JEE Mathematics Coordinates and Straight Lines MCQs Set 11

Answer: (d) \( (\sqrt2,\sqrt2) \)

Explanation:
1. Let the point be \( (t,t) \)
2. \( \sqrt{2t^2}=2\Rightarrow t=\pm\sqrt2 \)
3. In the options only \( (\sqrt2,\sqrt2) \) is on the line at this distance

Teacher's Note:
1. A point on y = x has equal co-ordinates.
2. Check that the point is on the line before measuring the distance.

Answer: (c) \( \sqrt3 \)

Explanation:
1. Side \( =\) distance between \( (1,1) \) and \( (-1,-1) \) \( =2\sqrt2 \), so side squared is 8
2. \( (1+\sqrt3)^2+(1-k)^2=8\Rightarrow k=\sqrt3 \) or \( 2-\sqrt3 \)
3. \( (-1+\sqrt3)^2+(-1-k)^2=8\Rightarrow k=\sqrt3 \) or \( -2-\sqrt3 \)
4. The common value is \( k=\sqrt3 \)

Teacher's Note:
1. Both remaining sides must equal the known side.
2. Take the value of k that satisfies both equations.

Answer: (b) \( \frac{ac+bd}{\sqrt{(a^2+b^2)(c^2+d^2)}} \)

Explanation:
1. \( \cos\theta=\frac{\overrightarrow{OA}\cdot\overrightarrow{OB}}{
2. OA
3.
4. OB
5. } \)
6. \( \overrightarrow{OA}\cdot\overrightarrow{OB}=ac+bd \)
7. \( \cos\theta=\frac{ac+bd}{\sqrt{(a^2+b^2)(c^2+d^2)}} \)

Teacher's Note:
1. Use the dot product formula for the angle between two vectors.
2. Match the x co-ordinates together and the y co-ordinates together.

Answer: (c) \( -\frac{11}{\sqrt{290}} \)

Explanation:
1. \( \overrightarrow{BA}=(-2,1) \), \( \overrightarrow{BC}=(7,3) \)
2. Dot product \( =-14+3=-11 \)
3. \(
4. BA
5. =\sqrt5 \), \(
6. BC
7. =\sqrt{58} \)
8. \( \cos B=\frac{-11}{\sqrt{290}} \)

Teacher's Note:
1. Form vectors from B to the other two vertices.
2. A negative cosine means angle B is obtuse.

Answer: (c) \( (2,\pm2\sqrt3) \)

Explanation:
1. The third vertex lies on the perpendicular bisector \( x=2 \) of the base
2. Its height is \( \frac{\sqrt3}{2}\times4=2\sqrt3 \)
3. So the third vertex is \( (2,\pm2\sqrt3) \)

Teacher's Note:
1. Use the height of an equilateral triangle, root 3 over 2 times the side.
2. There are two positions, one on each side of the base.

Answer: (c) \( \frac12(x_2^2-x_1^2+y_2^2-y_1^2) \)

Explanation:
1. \( (x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2 \)
2. Expanding and collecting: \( 2(x_1-x_2)x+2(y_1-y_2)y+x_2^2+y_2^2-x_1^2-y_1^2=0 \)
3. Dividing by 2 gives \( c=\frac12(x_2^2-x_1^2+y_2^2-y_1^2) \)

Teacher's Note:
1. The result is the perpendicular bisector of the two points.
2. Divide by 2 so that the coefficient of x matches the stated form.

Answer: (c) a right angled triangle

Explanation:
1. Squared sides: \( AB^2=100+16=116 \), \( BC^2=144+1=145 \), \( CA^2=4+25=29 \)
2. \( 116+29=145 \), so the angle at A is a right angle

Teacher's Note:
1. Compare the sum of the two smaller squared sides with the largest.
2. All three sides are different, so it is scalene.

Answer: (b) rectangle

Explanation:
1. Mid-point of AC \( =(0,-\frac12) \) and of BD \( =(0,-\frac12) \), so ABCD is a parallelogram
2. \( AC^2=65=BD^2 \), so the diagonals are equal
3. \( AB^2=13\ne BC^2=52 \), so it is not a rhombus
4. Hence it is a rectangle

Teacher's Note:
1. Equal diagonals turn a parallelogram into a rectangle.
2. Unequal adjacent sides rule out a rhombus or square.

Answer: (b) \( (4,4) \)

Explanation:
1. The three possible fourth vertices are \( (6,-2) \), \( (-2,2) \) and \( (4,4) \)
2. Taking \( A(1,3) \), \( B(2,0) \), \( C(5,1) \) with B as the vertex between A and C: \( D=A+C-B=(4,4) \)
3. Only \( (4,4) \) appears among the options

Teacher's Note:
1. Fourth vertex = sum of the two vertices on a diagonal minus the third.
2. Three different parallelograms can be formed from three points.

Answer: (b) parallelogram

Explanation:
1. Mid-point of the diagonal joining \( (-1,1) \) and \( (5,2) \) is \( \left(2,\frac32\right) \)
2. Mid-point of the diagonal joining \( (0,-3) \) and \( (4,6) \) is \( \left(2,\frac32\right) \)
3. Diagonals bisect each other; their lengths are \( \sqrt{37} \) and \( \sqrt{97} \), which are unequal, so it is only a parallelogram

Teacher's Note:
1. Check the mid-points of the diagonals first.
2. Unequal diagonals rule out a rectangle or square.

Answer: (c) an equilateral triangle

Explanation:
1. \( AB=2a \)
2. \( BC^2=3a^2+a^2=4a^2 \), so \( BC=2a \)
3. \( AC^2=3a^2+a^2=4a^2 \), so \( AC=2a \)
4. All sides are equal, so it is equilateral

Teacher's Note:
1. Compute all three side lengths.
2. Equal sides give an equilateral triangle, with all angles 60 degrees.

Answer: (b) right angled

Explanation:
1. \( PQ^2=68 \), \( QR^2=85 \), \( PR^2=17 \)
2. \( PQ^2+PR^2=85=QR^2 \), so the angle at P is a right angle
3. Sides are unequal, but the most specific description among the options is right angled

Teacher's Note:
1. Test the Pythagoras relation on the squared sides.
2. A right triangle with unequal sides is also scalene, but right angled is the more informative answer.

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

Explanation:
1. Let \( A(2,-2) \), \( B(8,4) \), \( C(5,7) \), \( D(x,y) \)
2. Mid-point of AC \( =\) mid-point of BD: \( \frac{x+8}{2}=\frac{7}{2} \), \( \frac{y+4}{2}=\frac52 \)
3. \( D=(-1,1) \)

Teacher's Note:
1. The diagonals of a rectangle bisect each other.
2. Check that AB is perpendicular to BC to confirm the right angle.

Answer: (d) rhombus

Explanation:
1. \( AB=BC=CD=DA=5 \)
2. \( AC^2=98 \) and \( BD^2=2 \), so the diagonals are unequal
3. Equal sides with unequal diagonals give a rhombus

Teacher's Note:
1. Four equal sides mean a rhombus.
2. It is a square only if the diagonals are also equal.

Answer: (c) square

Explanation:
1. All four sides have length \( \sqrt8 \)
2. Diagonals \( (1,-2)\)-\( (1,2) \) and \( (3,0)\)-\( (-1,0) \) both have length 4, so they are equal
3. Equal sides and equal diagonals give a square

Teacher's Note:
1. A square is a parallelogram, rectangle and rhombus all at once.
2. Choose the most specific name.

Answer: (d) none of these

Explanation:
1. Equal determinants mean equal areas
2. Triangles with the same area need not be similar or congruent, nor are they forbidden from being congruent
3. So none of the given statements must hold

Teacher's Note:
1. The determinant only measures area.
2. Equal area does not imply similar shape.

Answer: (c) rectangle

Explanation:
1. The sides are parallel to the axes with lengths 3 and 5
2. Adjacent sides are perpendicular and unequal
3. So it is a rectangle (most specific)

Teacher's Note:
1. Sides parallel to the axes meet at right angles.
2. Unequal adjacent sides rule out a square.

Answer: (b) parallelogram

Explanation:
1. By the mid-point theorem, PQ and SR are both parallel to AC and half its length
2. Similarly QR and PS are parallel to BD
3. So PQRS is always a parallelogram

Teacher's Note:
1. This is Varignon's theorem.
2. It becomes a rectangle only if the diagonals of ABCD are perpendicular.

Answer: (c) \( a=2,\ b=3 \)

Explanation:
1. Diagonals PR and QS bisect each other
2. \( \frac{1+5}{2}=\frac{4+a}{2}\Rightarrow a=2 \)
3. \( \frac{2+7}{2}=\frac{6+b}{2}\Rightarrow b=3 \)

Teacher's Note:
1. In PQRS the diagonals are PR and QS.
2. Equate the mid-points of the diagonals.

Answer: (b) obtuse angled

Explanation:
1. \( (3x+4y)^2+(4x+3y)^2=25x^2+48xy+25y^2 \)
2. \( (5x+5y)^2=25x^2+50xy+25y^2 \), which is larger
3. The square of the longest side exceeds the sum of squares of the other two, so the triangle is obtuse

Teacher's Note:
1. Compare the square of the longest side with the sum of the other two squares.
2. Larger means obtuse, equal means right, smaller means acute.

Answer: (c) \( D\left(\frac92,\frac12\right),\ B\left(-\frac12,\frac52\right) \)

Explanation:
1. The centre is the mid-point \( \left(2,\frac32\right) \); the half-diagonal vector is \( \left(1,\frac52\right) \)
2. Rotating it by \( 90^\circ \) gives \( \left(-\frac52,1\right) \)
3. The other vertices are \( \left(2,\frac32\right)\pm\left(\frac52,-1\right) \), that is \( \left(\frac92,\frac12\right) \) and \( \left(-\frac12,\frac52\right) \)

Teacher's Note:
1. The diagonals of a square are equal and perpendicular bisectors of each other.
2. Rotate the half-diagonal by 90 degrees to find the other diagonal.

Answer: (c) rhombus

Explanation:
1. The four lines form two pairs of parallel lines, \( ax+by=\pm c \) and \( ax-by=\pm c \)
2. Each pair is a distance \( \frac{2c}{\sqrt{a^2+b^2}} \) apart
3. The common distance between the pairs of parallel sides means all four sides are equal, so the figure is a rhombus (in general not a square)

Teacher's Note:
1. A parallelogram whose two heights are equal has equal sides.
2. So the figure is a rhombus.

Answer: (c) 3 : 1

Explanation:
1. Let the ratio be \( k:1 \); the x co-ordinate of the point is 0
2. \( \frac{k(1)+(-3)}{k+1}=0\Rightarrow k=3 \)
3. The ratio is \( 3:1 \)

Teacher's Note:
1. On the y-axis the x co-ordinate is zero.
2. The ratio is measured from the first point.

Answer: (a) \( (3,4),\ (6,8) \)

Explanation:
1. The points divide the segment in ratios \( 1:2 \) and \( 2:1 \)
2. \( \left(\frac{9}{3},\frac{12}{3}\right)=(3,4) \)
3. \( \left(\frac{18}{3},\frac{24}{3}\right)=(6,8) \)

Teacher's Note:
1. Trisection points are one-third and two-thirds of the way along.
2. Multiply the difference of co-ordinates by 1/3 and 2/3.