JEE Mathematics Coordinates and Straight Lines MCQs Set 03

Answer: (c) at least one coordinate irrational

Explanation:
1. Let the integral vertices be (0, 0) and (p, q). The third vertex is \( \left(\frac{p\mp q\sqrt3}{2},\frac{q\pm p\sqrt3}{2}\right) \).
2. Since p and q are not both zero, at least one coordinate contains \( \sqrt3 \) and is irrational (the other may be rational, e.g. when q = 0).

Teacher's Note:
1. Rotate the side vector through 60 degrees.
2. The factor root 3 makes at least one coordinate irrational.

Answer: (a) \( \frac{41}{7} \)

Explanation:
1. The dividing point is \( \left(\frac{3(-1)+4(2)}{7},\frac{3(2)+4(3)}{7}\right)=\left(\frac57,\frac{18}{7}\right) \).
2. It lies on the line, so \( k=\frac57+2\cdot\frac{18}{7}=\frac{41}{7} \).

Teacher's Note:
1. Use the section formula for internal division.
2. Substitute the point in the given line.

Answer: (c) equilateral

Explanation:
1. In Cartesian form the vertices are O(0, 0), A(0, 3) and \( B\left(\frac{3\sqrt3}{2},\frac32\right) \).
2. OA = 3, OB = 3 and \( AB=\sqrt{\frac{27}{4}+\frac94}=3 \).
3. All sides are equal, so the triangle is equilateral.

Teacher's Note:
1. Convert polar to Cartesian using x = r cos θ and y = r sin θ.
2. Compare the three side lengths.

Answer: (b) collinear

Explanation:
1. Each point has the form (t, S − t) where S = a + b + c.
2. So all three lie on the line x + y = a + b + c, and are collinear.

Teacher's Note:
1. Look for a common relation between x and y.
2. The sum of coordinates is constant.

Answer: (d) \( \left(\frac{ab}{a+b+\sqrt{a^2+b^2}},\frac{ab}{a+b+\sqrt{a^2+b^2}}\right) \)

Explanation:
1. The triangle has legs a and b and hypotenuse \( \sqrt{a^2+b^2} \).
2. For a right triangle, inradius \( r=\frac{a+b-\sqrt{a^2+b^2}}{2}=\frac{ab}{a+b+\sqrt{a^2+b^2}} \).
3. The incentre is at (r, r) from the right-angle vertex at the origin.

Teacher's Note:
1. The incentre of a right triangle at the origin is (r, r).
2. Use r = area divided by semi-perimeter.

Answer: (b) \( \left(\frac53,\frac1{\sqrt3}\right) \)

Explanation:
1. The midpoint of BC is (2, 0), so C = (4, 0).
2. A lies at distance 2 from B at 60 degrees: \( A=(2\cos60^\circ,2\sin60^\circ)=(1,\sqrt3) \).
3. Centroid = \( \left(\frac{0+4+1}{3},\frac{0+0+\sqrt3}{3}\right)=\left(\frac53,\frac1{\sqrt3}\right) \).

Teacher's Note:
1. Find C from the midpoint, A from the polar form.
2. Average the three vertices.

Answer: (b) (4, 6)

Explanation:
1. With A(1, 3), B(−1, 2), C(2, 5) consecutive, the fourth vertex is D = A + C − B.
2. D = (1 + 2 + 1, 3 + 5 − 2) = (4, 6).

Teacher's Note:
1. Diagonals of a parallelogram bisect each other.
2. The midpoint of AC equals the midpoint of BD.

Answer: (a) 30 unit\( ^2 \)

Explanation:
1. Use the shoelace formula in the given order: sum of (x_i y_i+1 − x_i+1 y_i).
2. The sum is 60, so the area is 60/2 = 30 square units.

Teacher's Note:
1. List the vertices in order and repeat the first at the end.
2. Half the absolute value of the sum gives the area.

Answer: (a) 18 unit\( ^2 \)

Explanation:
1. Taking the lines as the axes, the locus is @
2. x
3. +
4. y
5. =3@, a square with vertices (±3, 0) and (0, ±3).
6. Its diagonal is 6, so the area is \( \frac{6^2}{2}=18 \).

Teacher's Note:
1. Write the condition as
2. x
3. +
4. y
5. = 3.
6. The area of a square is half the square of the diagonal.

Answer: (b) 4

Explanation:
1. The condition says angle ACB is a right angle, so C lies on the circle with AB as diameter: radius \( \sqrt2 \), centre (2, 3).
2. Area 1 with \( AB=2\sqrt2 \) gives height \( \frac{1}{\sqrt2} \) from the line AB.
3. The line AB is at distance 0 from the centre, so there are 2 points on each side of AB: 4 positions.

Teacher's Note:
1. The condition is the diameter form of a circle.
2. Count the intersection points of each parallel line with the circle.

Answer: (d) concyclic

Explanation:
1. The four points are the corners of an axis-parallel rectangle with sides @
2. \gamma-\alpha
3. \( and \)
4. \delta-\beta
5. @.
6. The sides are unequal, so it is not a square or rhombus.
7. The corners of a rectangle are concyclic (they lie on the circle through the diagonals' common midpoint).

Teacher's Note:
1. A rectangle has equal diagonals that bisect each other.
2. The midpoint of the diagonals is equidistant from all corners.

Answer: (d) rhombus

Explanation:
1. The slopes of the diagonals are \( -\frac13 \) and 3, whose product is −1, so the diagonals are perpendicular.
2. A parallelogram with perpendicular diagonals is a rhombus.

Teacher's Note:
1. Check the product of slopes.
2. A square is a special case, but we cannot say it must be one.

Answer: (c) x − y + 3 = 0

Explanation:
1. The quadrilateral is a trapezium with parallel sides 2 and 4 and height 2, so its area is 6 and each half must be 3.
2. The line x − y + 3 = 0 passes through (−1, 2) and meets the top side y = 4 at (1, 4).
3. It cuts off the triangle with vertices (−2, 4), (−1, 2) and (1, 4), of base 3 along y = 4 and height 2, so its area is 3, half of 6.
4. The other options cut off areas 5 and 6 of the quadrilateral, so only (c) bisects the area.

Teacher's Note:
1. Compute the total area and test each line.
2. A line through the vertex must meet the opposite side.

Answer: (a) 9x − 20y + 96 = 0

Explanation:
1. Let the intercepts be A(p, 0) and B(0, q). The point divides AB as 5 : 3, so \( \left(\frac{3p}{8},\frac{5q}{8}\right)=(-4,3) \).
2. Hence \( p=-\frac{32}{3} \) and \( q=\frac{24}{5} \).
3. The line \( \frac xp+\frac yq=1 \) becomes \( -\frac{3x}{32}+\frac{5y}{24}=1 \), i.e. \( 9x-20y+96=0 \).

Teacher's Note:
1. Use the section formula with unknown intercepts.
2. Then write the intercept form.

Answer: (b) y = 1

Explanation:
1. The intercept of x + y = 2 between the axes joins (2, 0) and (0, 2), with midpoint (1, 1).
2. The intercept of 2x + 3y = 6 joins (3, 0) and (0, 2), with midpoint (3/2, 1).
3. The line through both midpoints is y = 1.

Teacher's Note:
1. Find the midpoint of each intercept segment.
2. Both midpoints have the same y-coordinate.

Answer: (c) x − y + 5 = 0

Explanation:
1. Equal length intercepts mean slope +1 or −1.
2. Slope −1 gives x + y = 1 (through (−2, 3)); slope +1 gives y − 3 = x + 2, i.e. x − y + 5 = 0.
3. Among the options only x − y + 5 = 0 is such a line.

Teacher's Note:
1. Equal lengths mean the intercepts are equal or opposite.
2. Check that the given point satisfies the equation.

Answer: (d) 9 : 1

Explanation:
1. A = (λ/3, 0) and B = (0, λ). In right triangle AOB, OC is the altitude to the hypotenuse.
2. Then \( \frac{BC}{CA}=\frac{OB^2}{OA^2}=\frac{\lambda^2}{\lambda^2/9}=9 \).

Teacher's Note:
1. In a right triangle the altitude divides the hypotenuse in the ratio of the squares of the adjacent sides.
2. Find the intercepts first.

Answer: (a) \( \sqrt2 \)

Explanation:
1. Points in the direction of x + y = 1 from (1, 1) are (1 + s, 1 − s), at distance
2. s
3. √2.
4. This point is on 2x − 3y = 4 when 2(1 + s) − 3(1 − s) = 4, so 5s = 5 and s = 1.
5. Distance = √2.

Teacher's Note:
1. Use the parametric form along a given direction.
2. Substitute into the second line.

Answer: (a) x − 4y + 5 = 0

Explanation:
1. The sides are x = 0, x = 2, y = 0, y = 3, a rectangle with centre (1, 3/2).
2. A line through the centre bisects the area of a rectangle.
3. The line with slope 1/4 through (1, 3/2): 4y − 6 = x − 1, i.e. x − 4y + 5 = 0.

Teacher's Note:
1. Any line through the centre of a rectangle halves its area.
2. Use the given slope.

Answer: (c) \( x+\sqrt3y=4 \)

Explanation:
1. The hexagon has side 1 and lies above AB: C = \( \left(\frac52,\frac{\sqrt3}{2}\right) \) and E = \( (1,\sqrt3) \).
2. Slope of CE = \( \frac{\sqrt3-\sqrt3/2}{1-5/2}=-\frac1{\sqrt3} \).
3. Line: \( x+\sqrt3y=\frac52+\frac32=4 \).

Teacher's Note:
1. Find the vertices using 60 degree turns.
2. Use the two-point form for CE.

Answer: (a) \( x+\sqrt3y=3 \)

Explanation:
1. With \( \angle A=120^\circ \) and BC = \( 4\sqrt3 \), AB = AC = 4 (since \( BC=2\cdot AB\sin60^\circ \)).
2. B = (3, 0) and C = (−1 + 4cos120°, 4 sin120°) = \( (-3,2\sqrt3) \) (taking C above the axis).
3. Slope BC = \( \frac{2\sqrt3}{-6}=-\frac1{\sqrt3} \), so the line is \( x+\sqrt3y=3 \).

Teacher's Note:
1. Find the equal sides from the triangle's angle.
2. The question takes AB along the positive x-direction from A.

Answer: (d) straight line passing through the point \( \left(\frac\pi2,-\sin^21\right) \) and parallel to the x-axis

Explanation:
1. \( \cos x\cos(x+2)=\frac12[\cos(2x+2)+\cos2] \) and \( \cos^2(x+1)=\frac12[1+\cos(2x+2)] \).
2. The difference is \( \frac{\cos2-1}{2}=-\sin^21 \), a constant.
3. So the graph is a horizontal line y = −sin²1, passing through \( (\pi/2,-\sin^21) \).

Teacher's Note:
1. Use the product-to-sum identity.
2. A constant function gives a line parallel to the x-axis.

Answer: (b) 1

Explanation:
1. The line through the first two points has slope \( \frac1{\sqrt3} \): \( x-\sqrt3y+2=0 \).
2. Its distance from the origin is \( \frac{2}{2}=1 \), so it touches the unit circle at exactly one point.
3. So exactly one value of θ works.

Teacher's Note:
1. The third point lies on the unit circle.
2. Compare the line's distance from the origin with the radius 1.

Answer: (a) (−5, 4)

Explanation:
1. At c = 1 the second line becomes 2x + 3y = 2.
2. Solving 3x + 4y = 1 and 2x + 3y = 2: x = −5 and y = 4.

Teacher's Note:
1. The intersection point is continuous in c.
2. Substitute c = 1 and solve the pair.

Answer: (d) none of these

Explanation:
1. Let the point be (x, 0): (x + 3)² + 16 = (x − 2)² + 25.
2. 6x + 25 = −4x + 29, so x = 2/5.
3. The point is (2/5, 0), which is not among the options.

Teacher's Note:
1. Set the squared distances equal.
2. Check the final value against each option.

Answer: (b) \( \frac{33}{10} \)

Explanation:
1. Write the second line as 3x + 4y + 7.5 = 0 and the first as 3x + 4y − 9 = 0.
2. Distance =
3. 7.5 + 9
4. /5 = 16.5/5 = 33/10.

Teacher's Note:
1. Make the coefficients of x and y equal first.
2. Use the distance formula for parallel lines.

Answer: (a) \( \left(\frac13,\frac13\right) \)

Explanation:
1. The foot of the perpendicular from the origin to x + y = 1 is (1/2, 1/2), the midpoint of the opposite side.
2. In an equilateral triangle the orthocentre is the centroid, at 2/3 of the way from the vertex along the median.
3. Orthocentre = (2/3)(1/2, 1/2) = (1/3, 1/3).

Teacher's Note:
1. All centres coincide in an equilateral triangle.
2. The centroid divides the median 2 : 1.