JEE Mathematics Coordinates and Straight Lines MCQs Set 07

Answer: (a) can never be collinear

Explanation:
1. Area \( = \tfrac{1}{2}|a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2)| = \tfrac{1}{2}|(a - b)(b - c)(a - c)| \) up to sign.
2. This is non-zero when a, b, c are distinct, so the points are never collinear (they lie on a parabola).

Teacher's Note:
1. A parabola cannot contain three collinear points.
2. Collinear needs zero area.

Answer: (d) \( x = \dfrac{11}{8} \) or \( x = -\dfrac{3}{8} \)

Explanation:
1. \( \Delta ABC = \dfrac{49}{2} \).
2. \( \Delta DBC = \tfrac{1}{2}|x(5 + 2) + (-3)(-2 - 3x) + 4(3x - 5)| = \tfrac{1}{2}|7x + 6 + 9x + 12x - 20| = \tfrac{1}{2}|28x - 14| \).
3. \( \tfrac{1}{2}|28x - 14| = \dfrac{49}{4} \Rightarrow |28x - 14| = \dfrac{49}{2} \Rightarrow 28x - 14 = \pm\dfrac{49}{2} \Rightarrow x = \dfrac{11}{8} \) or \( -\dfrac{3}{8} \).

Teacher's Note:
1. Compute the area of ABC first.
2. Expect two solutions from the modulus.

Answer: (c) 5

Explanation:
1. Area zero: \( a(-1 - 4) + 1(4 - 1) + 11(1 + 1) = 0 \).
2. \( -5a + 3 + 22 = 0 \Rightarrow a = 5 \).

Teacher's Note:
1. Equate the area expression to zero.
2. The same answer follows from equal slopes.

Answer: (a) \( \sqrt{a^2 + b^2} \)

Explanation:
1. \( d^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 \).
2. \( = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta) = a^2 + b^2 \).

Teacher's Note:
1. The cross terms cancel.
2. The result is independent of theta.

Answer: (b) \( 2 + \sqrt{2} \)

Explanation:
1. Two sides have length 1 and the hypotenuse is \( \sqrt{2} \).
2. Perimeter \( = 1 + 1 + \sqrt{2} = 2 + \sqrt{2} \).

Teacher's Note:
1. It is a right isosceles triangle.
2. Add all three sides.

Answer: (a) \( \sqrt{85} \)

Explanation:
1. The mid-point of AB is \( (-1, -1) \).
2. \( CM = \sqrt{(5 + 1)^2 + (-8 + 1)^2} = \sqrt{36 + 49} = \sqrt{85} \).

Teacher's Note:
1. The median goes to the mid-point of the opposite side.
2. Find the mid-point of AB first.

Answer: (c) \( -\dfrac{1}{3} \)

Explanation:
1. Area zero: \( k(3k - 1) + 3k(1 - 2k) + 3(2k - 3k) = 0 \).
2. \( 3k^2 - k + 3k - 6k^2 - 3k = -3k^2 - k = 0 \Rightarrow k(3k + 1) = 0 \). With \( k \ne 0 \), \( k = -\dfrac{1}{3} \).

Teacher's Note:
1. k = 0 makes two of the points coincide.
2. Factorise after setting the area to zero.

Answer: (b) \( \left(\dfrac{2}{5}, 0\right) \)

Explanation:
1. \( (x + 3)^2 + 16 = (x - 2)^2 + 25 \Rightarrow 6x + 25 = -4x + 29 \).
2. \( 10x = 4 \Rightarrow x = \dfrac{2}{5} \).

Teacher's Note:
1. The point has the form \( (x, 0) \).
2. The squares of x cancel.

Answer: (d) \( -4 \) or 2

Explanation:
1. Right angle at P: \( PA^2 + PB^2 = AB^2 \), or equivalently \( \vec{PA}\cdot\vec{PB} = 0 \).
2. \( (5 - 12)(11 - 12) + (3 - y)(-5 - y) = 0 \Rightarrow 7 + y^2 + 2y - 15 = 0 \Rightarrow y^2 + 2y - 8 = 0 \Rightarrow y = 2 \) or \( -4 \).

Teacher's Note:
1. Use the dot product of the two sides at P.
2. A quadratic gives two positions of P.

Answer: (b) \( -63 \)

Explanation:
1. Slope between \( (-3,-4) \) and \( (7,-5) \) is \( \dfrac{-1}{10} \).
2. \( \dfrac{2 + 4}{x + 3} = -\dfrac{1}{10} \Rightarrow x + 3 = -60 \Rightarrow x = -63 \).

Teacher's Note:
1. Use equal slopes.
2. Check the sign of the slope.

Answer: (a) \( \dfrac{4}{3} \)

Explanation:
1. The line through \( (-2, 6) \) and \( (3, 1) \) is \( x + y = 4 \).
2. \( t + 2t = 4 \Rightarrow t = \dfrac{4}{3} \).

Teacher's Note:
1. Find the equation of the line first.
2. Substitute the third point.

Answer: (c) 1

Explanation:
1. The line through the intercepts is \( \dfrac{x}{a} + \dfrac{y}{b} = 1 \).
2. The point \( (1, 1) \) lies on it, so \( \dfrac{1}{a} + \dfrac{1}{b} = 1 \).

Teacher's Note:
1. Use the intercept form of the line.
2. Substitute the third point.

Answer: (b) \( \dfrac{183}{2} \) square units

Explanation:
1. The third vertex is \( (3(1) - 4 + 9,\ 3(4) + 3 - 7) = (8, 8) \).
2. Area \( = \tfrac{1}{2}|4(7 - 8) + (-9)(8 + 3) + 8(-3 - 7)| = \tfrac{1}{2}|-4 - 99 - 80| = \dfrac{183}{2} \).

Teacher's Note:
1. Find the third vertex from the centroid first.
2. Then apply the area formula.

Answer: (a) 3 : 1

Explanation:
1. On the y-axis \( x = 0 \). The ratio is the ratio of the x-distances: \( 3 : 1 \).
2. Check: \( x = \dfrac{3(1) + 1(-3)}{4} = 0 \).

Teacher's Note:
1. The ratio is \( |x_1| : |x_2| \) for the y-axis.
2. Both points lie on opposite sides, so the division is internal.

Answer: (d) 2 : 3

Explanation:
1. Using x: \( \dfrac{2 - 4}{4 - 7} \) magnitude: the distances along x are 2 and 3.
2. The ratio is \( (4 - 2) : (7 - 4) = 2 : 3 \).

Teacher's Note:
1. The ratio is the ratio of distances along either coordinate.
2. The point is nearer to the first point.

Answer: (c) 2 : 1

Explanation:
1. On the x-axis \( y = 0 \), so the ratio is \( |6| : |-3| = 2 : 1 \).
2. Check: \( y = \dfrac{2(-3) + 1(6)}{3} = 0 \).

Teacher's Note:
1. The ratio is the ratio of the y-distances.
2. The x-axis cuts between the two points.

Answer: (b) \( 3abc \)

Explanation:
1. Centroid at the origin gives \( a + b + c = 0 \).
2. When \( a + b + c = 0 \), \( a^3 + b^3 + c^3 = 3abc \).

Teacher's Note:
1. Use the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(\ldots) \).
2. The right factor is zero here.

Answer: (d) \( (4, -1) \)

Explanation:
1. \( AP : PB = 2 : 3 \).
2. \( y = \dfrac{2(5) + 3(-5)}{5} = -1 \), \( x = 4 \), so \( P = (4, -1) \).

Teacher's Note:
1. AP : AB = 2 : 5 means AP : PB = 2 : 3.
2. The segment is vertical, so x stays 4.

Answer: (c) \( \left(1, -\dfrac{3}{2}\right) \)

Explanation:
1. The centre is the mid-point of the diameter.
2. \( \left(\dfrac{-3 + 5}{2}, \dfrac{1 - 4}{2}\right) = \left(1, -\dfrac{3}{2}\right) \).

Teacher's Note:
1. The centre of a circle is the mid-point of a diameter.
2. Average each coordinate.

Answer: (b) \( a = 1,\ b = 4 \)

Explanation:
1. \( \dfrac{a + 3}{2} = 2 \Rightarrow a = 1 \).
2. \( \dfrac{2 + 6}{2} = b \Rightarrow b = 4 \).

Teacher's Note:
1. Match the mid-point coordinates one at a time.
2. Substitute back to check.

Answer: (a) \( a = 2,\ b = 2 \)

Explanation:
1. \( \dfrac{2a - 2}{2} = 1 \Rightarrow a = 2 \).
2. \( \dfrac{4 + 3b}{2} = 2a + 1 = 5 \Rightarrow 3b = 6 \Rightarrow b = 2 \).

Teacher's Note:
1. Use the x-coordinate to get a.
2. Then use a in the y-coordinate to get b.

Answer: (d) \( (3, -10) \)

Explanation:
1. The centre is the mid-point, so the other end is \( 2C - A \).
2. \( (2(2) - 1,\ 2(-3) - 4) = (3, -10) \).

Teacher's Note:
1. Double the centre and subtract the known end.
2. Check by averaging the two ends.

Answer: (d) \( a = 3 \) and \( P = \left(0, \dfrac{5}{2}\right) \)

Explanation:
1. On the y-axis the x-coordinate is zero: \( \dfrac{1(a) + 3(-1)}{4} = 0 \Rightarrow a = 3 \).
2. \( y = \dfrac{1(5) + 3\left(\tfrac{5}{3}\right)}{4} = \dfrac{10}{4} = \dfrac{5}{2} \).

Teacher's Note:
1. Set the x-coordinate of the point of division to zero.
2. Then find y with the same ratio.

Answer: (c) 5 : 1, \( \left(0, -\dfrac{13}{3}\right) \)

Explanation:
1. On the y-axis \( x = 0 \): ratio \( = 5 : 1 \) (distances 5 and 1).
2. \( y = \dfrac{5(-4) + 1(-6)}{6} = -\dfrac{26}{6} = -\dfrac{13}{3} \).

Teacher's Note:
1. Distances from the y-axis give the ratio.
2. The first point has the larger distance, so it gets the larger part.

Answer: (b) 3 : 5, \( \left(\dfrac{21}{4}, 2\right) \)

Explanation:
1. Distances from \( y = 2 \): A is 3 above and B is 5 below, so the ratio is 3 : 5.
2. \( x = \dfrac{3(4) + 5(6)}{8} = \dfrac{42}{8} = \dfrac{21}{4} \).

Teacher's Note:
1. The ratio equals the ratio of distances from the line.
2. Use the same ratio for x.

Answer: (a) 2 : 9, \( \left(\dfrac{24}{11}, -\dfrac{4}{11}\right) \)

Explanation:
1. Values of \( 2x + y - 4 \): at A it is \( 4 - 2 - 4 = -2 \); at B it is \( 6 + 7 - 4 = 9 \).
2. The ratio is \( 2 : 9 \). Point: \( \left(\dfrac{2(3) + 9(2)}{11}, \dfrac{2(7) + 9(-2)}{11}\right) = \left(\dfrac{24}{11}, -\dfrac{4}{11}\right) \).

Teacher's Note:
1. The ratio is the ratio of the absolute values of the line's expression at the ends.
2. They have opposite signs, so the division is internal.

Answer: (d) \( L(8, 1) \), \( M(6, 3) \), LM : BC = 1 : 2

Explanation:
1. \( L = \left(\dfrac{10 + 6}{2}, \dfrac{5 - 3}{2}\right) = (8, 1) \) and \( M = \left(\dfrac{10 + 2}{2}, \dfrac{5 + 1}{2}\right) = (6, 3) \).
2. \( LM = \sqrt{4 + 4} = 2\sqrt{2} \) and \( BC = \sqrt{16 + 16} = 4\sqrt{2} \), so \( LM = \dfrac{1}{2}BC \).

Teacher's Note:
1. This is the mid-point theorem.
2. The segment joining mid-points of two sides is half the third side.

Answer: (c) \( (1, 2) \) or \( (-5, 4) \)

Explanation:
1. If the mid-point is of the side from \( (-1, 4) \) to the third vertex, then the vertex is \( 2(0, 3) - (-1, 4) = (1, 2) \).
2. If it is of the side from \( (5, 2) \), the vertex is \( 2(0, 3) - (5, 2) = (-5, 4) \). Both are possible.

Teacher's Note:
1. The given mid-point can belong to either side through the unknown vertex.
2. The third side cannot have that mid-point because its ends are both given.

Answer: (b) \( A(-4, 0) \) and \( B(0, 6) \)

Explanation:
1. Let \( A = (a, 0) \) and \( B = (0, b) \). The mid-point is \( \left(\dfrac{a}{2}, \dfrac{b}{2}\right) = (-2, 3) \).
2. So \( a = -4 \) and \( b = 6 \).

Teacher's Note:
1. A point on the x-axis has zero ordinate.
2. Double the mid-point's coordinates.

Answer: (a) \( \left(\dfrac{x_1 + x_2 + x_3}{3},\ \dfrac{y_1 + y_2 + y_3}{3}\right) \)

Explanation:
1. The centroid divides each median in the ratio 2 : 1 from the vertex.
2. The median from \( (x_1, y_1) \) meets the mid-point of the opposite side, and the section formula gives \( \dfrac{2\cdot\frac{x_2 + x_3}{2} + x_1}{3} = \dfrac{x_1 + x_2 + x_3}{3} \), similarly for y.

Teacher's Note:
1. The centroid is the average of the three vertices.
2. Dividing by 2 gives a mid-point, not a centroid.