \( \left(\tfrac9{17},-\tfrac6{17}\right) \) (from \( 2x+3y=0 \), \( 3x-4y=3 \)) and \( \left(\tfrac{20}{17},\tfrac{15}{17}\right) \) (from \( 2x+3y=5 \), \( 3x-4y=0 \)).
2. The diagonal joining these two has slope \( \dfrac{21/17}{11/17}=\tfrac{21}{11} \) and its equation is \( 21x-11y-15=0 \).

Teacher's Note:
1. The diagonal that misses the origin joins the two vertices that are not O and not opposite to O.
2. Find those two vertices by solving pairs of sides.

Answer: (d) N

Explanation:
1. Centroid \( (X,Y)=\left(\tfrac{6+h}3,\tfrac{-6+k}3\right) \) where \( R=(h,k) \), so \( h=3X-6 \), \( k=3Y+6 \).
2. Substituting in \( 9h+7k+4=0 \): \( 27X+21Y-54+42+4=0 \), i.e. \( 27X+21Y-8=0 \), which has the same direction as N.

Teacher's Note:
1. Express R in terms of the centroid.
2. Substitute into the line N.

Answer: (d) \( \left(-\tfrac52,\tfrac32\right) \)

Explanation:
1. Along AB the unit vector is \( \tfrac1{\sqrt5}(2,1) \) and along AC it is \( \tfrac1{\sqrt5}(1,2) \). Their sum is parallel to \( (1,1) \), so the bisector is \( y=x+4 \).
2. The mid-point of AB is \( (-2,1) \), so the line from \( C(-3,2) \) to it is \( x+y=-1 \).
3. Solving: \( 2x+4=-1 \), so \( x=-\tfrac52 \), \( y=\tfrac32 \).

Teacher's Note:
1. Add the unit vectors along the two arms to get the bisector direction.
2. Intersect the bisector with the median line.

Answer: (c) \( (0,-5) \)

Explanation:
1. A line bisects a rectangle when it passes through the centre \( (2,1) \).
2. With slope 3: \( y-1=3(x-2) \), so \( y=3x-5 \) and the y-intercept is \( (0,-5) \).

Teacher's Note:
1. Any line through the centre of a rectangle bisects its area.
2. Use slope 3 through (2,1).

Answer: (b) 3

Explanation:
1. The right-angle vertex lies on the circle with diameter joining \( (2,-3) \) and \( (4,1) \): centre \( (3,-1) \), \( r^2=5 \).
2. With \( y=10-2x \): \( (x-3)^2+(11-2x)^2=5 \), so \( x^2-10x+25=0 \) and \( x=5 \), giving \( P=(5,0) \).
3. Area \( =\tfrac12\left|(4-2)(0+3)-(5-2)(1+3)\right|=3 \).

Teacher's Note:
1. Thales: the right angle lies on the circle on the hypotenuse as diameter.
2. Intersect with the given line and use the determinant formula.

Answer: (a) \( x+y=1 \)

Explanation:
1. Let \( B=(b,0) \). The slope of AB is \( \tfrac1{1-b} \), so AC has slope \( -(1-b) \) and meets \( x=0 \) at \( C=(0,\,2-b) \).
2. Mid-point \( P=\left(\tfrac b2,\tfrac{2-b}2\right) \), so \( x+y=1 \).

Teacher's Note:
1. Parametrise B on the x-axis.
2. Find C using perpendicular slopes.

Answer: (c) three

Explanation:
1. Intercept form with \( \tfrac2a+\tfrac3b=1 \) gives \( 2b+3a=ab \), and the area condition is \( |ab|=24 \).
2. Case \( ab=24 \): \( 3a^2-24a+48=0 \), a double root \( a=4 \): one line. Case \( ab=-24 \): \( a^2+8a-16=0 \): two distinct real roots: two lines.
3. Total three lines.

Teacher's Note:
1. Treat ab = 24 and ab = -24 separately.
2. Count distinct real values of a in each case.