JEE Mathematics Coordinates and Straight Lines MCQs Set 05

Answer: (a) \( \frac1{p^2}+\frac1{q^2}=\frac1{a^2}+\frac1{b^2} \)

Explanation:
1. The perpendicular distance d of the line from the origin does not change on rotation.
2. For intercepts a, b: \( \frac1{d^2}=\frac1{a^2}+\frac1{b^2} \), and likewise \( \frac1{d^2}=\frac1{p^2}+\frac1{q^2} \).
3. So \( \frac1{p^2}+\frac1{q^2}=\frac1{a^2}+\frac1{b^2} \).

Teacher's Note:
1. The distance from the origin is invariant under rotation about the origin.
2. Use the intercept form distance formula.

Answer: (c) a circle

Explanation:
1. Let A = (p, 0), B = (0, q) with \( p^2+q^2=d^2 \). The midpoint is \( (x,y)=\left(\frac p2,\frac q2\right) \).
2. So \( x^2+y^2=\frac{d^2}{4} \), a circle.

Teacher's Note:
1. Express the midpoint in terms of p and q.
2. Eliminate p and q using the constant length.

Answer: (c) \( (x-7y)^2+14(x-7y)-15=0 \)

Explanation:
1. The shoelace formula for (6, 1), (7, 2), (−1, 0), (x, y) gives the signed area \( \frac{x-7y+7}{2} \).
2. Setting it equal to ±4: \( x-7y=1 \) or \( x-7y=-15 \).
3. Combined locus: \( (x-7y-1)(x-7y+15)=0 \), i.e. \( (x-7y)^2+14(x-7y)-15=0 \).

Teacher's Note:
1. Express the area with the shoelace formula.
2. Both signs of the area are possible.

Answer: (c) \( \frac ax+\frac by=1 \)

Explanation:
1. Let A = (p, 0) and B = (0, q). The line AB passes through (a, b), so \( \frac ap+\frac bq=1 \).
2. The vertical line through A is x = p and the horizontal line through B is y = q, so P = (p, q).
3. Locus: \( \frac ax+\frac by=1 \).

Teacher's Note:
1. Write the intercept form passing through the given point.
2. P has the same coordinates as the intercepts.

Answer: (a) Only (i), (iii) and (iv)

Explanation:
1. The centroid is the average of the vertices, so it is rational.
2. The circumcentre is found by solving two linear equations with rational coefficients, so it is rational.
3. The orthocentre is also found from linear equations with rational coefficients, so it is rational.
4. The incentre uses the side lengths, which can be irrational, so it need not be rational.

Teacher's Note:
1. Rational points stay rational under linear operations.
2. Square roots of side lengths appear only in the incentre.

Answer: (d) Only (i) and (iii)

Explanation:
1. The given side has length 5, so the other side has length 2, perpendicular to (−3, −4): the vector is \( \pm\frac25(4,-3)=\pm\left(\frac85,-\frac65\right) \).
2. Adding to (−2, −1) and (1, 3): \( \left(-\frac25,-\frac{11}5\right),\left(\frac{13}5,\frac95\right) \).
3. Subtracting: \( \left(-\frac{18}5,\frac15\right),\left(-\frac35,\frac{21}5\right) \).
4. These are options (i) and (iii); (ii) and (iv) mix up vertices.

Teacher's Note:
1. Find the perpendicular vector of the required length.
2. Both sides of the given side are possible.

Answer: (b) Only (i) and (ii)

Explanation:
1. The centre is \( \left(\frac12,-1\right) \) and the half-diagonal vector is \( \left(\frac32,-2\right) \).
2. The other diagonal's half-vector is perpendicular and equal in length: \( \pm\left(2,\frac32\right) \).
3. The other vertices are \( \left(\frac52,\frac12\right) \) and \( \left(-\frac32,-\frac52\right) \).

Teacher's Note:
1. The diagonals of a square are equal, perpendicular and bisect each other.
2. Rotate the half-diagonal through 90 degrees.

Answer: (a) Only (i) and (iii)

Explanation:
1. A right triangle such as (0, 0), (1, 0), (0, 1) has integral vertices, and so does an isosceles triangle.
2. An equilateral triangle needs area \( \frac{\sqrt3}{4}s^2 \), which is irrational when \( s^2 \) is an integer, but the area of a lattice triangle is rational, so it is impossible.

Teacher's Note:
1. The area of a lattice triangle is half an integer.
2. Compare with the area of an equilateral triangle.

Answer: (c) Only (ii) and (iv)

Explanation:
1. The fourth vertex can be A + B − C for any labelling of the three given points.
2. (−1, 2) + (2, −1) − (3, 1) = (−2, 0).
3. (2, −1) + (3, 1) − (−1, 2) = (6, −2).
4. (−1, 2) + (3, 1) − (2, −1) = (0, 4).

Teacher's Note:
1. Three positions are possible for the fourth vertex.
2. Check each given option against the list.

Answer: (d) Only (ii) and (iv)

Explanation:
1. The first diagonal has slope −1 and meets \( x-y=\alpha \) at the centre \( C\left(\alpha+\frac\beta2,\frac\beta2\right) \).
2. The half-diagonal is \( \left(\frac\beta2,-\frac\beta2\right) \), so the other vertices are \( C\pm\left(\frac\beta2,\frac\beta2\right)=(\alpha+\beta,\beta) \) and \( (\alpha,0) \).

Teacher's Note:
1. The diagonals of a square are perpendicular and bisect each other.
2. Find the centre first.

Answer: (c) Only (i) and (ii)

Explanation:
1. The line is \( 3x+4y-12=0 \). For (t, t): @\frac{
2. 7t-12
3. }{5}=4\( , so \)7t-12=\pm20@.
4. t = 32/7 or t = −8/7.

Teacher's Note:
1. Use the distance formula with the point (t, t).
2. Both signs of the modulus give valid points.

Answer: (d) slope of the line \( =\tan^{-1}3 \)

Explanation:
1. The direction ratios are \( 1:3 \), so the slope is 3 and the inclination is \( \tan^{-1}3 \).
2. The line is \( y-1=3(x+2) \), i.e. \( y=3x+7 \): x-intercept \( -\frac73 \) and y-intercept 7.
3. Only the statement about the inclination \( \tan^{-1}3 \) is correct.

Teacher's Note:
1. Eliminate r to get the Cartesian equation.
2. The coefficient of r in x and y gives the direction.

Answer: (c) Only (i) and (iii)

Explanation:
1. The vertices are O, \( A(a\cos\alpha,a\sin\alpha) \), \( B(a(\cos\alpha-\sin\alpha),a(\sin\alpha+\cos\alpha)) \) and \( D(-a\sin\alpha,a\cos\alpha) \).
2. Diagonal OB: \( y(\cos\alpha-\sin\alpha)=x(\cos\alpha+\sin\alpha) \) (statement i).
3. Diagonal AD: \( y(\sin\alpha+\cos\alpha)-x(\sin\alpha-\cos\alpha)=a \) (statement iii).

Teacher's Note:
1. Find all four vertices first.
2. Write each diagonal in two-point form.

Answer: (a) Only (i) and (iii)

Explanation:
1. The side x = k is vertical, so the hypotenuse makes 45° with the vertical.
2. The slope of x + my = 1 is −1/m, so
3. −1/m
4. = 1 and m = ±1.
5. Hence m = 1 or m = −1.

Teacher's Note:
1. The angle between the hypotenuse and a leg is 45 degrees.
2. A vertical line has no slope, so use angle with the vertical.

Answer: (b) Only (i) and (iii)

Explanation:
1. The vector from the centroid to the vertex is (0, 1).
2. The other vertices are the centroid plus this vector rotated by ±120°: \( (1,1)+\left(\mp\frac{\sqrt3}2,-\frac12\right) \).
3. They are \( \left(\frac{2-\sqrt3}{2},\frac12\right) \) and \( \left(\frac{2+\sqrt3}{2},\frac12\right) \).

Teacher's Note:
1. The vertices are spaced 120 degrees apart around the centroid.
2. Rotate the vector from the centroid.

Answer: (b) Only (i) and (ii)

Explanation:
1. A vertex on \( x=\sqrt3y \) at distance 2 is \( (\sqrt3,1) \) (or \( (-\sqrt3,-1) \)).
2. Rotating \( (\sqrt3,1) \) by +60° gives (0, 2); by −60° gives \( (\sqrt3,-1) \).
3. These are statements (i) and (ii).

Teacher's Note:
1. The third vertex is a 60 degree rotation of the second.
2. Check both rotation directions.

Answer: (d) Only (iii) and (iv)

Explanation:
1. The given line has slope 3. A line with slope m making 45° with it satisfies @\left
2. \frac{m-3}{1+3m}\right
3. =1@.
4. This gives \( m=\frac12 \) or \( m=-2 \).
5. Lines: y = x/2, i.e. x = 2y, and y = −2x, i.e. y + 2x = 0.

Teacher's Note:
1. Use the tangent of the angle between two lines.
2. Both signs of the modulus give valid lines.

Answer: (c) Only (i) and (ii)

Explanation:
1. The lines
2. x
3. =
4. y
5. are y = x and y = −x. Points equidistant from them lie on their angle bisectors, the coordinate axes.
6. On x + y = 3 the axes are met at (3, 0) and (0, 3).

Teacher's Note:
1. Equidistant from two lines means on a bisector.
2. The bisectors of y = ±x are the axes.

Answer: (c) Only (ii) and (iii)

Explanation:
1. The perpendicular line is \( 2x+3y=c \), and the x-intercept 3 gives \( c=6 \): \( 2x+3y=6 \).
2. Slope \( =-\frac23 \), y-intercept = 2, and area \( =\frac12\cdot3\cdot2=3 \).
3. So statements (ii) and (iii) are correct.

Teacher's Note:
1. A perpendicular line swaps the coefficients of x and y with a sign change.
2. Use the intercept to fix the constant.

Answer: (c) Only (ii) and (iii)

Explanation:
1. The diagonal joins (2, 0) and \( (0,2\sqrt3) \), with centre \( (1,\sqrt3) \) and half-diagonal \( (-1,\sqrt3) \).
2. The other diagonal has half-vector \( \pm(\sqrt3,1) \), so its ends are \( (1+\sqrt3,\sqrt3+1) \) and \( (1-\sqrt3,\sqrt3-1) \).

Teacher's Note:
1. Rotate the half-diagonal by 90 degrees about the centre.
2. Add and subtract it.

Answer: (c) Only (ii) and (iv)

Explanation:
1. The intercepts are \( \frac ab \) and \( \frac ac \), so the area is \( \frac{a^2}{2bc}=\frac18 \), giving \( 4a^2=bc \).
2. Then \( (2a)^2=bc \) and \( (-2a)^2=bc \), so b, 2a, c are in GP, and so are b, −2a, c.
3. b, a, c are not in GP and b, a/2, c are not in AP.

Teacher's Note:
1. Equate the area to 1/8.
2. The GP condition is that the middle term squared equals the product of the others.

Answer: (b) Only (ii) and (iii)

Explanation:
1. The vertices are (7/8, 1/8), (0, 0) and ((3 + √3)/2, −(1 + √3)/2), about (2.37, −1.37); the triangle is obtuse, so the circumcentre and the orthocentre lie outside it.
2. The centroid and the incentre are always inside.
3. Numerically the circumcentre is about (0.67, −1.57) and the orthocentre about (1.9, 1.9), both outside.

Teacher's Note:
1. Centroid and incentre are always interior.
2. In an obtuse triangle circumcentre and orthocentre are outside.

Answer: (d) Only (ii) and (iv)

Explanation:
1. y = 2x meets the first line where \( 2x+2+2x=0 \), i.e. \( x=-\frac12 \) (y = −1).
2. It meets the second line where \( x+6x-3=0 \), i.e. \( x=\frac37 \) (y = 6/7).
3. So \( x\in\left(-\frac12,\frac37\right) \) and \( y\in\left(-1,\frac67\right) \).

Teacher's Note:
1. Find the two intersection points on the line.
2. Convert the x-range into the y-range with y = 2x.

Answer: (d) Only (ii) and (iv)

Explanation:
1. The vertices are (1/2, 1/2), (1/4, 1/4) and (3/2, −1/2).
2. Two altitudes meet at the orthocentre (1/2, 1/2) (the triangle is right angled at (1/2, 1/2), since x + y = 1 and x − y = 0 are perpendicular).
3. Check: 3x − 5y + 1 = 0 and 5x − 3y = 1 are both satisfied by (1/2, 1/2).

Teacher's Note:
1. A right triangle has its orthocentre at the right-angle vertex.
2. Substitute the point in each option.

Answer: (d) Only (ii) and (iii)

Explanation:
1. The mirror is a bisector of the incident and reflected lines: (3x − 4y − 5)/5 = ±(5x + 12y − 13)/13.
2. With + sign: 39x − 52y − 65 = 25x + 60y − 65, so x = 8y.
3. With − sign: 39x − 52y − 65 = −25x − 60y + 65, so 32x + 4y = 65.

Teacher's Note:
1. The reflecting line bisects the angle between the two rays.
2. Both bisectors can act as the mirror.

Answer: (a) \( x+(2-\sqrt3)y=1 \)

Explanation:
1. The incident ray meets the x-axis at (1, 0) and makes 45° with the axis.
2. Turning away from the axis by 30° makes the angle 75°, so the slope is \( -\tan75^\circ=-(2+\sqrt3) \).
3. The line \( y=-(2+\sqrt3)(x-1) \) is \( x+(2-\sqrt3)y=1 \) after multiplying by \( (2-\sqrt3) \).

Teacher's Note:
1. Find the point of incidence and the new inclination.
2. Rationalise the slope to match the options.