JEE Mathematics Coordinates and Straight Lines MCQs Set 06

Answer: (b) 2, 5, (2, 5)

Explanation:
1. The abscissa is the distance from the y-axis, taken with sign, so it is 2.
2. The ordinate is the distance from the x-axis, so it is 5. The coordinates are (abscissa, ordinate) = (2, 5).

Teacher's Note:
1. The abscissa always comes first in the pair.
2. A point to the right of the y-axis has a positive abscissa.

Answer: (a) \( (-5, 2) \)

Explanation:
1. Left of the y-axis means the abscissa is negative: \( -5 \).
2. Above the x-axis means the ordinate is positive: 2. So P is \( (-5, 2) \).

Teacher's Note:
1. Left is negative x and below is negative y.
2. This point lies in the second quadrant.

Answer: (b) \( 2\sqrt{2}\,|b| \)

Explanation:
1. The x-difference is \( (a+b) - (a-b) = 2b \) and the y-difference is \( (b+c) - (c-b) = 2b \).
2. \( d = \sqrt{4b^2 + 4b^2} = 2\sqrt{2}\,|b| \).

Teacher's Note:
1. Collect the terms in a, b and c separately in each difference.
2. Equal differences in x and y mean a 45 degree line.

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

Explanation:
1. The points lie on the axes, forming a right triangle with legs |a| and |b| with the origin.
2. \( d = \sqrt{a^2 + b^2} \).

Teacher's Note:
1. This is the length of the intercept of a line between the axes.
2. Do not add the legs.

Answer: (c) \( (3, 0) \)

Explanation:
1. Let the point be \( (x, 0) \). Then \( (x-5)^2 + 81 = (x+4)^2 + 36 \).
2. \( -10x + 25 + 81 = 8x + 16 + 36 \Rightarrow 54 = 18x \Rightarrow x = 3 \).

Teacher's Note:
1. A point on the x-axis has ordinate zero.
2. Equate the squared distances to avoid roots.

Answer: (b) \( (-b,\ b) \)

Explanation:
1. Diagonals bisect each other, so \( D = A + C - B \).
2. \( x = (a+b) + (a-b) - (2a+b) = -b \) and \( y = (a-b) + (a+b) - (2a-b) = b \).

Teacher's Note:
1. In a parallelogram \( A + C = B + D \).
2. Take the vertices in order around the figure.

Answer: (c) \( (-4, -7) \)

Explanation:
1. \( \dfrac{1+3+x}{3} = 0 \Rightarrow x = -4 \) and \( \dfrac{2+5+y}{3} = 0 \Rightarrow y = -7 \).

Teacher's Note:
1. The coordinates of the three vertices sum to three times the centroid.
2. Here the sums are zero.

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

Explanation:
1. The triangle formed by the mid-points has the same centroid as the original triangle.
2. \( x = \dfrac{-2+4+4}{3} = 2 \) and \( y = \dfrac{3-3+5}{3} = \dfrac{5}{3} \).

Teacher's Note:
1. The medial triangle shares the centroid with the main triangle.
2. There is no need to find the vertices.

Answer: (d) \( a^2 \)

Explanation:
1. Area \( = \tfrac{1}{2}|a(c-(c-a)) + a((c-a)-(c+a)) + (-a)((c+a)-c)| \).
2. \( = \tfrac{1}{2}|a\cdot a + a(-2a) + (-a)(a)| = \tfrac{1}{2}|a^2 - 2a^2 - a^2| = a^2 \).

Teacher's Note:
1. Two vertices share the x-coordinate a, so the base is vertical with length a.
2. The height is the horizontal distance 2a, giving \( \tfrac{1}{2}\cdot a\cdot 2a \).

Answer: (c) \( (2a, 2a) \)

Explanation:
1. B is at \( (2a, 0) \) and D at \( (0, 2a) \).
2. C is diagonally opposite to A, so it is \( (2a, 2a) \).

Teacher's Note:
1. Place the figure so the sides lie along the axes.
2. Opposite vertex of a square from the origin has both coordinates equal to the side.

Answer: (b) \( (-a, -a) \) and \( (a, a) \)

Explanation:
1. The half side is a, so the vertices are at distance a from the axes.
2. A is \( (-a, -a) \) and C is \( (a, a) \).

Teacher's Note:
1. Each vertex is at \( (\pm a, \pm a) \).
2. The centre is the mid-point of each diagonal.

Answer: (a) rectangle which is not a square

Explanation:
1. Mid-point of the first diagonal \( (-4,-1),(4,0) \) is \( (0, -\tfrac{1}{2}) \); of the second \( (-2,-4),(2,3) \) is \( (0, -\tfrac{1}{2}) \), so it is a parallelogram.
2. Both diagonals have length \( \sqrt{65} \), so it is a rectangle. The sides are \( \sqrt{13} \) and \( 2\sqrt{13} \), unequal, so it is not a square.

Teacher's Note:
1. Equal diagonals make a parallelogram a rectangle.
2. A square also needs equal adjacent sides.

Answer: (d) parallelogram

Explanation:
1. Mid-point of AC is \( (3, 4) \) and mid-point of BD is \( (3, 4) \). The diagonals bisect each other.
2. Sides: \( AB = \sqrt{4+64} = \sqrt{68} \), \( BC = \sqrt{4+16} = \sqrt{20} \), unequal, and \( AC \ne BD \), so it is only a parallelogram.

Teacher's Note:
1. Bisecting diagonals prove a parallelogram.
2. Unequal sides rule out a rhombus and unequal diagonals rule out a rectangle.

Answer: (b) right angled isosceles triangle

Explanation:
1. Sides: \( (3,0)\text{-}(6,4) = 5 \), \( (3,0)\text{-}(-1,3) = 5 \), \( (6,4)\text{-}(-1,3) = \sqrt{50} \).
2. Two sides are equal and \( 5^2 + 5^2 = 50 \), so the angle between them is a right angle.

Teacher's Note:
1. Check the Pythagoras relation on the longest side.
2. Two equal sides give an isosceles triangle.

Answer: (c) are collinear and do not form a triangle

Explanation:
1. Area \( = \tfrac{1}{2}\left|2(-6-\tfrac{3}{2}) + (-4)(\tfrac{3}{2}-3) + 1(3+6)\right| = \tfrac{1}{2}|-15 + 6 + 9| = 0 \).
2. Zero area means the points are collinear.

Teacher's Note:
1. Collinear points give zero area.
2. No triangle can be formed from collinear points.

Answer: (b) \( \left(\dfrac{1 - \sqrt{3}}{2},\ \dfrac{7 + 5\sqrt{3}}{2}\right) \)

Explanation:
1. The side joining the two given points has direction \( (5, 1) \) and length \( \sqrt{26} \); its mid-point is \( \left(\tfrac{1}{2}, \tfrac{7}{2}\right) \). The third vertex lies on the perpendicular bisector at distance \( \dfrac{\sqrt{3}}{2}\sqrt{26} \) from the mid-point, along the unit direction \( \pm\dfrac{(-1, 5)}{\sqrt{26}} \).
2. The third vertex is \( \left(\tfrac{1}{2}, \tfrac{7}{2}\right) \pm \dfrac{\sqrt{3}}{2}(-1, 5) \), which gives \( \left(\dfrac{1 - \sqrt{3}}{2}, \dfrac{7 + 5\sqrt{3}}{2}\right) \) and \( \left(\dfrac{1 + \sqrt{3}}{2}, \dfrac{7 - 5\sqrt{3}}{2}\right) \). The first is one of the two valid answers, and it is the only one of that form among the options.

Teacher's Note:
1. There are two possible third vertices, one on each side of the given side.
2. Only one of them appears in the options.

Answer: (d) \( \left(2 + \dfrac{\sqrt{11}}{2},\ \dfrac{5}{2}\right) \)

Explanation:
1. The third vertex is equidistant (3) from both points, so it lies on the perpendicular bisector \( y = \tfrac{5}{2} \).
2. \( (x-2)^2 + \tfrac{25}{4} = 9 \Rightarrow (x-2)^2 = \tfrac{11}{4} \Rightarrow x = 2 \pm \dfrac{\sqrt{11}}{2} \). The option with the plus sign is correct.

Teacher's Note:
1. The apex lies on the perpendicular bisector of the base.
2. There are two symmetric answers.

Answer: (b) \( \left(\dfrac{8}{7}, -\dfrac{9}{7}\right) \)

Explanation:
1. Section formula: \( x = \dfrac{3(4) + 4(-1)}{7} = \dfrac{8}{7} \).
2. \( y = \dfrac{3(-7) + 4(3)}{7} = -\dfrac{9}{7} \).

Teacher's Note:
1. The ratio m : n attaches m to the second point.
2. Check that the answer lies between the two points.

Answer: (a) \( \left(1, -\dfrac{7}{3}\right) \) and \( \left(-3, \dfrac{4}{3}\right) \)

Explanation:
1. Divide in the ratios 1 : 2 and 2 : 1.
2. First point: \( \left(\dfrac{1(-7) + 2(5)}{3}, \dfrac{1(5) + 2(-6)}{3}\right) = \left(1, -\dfrac{7}{3}\right) \).
3. Second point: \( \left(\dfrac{2(-7) + 1(5)}{3}, \dfrac{2(5) + 1(-6)}{3}\right) = \left(-3, \dfrac{4}{3}\right) \).

Teacher's Note:
1. Trisection points divide the segment in 1 : 2 and 2 : 1.
2. Both points lie between the end points.

Answer: (c) \( (1, 2) \)

Explanation:
1. With B the middle vertex, \( D = A + C - B = (-2 + 4 - 1,\ -1 + 3 - 0) \).
2. \( D = (1, 2) \).

Teacher's Note:
1. The middle vertex is the one that is adjacent to both others.
2. The diagonals bisect each other.

Answer: (b) 5

Explanation:
1. The mid-point of BC is \( (3, 0) \).
2. \( AD = \sqrt{(3+1)^2 + (0-3)^2} = \sqrt{16 + 9} = 5 \).

Teacher's Note:
1. A median joins a vertex to the mid-point of the opposite side.
2. Find the mid-point first.

Answer: (a) \( -15 \)

Explanation:
1. Mid-point \( = \left(\dfrac{3 + k}{2}, \dfrac{11}{2}\right) \).
2. \( (3 + k) + 11 + 1 = 0 \Rightarrow k = -15 \).

Teacher's Note:
1. Substitute the mid-point into the equation of the line.
2. The coefficient 2 cancels the halves.

Answer: (c) 3 : 2 and \( m = -\dfrac{2}{5} \)

Explanation:
1. Let the ratio be \( k : 1 \). From the y-coordinate: \( \dfrac{8k + 3}{k + 1} = 6 \Rightarrow 2k = 3 \Rightarrow k = \dfrac{3}{2} \).
2. \( m = \dfrac{2k - 4}{k + 1} = \dfrac{3 - 4}{5/2} = -\dfrac{2}{5} \).

Teacher's Note:
1. Use the known coordinate to find the ratio.
2. Then use the ratio to find the unknown.

Answer: (a) 80 square units

Explanation:
1. Split along a diagonal. Triangle \( (-3,2),(5,4),(7,-6) \) has area \( \tfrac{1}{2}|-3(10) + 5(-8) + 7(-2)| = \tfrac{1}{2}|-84| = 42 \).
2. Triangle \( (-3,2),(7,-6),(-5,-4) \) has area \( \tfrac{1}{2}|-3(-2) + 7(-6) + (-5)(8)| = \tfrac{1}{2}|6 - 42 - 40| = 38 \).
3. Total \( = 42 + 38 = 80 \).

Teacher's Note:
1. Divide a quadrilateral into two triangles.
2. Take the vertices in order.

Answer: (d) \( \dfrac{11}{2} \) square units

Explanation:
1. Triangle \( (1,2),(6,2),(5,3) \): area \( \tfrac{1}{2}\cdot 5\cdot 1 = \dfrac{5}{2} \).
2. Triangle \( (1,2),(5,3),(3,4) \): area \( \tfrac{1}{2}|1(-1) + 5(2) + 3(-1)| = \tfrac{1}{2}(6) = 3 \).
3. Total \( = \dfrac{5}{2} + 3 = \dfrac{11}{2} \).

Teacher's Note:
1. Split into two triangles from one vertex.
2. The base of the first triangle is horizontal.

Answer: (b) \( (1, -1),\ (2, 1),\ (4, 5) \)

Explanation:
1. For \( (1,-1),(2,1),(4,5) \): slopes are \( \dfrac{1+1}{2-1} = 2 \) and \( \dfrac{5-1}{4-2} = 2 \). They are equal, so the points are collinear.
2. For \( (2,5),(4,6),(8,9) \) the slopes are \( \tfrac{1}{2} \) and \( \tfrac{3}{4} \); the others also have unequal slopes.

Teacher's Note:
1. Equal slopes between consecutive points mean collinear.
2. Zero area is the equivalent test.

Answer: (a) a parallelogram

Explanation:
1. Mid-point of the diagonal joining \( (3,-2) \) and \( (6,-3) \) is \( \left(\dfrac{9}{2}, -\dfrac{5}{2}\right) \). The mid-point of the other diagonal \( (4,0),(5,-5) \) is also \( \left(\dfrac{9}{2}, -\dfrac{5}{2}\right) \).
2. The diagonals bisect each other, so it is a parallelogram. Sides \( \sqrt{5} \) and \( \sqrt{10} \) differ, so it is not a rhombus.

Teacher's Note:
1. Same mid-point of the diagonals means a parallelogram.
2. Adjacent sides are unequal.

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

Explanation:
1. The centre is equidistant from the three points. Equating distances from \( (2,-9) \) and \( (2,1) \): the centre lies on \( y = -4 \).
2. Equating distances from \( (5,-8) \) and \( (2,-9) \): \( (x-5)^2 + (y+8)^2 = (x-2)^2 + (y+9)^2 \Rightarrow 6x + 2y = 4 \). With \( y = -4 \) this gives \( x = 2 \).
3. The centre is \( (2, -4) \).

Teacher's Note:
1. The centre is the circumcentre of the triangle.
2. Perpendicular bisector of a vertical segment is horizontal.

Answer: (b) \( x = 5 \) or \( x = -3 \)

Explanation:
1. \( PQ^2 = 25 + 16 = 41 \) and \( QR^2 = (x - 1)^2 + 25 \).
2. \( (x - 1)^2 = 16 \Rightarrow x - 1 = \pm 4 \Rightarrow x = 5 \) or \( x = -3 \).

Teacher's Note:
1. A square root equation has two solutions.
2. Both positions of R satisfy PQ = QR.

Answer: (b) \( \left(\dfrac{7}{2}, \dfrac{13}{2}\right) \) or \( \left(-\dfrac{3}{2}, \dfrac{3}{2}\right) \)

Explanation:
1. Let the vertex be \( (x, x + 3) \). Area \( = \tfrac{1}{2}|2(-2 - x - 3) + 3(x + 3 - 1) + x(1 + 2)| \), and the bracket simplifies to \( 2(-x - 5) + 3(x + 2) + 3x = 4x - 4 \).
2. \( \tfrac{1}{2}|4x - 4| = 5 \Rightarrow 4x - 4 = \pm 10 \Rightarrow x = \dfrac{7}{2} \) or \( x = -\dfrac{3}{2} \).
3. The vertices are \( \left(\dfrac{7}{2}, \dfrac{13}{2}\right) \) and \( \left(-\dfrac{3}{2}, \dfrac{3}{2}\right) \).

Teacher's Note:
1. The modulus gives two values.
2. Both points lie on the given line, one on each side of the base.