JEE Mathematics Coordinates and Straight Lines MCQs Set 04

Answer: (a) (4, 0)

Explanation:
1. The vertices are (2, −1), (2, 1) and (6, −1).
2. The triangle is right angled at (2, −1), so the circumcentre is the midpoint of the hypotenuse joining (2, 1) and (6, −1): (4, 0).

Teacher's Note:
1. The circumcentre of a right triangle is the midpoint of the hypotenuse.
2. Find the vertices by pairwise intersection.

Answer: (a) (1, 1)

Explanation:
1. The signed distance sum is zero when the line passes through the centroid.
2. The centroid is \( \left(\frac{1+2+0}{3},\frac{1+0+2}{3}\right)=(1,1) \).

Teacher's Note:
1. The sum of signed distances is linear in the point.
2. It vanishes at the centroid.

Answer: (c) \( \frac{4\sqrt7}{\sqrt3} \)

Explanation:
1. Take B = (0, 0) and C = (p, 6). Rotating BC through 60° gives A with height \( \frac{\sqrt3}{2}p+3 \).
2. Setting this equal to 4: \( p=\frac{2}{\sqrt3} \), so \( s^2=p^2+36=\frac43+36=\frac{112}{3} \).
3. The side is \( \sqrt{\frac{112}{3}}=\frac{4\sqrt7}{\sqrt3} \).

Teacher's Note:
1. Set up coordinates with the parallel lines as horizontal.
2. Use rotation by 60 degrees to locate A.

Answer: (a) \( 4p^2+p'^2=a^2 \)

Explanation:
1. \( p=\frac{a}{\sqrt{\sec^2\theta+\csc^2\theta}}=a\sin\theta\cos\theta=\frac a2\sin2\theta \).
2. @p'=
3. a\cos2\theta
4. @.
5. \( 4p^2+p'^2=a^2\sin^22\theta+a^2\cos^22\theta=a^2 \).

Teacher's Note:
1. Use the perpendicular distance formula.
2. Combine with sin squared plus cos squared equals one.

Answer: (b) p = p'

Explanation:
1. p =
2. 24x + 7y − 20
3. /25 and p' =
4. 4x − 3y − 2
5. /5.
6. Writing 24x + 7y − 20 = 5(4x − 3y − 2) + (4x + 22y − 10) = 5(4x − 3y − 2) + 2(2x + 11y − 5), on the given line the second bracket vanishes.
7. So p = 5
8. 4x − 3y − 2
9. /25 = p'.

Teacher's Note:
1. The given line is the bisector of the two lines.
2. Points on a bisector are equidistant from both lines.

Answer: (d) none of these

Explanation:
1. Let the other line be 3x + 4y = c. It meets x + y = 1 at (4 − c, c − 3), and the first line meets it at (2, −1).
2. The distance between the points is @\sqrt2
3. c-2
4. =5\( , so \)c=2\pm\frac{5}{\sqrt2}@.
5. None of the given options matches.

Teacher's Note:
1. Find the two intersection points on the transversal.
2. Compare with each option.

Answer: (a) \( \left(-\infty,-\frac43\right)\cup\left(\frac43,\infty\right) \)

Explanation:
1. The line meets y = 2 at (2/m, 2) and y = 6 at (6/m, 6).
2. Length = \( \sqrt{\frac{16}{m^2}+16}\lt5 \), so \( 16(1+m^2)\lt25m^2 \).
3. \( m^2\gt\frac{16}{9} \), i.e. @
4. m
5. \gt\frac43@.

Teacher's Note:
1. Find both intersection points.
2. Square the inequality carefully.

Answer: (b) always passes through a fixed point

Explanation:
1. In an AP, 2b = a + c, i.e. a − 2b + c = 0.
2. This says that the point (1, −2) satisfies ax + by + c = 0 for all such lines.
3. So the line always passes through the fixed point (1, −2).

Teacher's Note:
1. Convert the AP condition into a linear relation.
2. Read the fixed point off the coefficients.

Answer: (c) forms a triangle with the axes whose area is constant

Explanation:
1. In a GP, \( c^2=ab \).
2. The intercepts are \( -\frac ca \) and \( -\frac cb \), so the area is @\frac12\cdot\frac{c²}{
3. ab
4. }=\frac12@.
5. The area is constant.

Teacher's Note:
1. Use the GP condition c squared equals ab.
2. Area = half the product of the intercepts.

Answer: (a) 0

Explanation:
1. The determinant \( \begin{vmatrix}1&-2&3\\k&3&1\\4&-k&2\end{vmatrix}=-3k^2+5k-38 \) must be zero.
2. The discriminant of \( 3k^2-5k+38=0 \) is \( 25-456\lt0 \), so there is no real k.

Teacher's Note:
1. Three lines are concurrent if the determinant of the coefficients is zero.
2. A negative discriminant means no real solutions.

Answer: (c) 12x + 33y = 7

Explanation:
1. Write it as \( (x+y)+\lambda(2x-y+1)=0 \), so every line passes through \( \left(-\frac13,\frac13\right) \).
2. The farthest line from (1, 4) is perpendicular to the join of the two points, whose direction is \( \left(\frac43,\frac{11}{3}\right) \), i.e. (4, 11).
3. Line: \( 4\left(x+\frac13\right)+11\left(y-\frac13\right)=0 \), which is \( 12x+33y=7 \).

Teacher's Note:
1. Find the fixed point of the family.
2. The maximum distance is the distance to the fixed point.

Answer: (a) (3, −1)

Explanation:
1. Rewrite as \( \lambda(x+2y-1)+\mu(x+y-2)=0 \).
2. The lines pass through the intersection of x + 2y = 1 and x + y = 2, which is (3, −1).

Teacher's Note:
1. A family λL1 + μL2 = 0 passes through the point L1 ∩ L2.
2. Solve the two lines.

Answer: (b) x − 3y + 4 = 0

Explanation:
1. B is the intersection of y − x = 2 and x + 2y = 1: B = (−1, 1).
2. The side CA has slope −3, so the altitude from B has slope 1/3.
3. y − 1 = (x + 1)/3, so x − 3y + 4 = 0.

Teacher's Note:
1. B is the vertex common to AB and BC.
2. The altitude is perpendicular to the opposite side CA.

Answer: (a) (0, 1)

Explanation:
1. The triangle has vertices (0, 0), (4, 0) and (2, 2).
2. On x = 1, the points inside lie between y = 0 (the x-axis) and y = x = 1, not including the boundaries.
3. So the ordinate lies in (0, 1).

Teacher's Note:
1. Find the vertices of the triangle.
2. Interior points exclude the boundary.

Answer: (c) |a| < 1

Explanation:
1. The lines are x + y = 2 and x + y = −2.
2. For (a, a):
3. 2a
4. < 2, so
5. a
6. < 1.

Teacher's Note:
1. Between the lines means
2. x + y
3. < 2.
4. Substitute x = y = a.

Answer: (a) \( \left(-\frac\pi4,\frac\pi4\right)\cup\left(\frac\pi4,\frac{3\pi}4\right) \)

Explanation:
1. Same side means \( \left(\sin\alpha-\frac1{\sqrt2}\right)\left(\frac1{\sqrt2}-\cos\alpha\right)\gt0 \).
2. Solving this product inequality over \( [-\pi,\pi] \) (and checking numerically) gives \( -\frac\pi4\lt\alpha\lt\frac{3\pi}{4} \) with \( \alpha\ne\frac\pi4 \).
3. So α is in \( \left(-\frac\pi4,\frac\pi4\right)\cup\left(\frac\pi4,\frac{3\pi}4\right) \).

Teacher's Note:
1. Two points are on the same side if their line values have the same sign.
2. At α = π/4 both points lie on the line.

Answer: (c) are concurrent

Explanation:
1. L1 and L2 meet where \( x=\frac4{25},\ y=\frac{22}{25} \).
2. Substituting in L3: \( \frac4{25}-\frac{154}{25}+6=0 \), so L3 passes through the same point.
3. The lines are concurrent (so they form no triangle).

Teacher's Note:
1. Check concurrency by substituting the meeting point of two lines.
2. Perpendicular lines alone do not make a triangle.

Answer: (b) x − y + 1 = 0

Explanation:
1. The bisectors are \( \frac{2x-y+4}{\sqrt5}=\pm\frac{x-2y-1}{\sqrt5} \), i.e. \( x+y+5=0 \) and \( x-y+1=0 \).
2. The slopes 2 and 1/2 make an acute angle (tan = 3/4) bisected by the line of slope 1, since the bisector of the acute angle lies between the slopes 1/2 and 2.
3. So the acute bisector is x − y + 1 = 0.

Teacher's Note:
1. Find both bisectors and test which lies between the two slopes.
2. The acute bisector has slope between the given slopes.

Answer: (c) \( \frac5{19} \)

Explanation:
1. Solving y − x = 5 and 3x + 4y = 1: x = −19/7 and y = 16/7.
2. Substituting in y = mx + 3: 16/7 = −19m/7 + 3, so m = 5/19.

Teacher's Note:
1. Find the common point of the first two lines.
2. Use it in the third line.

Answer: (b) \( \left[-\frac\pi2,\frac\pi2\right] \)

Explanation:
1. The lines y = x − 1 and y = 1 − x meet at (1, 0); the angle containing the origin is @x\lt1,\
2. y
3. \lt1-x@.
4. A point \( (\cos\theta,\sin\theta) \) lies in it when @
5. \sin\theta
6. \lt1-\cos\theta\( , i.e. \)\theta\in\left(\frac\pi2,\frac{3\pi}{2}\right)@.
7. It does not lie in it for \( \theta\in\left[-\frac\pi2,\frac\pi2\right] \).

Teacher's Note:
1. Describe the angle region by inequalities.
2. The boundary points are not inside the open angle.

Answer: (b) y = x

Explanation:
1. The midpoint of the points is (0, 0), which lies on y = x.
2. The join has slope −1, which is perpendicular to y = x.
3. So the points are mirror images in y = x.

Teacher's Note:
1. A mirror line is the perpendicular bisector of the segment.
2. Check both the midpoint and the slope.

Answer: (c) 7x − y = 6

Explanation:
1. The line y = x meets the mirror at (1, 1), which stays fixed.
2. The image of the origin in 2x − y − 1 = 0 is (4/5, −2/5).
3. The image line passes through (1, 1) and (4/5, −2/5), with slope 7: 7x − y − 6 = 0.

Teacher's Note:
1. Use the reflection of a second point on the line.
2. The intersection with the mirror stays fixed.

Answer: (a) \( \left(\frac53,0\right) \)

Explanation:
1. Reflect P in the x-axis to P'(1, −1).
2. The line P'Q has slope 3/2 and meets the x-axis at x = 1 + 2/3 = 5/3.

Teacher's Note:
1. P and Q lie on the same side of the x-axis, so use the reflection principle.
2. R is where the line P'Q crosses the axis.

Answer: (a) y = 0

Explanation:
1. The incident ray along x = 1 meets the mirror x + y = 1 at (1, 0).
2. For the mirror x + y = 1, a direction (dx, dy) is reflected to (−dy, −dx); the incident direction (0, −1) becomes (1, 0).
3. So the reflected ray travels along y = 0.

Teacher's Note:
1. Reflect the direction vector in the mirror.
2. The reflected ray passes through the point of incidence.

Answer: (d) (−3, 2)

Explanation:
1. A unit vector along x + y = 1 with increasing y is (−1, 1)/√2.
2. Q = (2, 1) + 3√2·(−1, 1)/√2 = (−1, 4).
3. The image of (x, y) in x + y = 1 is (1 − y, 1 − x), so the image of Q is (−3, 2).

Teacher's Note:
1. Move along the direction vector of the line.
2. Use the reflection formula for x + y = c.

Answer: (a) \( \left(\frac{3+\sqrt3}2,\frac{\sqrt3-1}2\right) \)

Explanation:
1. \( \vec{AB}=(1,1) \). Rotating clockwise by 30° gives \( (\cos30^\circ+\sin30^\circ,\ -\sin30^\circ+\cos30^\circ)=\left(\frac{\sqrt3+1}{2},\frac{\sqrt3-1}{2}\right) \).
2. B' = A + this = \( \left(\frac{3+\sqrt3}{2},\frac{\sqrt3-1}{2}\right) \).

Teacher's Note:
1. Use the rotation matrix with angle −30°.
2. Add the rotated vector to A.