JEE Mathematics Straight Lines MCQs Set 08

Answer: (d) (0, 0)

Explanation:
1. The image of (a, b) in y = −x is \( Q=(-b,-a) \).
2. The image of (−b, −a) in y = x is \( R=(-a,-b) \).
3. The mid-point of P(a, b) and R(−a, −b) is (0, 0).

Teacher's Note:
1. Reflection in y = −x sends (x, y) to (−y, −x); in y = x it swaps the coordinates.
2. R is the reflection of P in the origin.

Answer: (a) 7y + x − 6 = 0

Explanation:
1. The slope of the given line is \( m=\frac34 \). A clockwise rotation by 45° gives the new slope \( \tan(\theta-45^\circ)=\frac{\frac34-1}{1+\frac34}=-\frac17 \).
2. The new line passes through (−1, 1): \( y-1=-\frac17(x+1) \), i.e. \( x+7y-6=0 \).

Teacher's Note:
1. Clockwise rotation decreases the inclination of the line.
2. Use tan(θ − α) and the fixed point.

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

Explanation:
1. The vertices are (0,0), \( \left(\frac1{n-m},\frac{n}{n-m}\right) \), \( \left(\frac{1}{m-n},\frac{m}{m-n}\right) \) and (0, 1) ... a direct way: area = (product of the distances between the two pairs of parallel lines) divided by sin of the angle between them.
2. The distances are \( \frac1{\sqrt{1+m^2}} \) and \( \frac1{\sqrt{1+n^2}} \), and \( \sin\theta=\frac{|m-n|}{\sqrt{1+m^2}\sqrt{1+n^2}} \).
3. Area \( =\frac{1}{|m-n|} \).

Teacher's Note:
1. Area of a parallelogram = (d₁ × d₂)/sin θ where d₁, d₂ are the distances between opposite sides.
2. sin θ follows from tan θ = |(m − n)/(1 + mn)|.

Answer: (b) 20

Explanation:
1. The line is \( ax+by+c=1 \). The condition \( 5a+4b+20c=20 \) can be written as \( \frac a4+\frac b5+c=1 \).
2. So (x, y) = (1/4, 1/5) satisfies the line for every such (a, b, c): a fixed point. Hence t = 20.

Teacher's Note:
1. Compare ax + by + c = 1 with a linear relation between a, b, c.
2. Divide the relation by its right-hand side to make it 1.

Answer: (c) \( (2-2\sqrt2,\ 1-2\sqrt2) \)

Explanation:
1. The line has direction \( \frac{1}{\sqrt2}(1,1) \). Moving 4 units along it gives \( A'=(2\pm2\sqrt2,\ 1\pm2\sqrt2) \).
2. The point with the minus sign, \( (2-2\sqrt2,\ 1-2\sqrt2)\approx(-0.83,-1.83) \), lies in the third quadrant.

Teacher's Note:
1. The direction vector of the line x − y = 3 is (1, 1)/√2.
2. Test the sign of both coordinates.

Answer: (a) \( \left(\frac{13}5,0\right) \)

Explanation:
1. Reflect (1, 2) in the x-axis to get (1, −2). The reflected ray appears to come from (1, −2) and passes through A and (5, 3).
2. The line through (1, −2) and (5, 3) has slope \( \frac54 \) and meets y = 0 at \( x=1+\frac{2}{5/4}=\frac{13}5 \).

Teacher's Note:
1. Use the image method for reflection.
2. A, the image point and the final point are collinear.

Answer: (a) I quadrant

Explanation:
1. The vertices are A(−3, 4) (first two lines), \( B\left(-\frac35,\frac85\right) \) (first and third) and \( C\left(-\frac37,\frac{16}7\right) \) (second and third).
2. The orthocentre, found from two altitudes, is \( \left(\frac37,\frac{22}7\right) \), which lies in the first quadrant.

Teacher's Note:
1. Find the three vertices, then two altitudes.
2. Check the signs of the coordinates of the orthocentre.

Answer: (b) \( \left(\frac13,\frac23\right) \)

Explanation:
1. The centroid divides the line joining the circumcentre O and the orthocentre H in the ratio 1 : 2 from O: \( G=\frac{2O+H}{3} \).
2. So \( G=\left(\frac13,\frac23\right) \).

Teacher's Note:
1. Euler line: O, G, H are collinear with OG : GH = 1 : 2.
2. G = (2·O + H)/3.

Answer: (c) |a| < 1

Explanation:
1. Between the lines means \( |x+y|<2 \).
2. For (a, a): \( |2a|<2 \), so |a| < 1.

Teacher's Note:
1. Substitute x = y = a in the inequality describing the strip.
2. The point lies between the lines if |x + y| < 2.

Answer: (c) \( \frac13f(6) \) sq unit

Explanation:
1. From the functional equation with f(1) = 2: \( f(x)=2^x \) (so f(6) = 64).
2. The region is a rhombus with vertices \( \left(\pm\frac83,0\right) \) and (0, ±4); its diagonals are 16/3 and 8.
3. Area \( =\frac12\cdot\frac{16}3\cdot8=\frac{64}3=\frac13f(6) \).

Teacher's Note:
1. Find f first: f(x + y) = f(x)f(y) means an exponential function.
2. Area of a rhombus = half the product of the diagonals.