JEE Mathematics Coordinates and Straight Lines MCQs Set 10

Answer: (d) \( bx-ay=0 \)

Explanation:
1. Let \( P(x,y) \), \( A(a+b,b-a) \), \( B(a-b,a+b) \) and put \( PA^2=PB^2 \)
2. \( (a+b-x)^2+(b-a-y)^2=(a-b-x)^2+(a+b-y)^2 \)
3. Expanding, the squared terms cancel and \( -4bx=-4ay \)
4. So \( bx-ay=0 \)

Teacher's Note:
1. Equidistant from two points means the point lies on their perpendicular bisector.
2. Equate the squared distances so that no square roots appear.

Answer: (b) \( (-1,0) \)

Explanation:
1. Diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD
2. Mid-point of \( A(3,-4) \), \( C(-6,5) \) is \( \left(-\frac32,\frac12\right) \)
3. With \( B(-2,1) \) and \( D(x,y) \): \( \frac{x-2}{2}=-\frac32 \), \( \frac{y+1}{2}=\frac12 \)
4. So \( x=-1 \), \( y=0 \)

Teacher's Note:
1. Equal mid-points of the diagonals is the quickest test for a parallelogram.
2. The fourth vertex is the sum of the opposite pair minus the third vertex.

Answer: (a) isosceles and right angled

Explanation:
1. Let \( A(4,0) \), \( B(-1,-1) \), \( C(3,5) \)
2. \( AB^2=26 \), \( AC^2=26 \), \( BC^2=52 \)
3. \( AB=AC \) so it is isosceles
4. \( AB^2+AC^2=BC^2 \), so the right angle is at A

Teacher's Note:
1. Compute the three squared sides and compare.
2. If the sum of two equals the third the triangle is right angled.

Answer: (c) \( \left(\frac{19}{3},\frac83\right) \)

Explanation:
1. \( x=\frac{7(8)+5(4)}{12}=\frac{76}{12}=\frac{19}{3} \)
2. \( y=\frac{7(6)+5(-2)}{12}=\frac{32}{12}=\frac83 \)

Teacher's Note:
1. In the ratio \( m:n \) the first point's co-ordinates are multiplied by \( n \) and the second's by \( m \).
2. Dividing by \( m+n \) at the end is the step most often forgotten.

Answer: (a) 2 : 5

Explanation:
1. Let the ratio be \( k:1 \); the dividing point has x co-ordinate 0
2. \( \frac{k(-5)+1(2)}{k+1}=0 \Rightarrow k=\frac25 \)
3. Ratio is \( 2:5 \)

Teacher's Note:
1. A point on the y-axis has x = 0, which gives one equation in the ratio.
2. Use \( k:1 \) to keep the algebra short.

Answer: (b) \( (4,7) \)

Explanation:
1. Let the third vertex be \( (x,y) \)
2. \( \frac{x+4-2}{3}=2 \Rightarrow x=4 \)
3. \( \frac{y+8+6}{3}=7 \Rightarrow y=7 \)

Teacher's Note:
1. The centroid is the average of the three vertices.
2. Multiply by 3 to find the missing sum.

Answer: (d) \( \left(1,\frac1{\sqrt3}\right) \)

Explanation:
1. All three sides equal 2, so the triangle is equilateral
2. In an equilateral triangle the incentre coincides with the centroid
3. \( \left(\frac{1+0+2}{3},\frac{\sqrt3+0+0}{3}\right)=\left(1,\frac1{\sqrt3}\right) \)

Teacher's Note:
1. Spot the special triangle first to avoid the long incentre formula.
2. In an equilateral triangle all four centres coincide.

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

Explanation:
1. Slope of the line from \( (0,3) \) to \( (3,0) \) is \( -1 \) and to \( (-3,0) \) is \( 1 \); the product is \( -1 \)
2. So the triangle is right angled at \( (0,3) \)
3. The orthocentre of a right angled triangle is the vertex of the right angle, \( (0,3) \)

Teacher's Note:
1. Check for a right angle with the product of slopes before using altitudes.
2. The orthocentre of a right triangle sits at the right-angled vertex.

Answer: (d) 0

Explanation:
1. Area \( =\frac12\begin{vmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{vmatrix} \)
2. Apply \( C_2\to C_2+C_1 \): the second column becomes \( (a+b+c) \) in every row
3. Two proportional columns make the determinant 0, so the points are collinear

Teacher's Note:
1. A column operation that makes two columns proportional shows the determinant vanishes.
2. Zero area means the three points lie on one line.

Answer: (a) \( \left|\frac{x+y-2}{7}\right| \)

Explanation:
1. Twice the area of ABC \( =
2. 6(5+2)-3(-2-3)+4(3-5)
3. =
4. 42+15-8
5. =49 \)
6. Twice the area of PBC \( =
7. x(5+2)-3(-2-y)+4(y-5)
8. =
9. 7x+7y-14
10. \)
11. Ratio \( =\frac{7
12. x+y-2
13. }{49}=\left
14. \frac{x+y-2}{7}\right
15. \)

Teacher's Note:
1. Both areas use the same determinant formula, so the common factor 1/2 cancels.
2. Expand each determinant along its first row.

Answer: (b) \( 0 \) and \( -4 \) only

Explanation:
1. Set the determinant of the three points equal to zero
2. Using \( R_1\to R_1-R_2 \) and then \( C_2\to C_1+C_2 \) it reduces to \( x^2(x+4)=0 \)
3. So \( x=0 \) or \( x=-4 \)

Teacher's Note:
1. Simplify the determinant with row and column operations before expanding.
2. A repeated root x = 0 here makes all three points coincide.

Answer: (b) \( \theta\ne\frac{n\pi}{2} \)

Explanation:
1. Collinearity needs \( \begin{vmatrix}1&1&1\\0&\sec^2\theta&1\\ \mathrm{cosec}^2\theta&0&1\end{vmatrix}=0 \)
2. Expanding: \( \sec^2\theta+\mathrm{cosec}^2\theta-\sec^2\theta\,\mathrm{cosec}^2\theta=0 \), which is the identity \( \frac{1}{\cos^2\theta}+\frac1{\sin^2\theta}=\frac{1}{\sin^2\theta\cos^2\theta} \)
3. It holds whenever the points exist, that is when \( \theta\ne\frac{n\pi}{2} \)

Teacher's Note:
1. The condition reduces to a trigonometric identity, so it is always true.
2. The only exclusions are angles where sec or cosec is undefined.

Answer: (b) \( \left(\frac{-1}{\sqrt2},\frac7{\sqrt2}\right) \)

Explanation:
1. Reflection in \( y=x \): \( (2,3)\to(3,2) \)
2. Translation by 2 along y: \( (3,4) \)
3. Rotation through \( 45^\circ \): \( (3\cos45^\circ-4\sin45^\circ,\ 3\sin45^\circ+4\cos45^\circ) \)
4. \( =\left(\frac{-1}{\sqrt2},\frac{7}{\sqrt2}\right) \)

Teacher's Note:
1. Apply the transformations strictly in the order given.
2. The rotation of a point uses \( x'=x\cos\theta-y\sin\theta \), \( y'=x\sin\theta+y\cos\theta \).

Answer: (a) \( \left(\frac{-1}{\sqrt2},\frac{-3}{\sqrt2}\right) \)

Explanation:
1. Reflection in the y-axis: \( (2,-1)\to(-2,-1) \)
2. Axes rotated through \( \theta=-45^\circ \): \( x'=x\cos\theta+y\sin\theta=-2\cdot\frac1{\sqrt2}+(-1)\left(-\frac1{\sqrt2}\right)=\frac{-1}{\sqrt2} \)
3. \( y'=-x\sin\theta+y\cos\theta=2\left(-\frac1{\sqrt2}\right)-\frac1{\sqrt2}=\frac{-3}{\sqrt2} \)

Teacher's Note:
1. Rotating the axes through theta changes the co-ordinates as if the point rotated through minus theta.
2. Substitute theta = -45 degrees carefully.

Answer: (d) \( 2x+3y=9 \)

Explanation:
1. Let \( C(\alpha,\beta) \); centroid \( =\left(\frac{\alpha}{3},\frac{\beta-2}{3}\right) \)
2. It lies on \( 2x+3y=1 \): \( \frac{2\alpha}{3}+(\beta-2)=1 \)
3. \( 2\alpha+3\beta=9 \), so the locus of C is \( 2x+3y=9 \)

Teacher's Note:
1. Write the moving point's co-ordinates, substitute in the given condition, then rename them x and y.
2. The locus of C is parallel to the locus of the centroid.

Answer: (c) \( 9x^2+36y^2=4l^2 \)

Explanation:
1. Let the ends be \( A(a,0) \) and \( B(0,b) \) and the point \( P(h,k) \) with \( AP:PB=1:2 \)
2. \( h=\frac{2a}{3} \), \( k=\frac{b}{3} \), so \( a=\frac{3h}{2} \), \( b=3k \)
3. \( a^2+b^2=l^2 \) gives \( \frac{9h^2}{4}+9k^2=l^2 \)
4. Locus: \( 9x^2+36y^2=4l^2 \)

Teacher's Note:
1. Express the two ends in terms of the moving point and use the fixed length of the rod.
2. The locus is an ellipse.

Answer: (a) 13

Explanation:
1. \( d^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2) \)
2. \( =289+50-2(17)(5\sqrt2)\cos45^\circ=339-170=169 \)
3. \( d=13 \)

Teacher's Note:
1. This is the cosine rule applied in the triangle formed with the pole.
2. The angle between the radii here is 45 degrees.

Answer: (c) \( (-r,\pi+\theta) \)

Explanation:
1. Changing the sign of r reverses the direction of the point
2. Reversing direction is equivalent to adding \( \pi \) to the angle
3. So \( (r,\theta)\equiv(-r,\pi+\theta) \)

Teacher's Note:
1. A negative r means moving in the opposite direction along the ray.
2. Adding pi to the angle gives the same reversal.

Answer: (d) \( 3\pm2\sqrt{15} \)

Explanation:
1. \( (a-3)^2+(2-4)^2=64 \)
2. \( (a-3)^2=60 \)
3. \( a=3\pm\sqrt{60}=3\pm2\sqrt{15} \)

Teacher's Note:
1. Square the distance formula to avoid the root.
2. Both signs give valid points, so keep the plus-minus.

Answer: (a) \( a(m_1-m_2)\sqrt{(m_1+m_2)^2+4} \)

Explanation:
1. \( d^2=a^2(m_1^2-m_2^2)^2+4a^2(m_1-m_2)^2 \)
2. \( =a^2(m_1-m_2)^2\left[(m_1+m_2)^2+4\right] \)
3. \( d=a(m_1-m_2)\sqrt{(m_1+m_2)^2+4} \) (taking \( m_1>m_2,\ a>0 \))

Teacher's Note:
1. Factor out the common (m1 - m2) squared.
2. The points lie on the parabola y squared = 4ax.

Answer: (d) \( 2a\sin\frac{\alpha-\beta}{2} \)

Explanation:
1. \( d^2=a^2\left[2-2\cos(\alpha-\beta)\right] \)
2. \( =4a^2\sin^2\frac{\alpha-\beta}{2} \)
3. \( d=2a\left
4. \sin\frac{\alpha-\beta}{2}\right
5. \), which is \( 2a\sin\frac{\alpha-\beta}{2} \) for \( \alpha>\beta \)

Teacher's Note:
1. Use \( 1-\cos x=2\sin^2\frac x2 \).
2. This is the length of a chord of a circle subtending the angle alpha - beta.

Answer: (b) \( \left(0,-\frac32\right) \)

Explanation:
1. Let the point be \( (0,y) \)
2. \( 9+(y-2)^2=1+(y-3)^2 \)
3. \( 13-4y=10-6y \Rightarrow y=-\frac32 \)

Teacher's Note:
1. A point on the y-axis has x = 0.
2. Equate squared distances.

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

Explanation:
1. Let the point be \( (x,x) \)
2. \( (x-1)^2+x^2=x^2+(x-3)^2 \)
3. \( -2x+1=-6x+9\Rightarrow x=2 \)

Teacher's Note:
1. Put both co-ordinates equal to x to get one equation.
2. Equate squared distances.

Answer: (b) 20

Explanation:
1. The segment joining the mid-points of two sides is parallel to the third side and half its length
2. Distance between \( (3,5) \) and \( (-3,-3) \) is \( \sqrt{36+64}=10 \)
3. \( BC=2\times10=20 \)

Teacher's Note:
1. Use the mid-point theorem.
2. The third side is twice the segment joining the two mid-points.