JEE Mathematics Straight Lines MCQs Set 11

Answer: (b) 5

Explanation:
1. \( d=\sqrt{(4-1)^2+(-1-3)^2}=\sqrt{9+16}=5 \)

Teacher's Note:
1. Use the distance formula directly.
2. 3, 4, 5 is a Pythagorean triple, a quick check.

Answer: (d) 41

Explanation:
1. \( d=\sqrt{40^2+9^2}=\sqrt{1681}=41 \)

Teacher's Note:
1. 9, 40, 41 is a Pythagorean triple.
2. The distance from the origin is simply the square root of x² + y².

Answer: (a) 2

Explanation:
1. \( d=\sqrt{4\cos^2\theta+4\sin^2\theta}=2 \)

Teacher's Note:
1. The point lies on a circle of radius 2 about the origin.
2. Use cos²θ + sin²θ = 1.

Answer: (b) \( \left(4,\tfrac53\right)\text{ and }\left(7,\tfrac73\right) \)

Explanation:
1. The points divide the segment in the ratios 1 : 2 and 2 : 1.
2. \( P=\left(\frac{10+2}{3},\frac{3+2}{3}\right)=\left(4,\tfrac53\right) \)
3. \( Q=\left(\frac{20+1}{3},\frac{6+1}{3}\right)=\left(7,\tfrac73\right) \)

Teacher's Note:
1. Trisection points divide the segment internally in the ratios 1 : 2 and 2 : 1.
2. Check that the three parts have equal length.

Answer: (d) 1 : 1

Explanation:
1. Let the ratio be k : 1. The y-coordinate of the dividing point is \( \frac{k(-1)+1(1)}{k+1}=0\Rightarrow k=1 \).
2. So the ratio is 1 : 1 (the x-axis cuts the segment at its midpoint (2, 0)).

Teacher's Note:
1. A point on the x-axis has y = 0; use that to find the ratio.
2. Opposite signs of y-coordinates mean the division is internal.

Answer: (c) (-1, 2), (1, -4), (3, 2)

Explanation:
1. Each vertex equals the sum of the two mid-points adjacent to it minus the third mid-point.
2. (1 + 0 - 2, 2 - 1 + 1) = (-1, 2); (0 + 2 - 1, -1 - 1 - 2) = (1, -4); (1 + 2 - 0, 2 - 1 + 1) = (3, 2).

Teacher's Note:
1. The medial triangle's vertices are the mid-points, and the original vertices form a triangle twice as large.
2. Verify by taking mid-points of the answer.

Answer: (d) (2, 10)

Explanation:
1. Diagonals bisect each other, so A + C = B + D.
2. D = A + C - B = (4 + 3 - 5, 1 + 7 + 2) = (2, 10).

Teacher's Note:
1. In parallelogram ABCD the diagonals AC and BD share a mid-point.
2. Be careful about which vertex is opposite which.

Answer: (c) 1

Explanation:
1. Collinear points give zero area, so \( \begin{vmatrix}k+1&2-k&1\\1-k&-k&1\\2+k&3-k&1\end{vmatrix}=0 \).
2. Expanding the determinant gives a linear equation in k with root k = 1.
3. Check at k = 1: points (2, 1), (0, -1), (3, 2) all lie on y = x - 1.

Teacher's Note:
1. Collinearity is the same as a zero determinant (zero area).
2. Substitute the answer back to confirm.

Answer: (d) 83 or -77

Explanation:
1. \( \text{Area}=\tfrac12\left|1(3-4)+2(4-2)+x(2-3)\right|=\tfrac12|3-x| \)
2. Setting this equal to 40 gives |3 - x| = 80, so x = -77 or x = 83.

Teacher's Note:
1. Area = half the modulus of the determinant, so there are two signs to consider.
2. Do not forget the modulus.

Answer: (a) 41/2 sq. units

Explanation:
1. Use the shoelace formula; the cross terms x_i y_(i+1) - x_(i+1) y_i are:
2. 1·4 - 3·1 = 1, 3·(-2) - 5·4 = -26, 5·(-7) - 4·(-2) = -27, 4·1 - 1·(-7) = 11.
3. Sum = 1 - 26 - 27 + 11 = -41, so the area is 41/2 sq. units.

Teacher's Note:
1. List the vertices in cyclic order before applying the shoelace formula.
2. Take half the modulus of the sum.

Answer: (b) (5, 5) or (5, -5)

Explanation:
1. P lies on the perpendicular bisector of AB: x = 5.
2. AB = 4, so the area is \( \tfrac12\cdot4\cdot|y|=10\Rightarrow |y|=5 \).
3. So P = (5, 5) or (5, -5).

Teacher's Note:
1. Equidistant from two points means P is on their perpendicular bisector.
2. The height of the triangle is |y| because AB lies on the x-axis.

Answer: (c) \( a^2\,|(t_1-t_2)(t_2-t_3)(t_3-t_1)| \)

Explanation:
1. Area = \( \tfrac12\begin{vmatrix}at_1^2&2at_1&1\\at_2^2&2at_2&1\\at_3^2&2at_3&1\end{vmatrix}=\tfrac12\cdot2a^2\,|\text{Vandermonde}| \).
2. The determinant reduces to the Vandermonde product, giving \( a^2|(t_1-t_2)(t_2-t_3)(t_3-t_1)| \).

Teacher's Note:
1. Take out a from the first column and 2a from the second to get a Vandermonde determinant.
2. Check with t = 0, 1, -1 on y² = 4x: area should be 2a².

Answer: (a) x² + y² - 2x = 0

Explanation:
1. Slopes of PA and PB are y/x and y/(x - 2); their product is -1.
2. y² = -x(x - 2), so x² + y² - 2x = 0 (circle on AB as diameter, excluding A and B).

Teacher's Note:
1. A right angle subtended by a segment means the point lies on the circle with that segment as diameter.
2. Use the product of slopes equal to -1.

Answer: (c) x² + y² = 4

Explanation:
1. Let A = (a, 0), B = (0, b) with a² + b² = 16 and mid-point (h, k) = (a/2, b/2).
2. Then 4h² + 4k² = 16, so x² + y² = 4.

Teacher's Note:
1. Express a and b in terms of h and k and substitute in a² + b² = 16.
2. The mid-point of the hypotenuse of right triangle AOB is at distance 2 from O.

Answer: (d) 9x² + 9y² = 16

Explanation:
1. Centroid (h, k) = (a/3, b/3), so a = 3h and b = 3k.
2. a² + b² = 16 gives 9h² + 9k² = 16.

Teacher's Note:
1. The centroid of A(a,0), B(0,b), O is (a/3, b/3).
2. Replace a and b in the length condition.

Answer: (c) \( x^2-\frac{y^2}{4}=1 \)

Explanation:
1. sec²θ - tan²θ = 1 gives \( x^2-\frac{y^2}{4}=1 \), a hyperbola.

Teacher's Note:
1. Use sec²θ - tan²θ = 1.
2. Here tan θ = y/2.

Answer: (d) y² = 4ax

Explanation:
1. t = y/(2a), so x = a·y²/(4a²) and y² = 4ax.

Teacher's Note:
1. Eliminate t by writing t = y/(2a).
2. This is the standard parabola.

Answer: (a) \( \frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1 \)

Explanation:
1. The difference of distances from two fixed points is constant, so P traces a hyperbola with foci (±c, 0) and transverse axis 2a.
2. Since b² = c² - a², the equation is \( \frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1 \).

Teacher's Note:
1. A constant difference of focal distances defines a hyperbola; a constant sum defines an ellipse.
2. The relation c² = a² + b² gives b² = c² - a².

Answer: (b) \( -\tfrac32 \)

Explanation:
1. \( m=\frac{0-9}{7-1}=-\frac96=-\tfrac32 \)

Teacher's Note:
1. Slope = change in y over change in x.
2. Mind the signs when subtracting.

Answer: (c) y = 2x + 1

Explanation:
1. y - 3 = 2(x - 1), so y = 2x + 1.

Teacher's Note:
1. Use the point-slope form y - y1 = m(x - x1).
2. Check that (1, 3) satisfies the answer.

Answer: (d) \( x-\sqrt3\,y-1=0 \)

Explanation:
1. The original line has slope 1, i.e. inclination 45°.
2. Clockwise rotation by 15° gives inclination 30°, slope \( \tfrac1{\sqrt3} \).
3. Through (1, 0): \( y=\frac{x-1}{\sqrt3} \), i.e. \( x-\sqrt3 y-1=0 \).

Teacher's Note:
1. Clockwise rotation decreases the inclination angle.
2. Convert the new slope into an equation through the fixed point.

Answer: (a) \( -\tfrac27,\ \tfrac32,\ \tfrac37 \)

Explanation:
1. Slope = -a/b = -2/7.
2. x-intercept: y = 0 gives x = 3/2. y-intercept: x = 0 gives y = 3/7.

Teacher's Note:
1. Put y = 0 for the x-intercept and x = 0 for the y-intercept.
2. Slope of ax + by + c = 0 is -a/b.

Answer: (b) x + 2y = 5

Explanation:
1. Let the y-intercept be b, then the x-intercept is 2b: x/(2b) + y/b = 1.
2. Through (1, 2): 1/(2b) + 2/b = 1, so 5/(2b) = 1 and b = 5/2.
3. The line is x/5 + 2y/5 = 1, i.e. x + 2y = 5.

Teacher's Note:
1. Use the double-intercept form with intercepts in the given ratio.
2. Substitute the given point to find the unknown intercept.

Answer: (c) 2x + y = 7

Explanation:
1. Let the x-intercept be a and the y-intercept 2a: x/a + y/(2a) = 1.
2. Through (2, 3): 2/a + 3/(2a) = 1, so 7/(2a) = 1 and a = 7/2.
3. The line is 2x/7 + y/7 = 1, i.e. 2x + y = 7.

Teacher's Note:
1. Double-intercept form with the stated ratio of intercepts.
2. Verify: 2(2) + 3 = 7.