JEE Mathematics Coordinates and Straight Lines MCQs Set 13

Answer: (d) the orthocentre of the triangle

Explanation:
1. The vertices are \( (0,0) \), \( \left(1,\frac13\right) \) and \( (-1,3) \)
2. The sides \( x-3y=0 \) and \( 3x+y=0 \) have slopes \( \frac13 \) and \( -3 \), which are perpendicular, so the orthocentre is the origin
3. The line \( 3x-4y=0 \) passes through the origin
4. The centroid \( \left(0,\frac{10}9\right) \) and circumcentre \( \left(0,\frac53\right) \) do not lie on it

Teacher's Note:
1. Find the centre that is simplest to locate.
2. Then test each centre in the given line.

Answer: (b) \( \left(-a,\ a(t_1+t_2+t_3+t_1t_2t_3)\right) \)

Explanation:
1. The side joining the first two vertices has direction proportional to \( (t_2,1) \), and the side joining the last two has direction proportional to \( (t_3,1) \)
2. Altitude from the third vertex: \( t_2(x-at_3t_1)+y-a(t_3+t_1)=0 \)
3. Altitude from the first vertex: \( t_3(x-at_1t_2)+y-a(t_1+t_2)=0 \)
4. Subtracting gives \( (t_2-t_3)(x+a)=0 \), so \( x=-a \)
5. Substituting back, \( y=a(t_1+t_2+t_3+t_1t_2t_3) \)

Teacher's Note:
1. Find the direction of each side first, then write two altitudes.
2. Subtracting the altitudes eliminates y at once.

Answer: (a) 14

Explanation:
1. Area \( =\frac12
2. x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)
3. \)
4. \( =\frac12
5. -4(2+3)+1(-3-1)+4(1-2)
6. \)
7. \( =\frac12
8. -20-4-4
9. =14 \)

Teacher's Note:
1. Use the determinant formula.
2. Take the modulus at the end.

Answer: (d) 7

Explanation:
1. B and C have the same x co-ordinate, so \( BC=14 \) is a vertical base
2. The height is the horizontal distance from A to \( x=3 \), which is 1
3. Area \( =\frac12\times14\times1=7 \)

Teacher's Note:
1. A vertical or horizontal side makes the area easy.
2. Area is half base times perpendicular distance.

Answer: (a) 2

Explanation:
1. The points \( (-1,1) \) and \( (-1,-1) \) lie on \( x=-1 \), giving a base of 2
2. The height is the distance from \( (1,-1) \) to \( x=-1 \), which is 2
3. Area \( =\frac12\times2\times2=2 \)

Teacher's Note:
1. Sketch the points to see the right angle.
2. Legs 2 and 2 give area 2.

Answer: (a) 1

Explanation:
1. The determinant equals \( 4(\lambda+1)(2\lambda-1) \), so the area is \( 2
2. (\lambda+1)(2\lambda-1)
3. \)
4. Setting it to 70 gives \( 2\lambda^2+\lambda-1=\pm35 \)
5. \( 2\lambda^2+\lambda-36=0 \) gives \( \lambda=4 \) or \( -\frac92 \), while \( 2\lambda^2+\lambda+34=0 \) has no real roots
6. Only \( \lambda=4 \) is an integer

Teacher's Note:
1. Expand the determinant and factorise.
2. Test each quadratic for integer roots.

Answer: (a) \( \frac{11}{8} \)

Explanation:
1. Twice the area of ABC \( =
2. 6(5+2)+(-3)(-2-3)+4(3-5)
3. =49 \)
4. Twice the area of DBC \( =
5. x(5+2)+(-3)(-2-3x)+4(3x-5)
6. =
7. 28x-14
8. \)
9. \( \frac{
10. 28x-14
11. }{49}=\frac12\Rightarrow 28x-14=\pm\frac{49}2 \)
12. \( x=\frac{11}8 \) (the other root \( -\frac38 \) is not among the options)

Teacher's Note:
1. Use the determinant formula for both areas.
2. Keep the plus-minus from the modulus.

Answer: (c) 7, \( \frac{31}{9} \)

Explanation:
1. \( A=\left(\frac{3k-5}{k+1},\frac{5k+1}{k+1}\right) \)
2. Expanding the determinant for A, B, C gives twice the area \( =\frac{2
3. 7k-33
4. }{
5. k+1
6. } \)
7. Setting area 2: \(
8. 7k-33
9. =2(k+1) \)
10. \( 7k-33=2k+2\Rightarrow k=7 \) and \( 33-7k=2k+2\Rightarrow k=\frac{31}9 \)

Teacher's Note:
1. Write A with the section formula and use the area determinant.
2. Keep both signs of the modulus.

Answer: (c) 14 sq. units

Explanation:
1. The vertices are \( (0,5) \), \( (2,-2) \) and \( (-2,-2) \)
2. Base on \( y=-2 \) has length 4 and height \( 5-(-2)=7 \)
3. Area \( =\frac12\times4\times7=14 \)

Teacher's Note:
1. Find the three intersection points first.
2. A horizontal side gives a simple base and height.

Answer: (c) \( \frac12ab(r-1)(s-1)(s-r) \)

Explanation:
1. \( x_1=ar \), \( x_2=ar^2 \), \( y_1=bs \), \( y_2=bs^2 \)
2. Area \( =\frac12\begin{vmatrix}a&b&1\\ar&bs&1\\ar^2&bs^2&1\end{vmatrix} \)
3. Expanding and factorising gives \( \frac12ab(r-1)(s-1)(s-r) \) in magnitude

Teacher's Note:
1. Substitute the G.P. terms and factorise the determinant.
2. The factors vanish when r or s equals 1 or when r equals s, where the points are collinear.

Answer: (b) 4

Explanation:
1. PQRS is a parallelogram because the lines through P and R are parallel to QR and PQ
2. Area of triangle \( PQR=\frac12
3. 2(-1-2)+4(2-1)+3(1+1)
4. =\frac12
5. -6+4+6
6. =2 \)
7. The parallelogram is made of two such triangles, so its area is \( 4 \)

Teacher's Note:
1. A diagonal splits a parallelogram into two equal triangles.
2. So the area is twice the triangle's area.

Answer: (b) \( \frac{3a^4}4 \)

Explanation:
1. The determinant equals twice the area, \( 2\cdot\frac{\sqrt3}4a^2=\frac{\sqrt3}2a^2 \)
2. Its square is \( \frac{3}{4}a^4 \)

Teacher's Note:
1. The determinant is twice the signed area.
2. Square it to remove the sign.

Answer: (a) 23

Explanation:
1. \( P=\left(\frac{3\lambda-5}{\lambda+1},\frac{5\lambda+1}{\lambda+1}\right) \)
2. Twice the area of PQR is \( \frac{6
3. \lambda-7
4. }{\lambda+1} \)
5. Setting the area to 2: \( 3
6. \lambda-7
7. =2(\lambda+1) \)
8. \( 3\lambda-21=2\lambda+2\Rightarrow\lambda=23 \) (the other root \( \frac{19}5 \) is not listed)

Teacher's Note:
1. Write P with the section formula and expand the area determinant.
2. Test each option or solve with both signs of the modulus.

Answer: (c) 2

Explanation:
1. Collinearity requires the area determinant to vanish
2. Slope between the first two \( =\frac{2}{p} \) and between the last two \( =\frac{2p-3}{1} \)
3. \( 2=p(2p-3)\Rightarrow2p^2-3p-2=0 \)
4. \( p=2 \) or \( p=-\frac12 \); only \( p=2 \) is among the options

Teacher's Note:
1. Equate the slopes of two pairs of points.
2. Solve the resulting quadratic.

Answer: (a) \( ab'=a'b \)

Explanation:
1. Expand \( \begin{vmatrix}a&b&1\\a'&b'&1\\a-a'&b-b'&1\end{vmatrix} \)
2. Apply \( R_3\to R_3-R_1+R_2 \): the third row becomes \( (0,0,1) \), and the determinant is \( ab'-a'b \)
3. Collinear points need this to be zero, so \( ab'=a'b \)

Teacher's Note:
1. Row operations simplify the determinant quickly.
2. The condition says the two position vectors are parallel.

Answer: (a) \( \frac12,\ -1 \)

Explanation:
1. Set the area determinant to zero
2. It factorises as \( 4(k+1)(2k-1)=0 \)
3. So \( k=-1 \) or \( k=\frac12 \)

Teacher's Note:
1. Expand the determinant and factorise.
2. Each factor gives one value of k.

Answer: (b) \( \frac52 \)

Explanation:
1. Slope between the first two \( =\frac{3}{4} \)
2. Slope between \( (2,-2) \) and \( (8,a) \) \( =\frac{a+2}6 \)
3. \( \frac{a+2}6=\frac34\Rightarrow a=\frac52 \)

Teacher's Note:
1. Equate the two slopes.
2. Remember a + 2, not a - 2, from the second point.

Answer: (c) 7

Explanation:
1. Slope between \( (5,5) \) and \( (-5,1) \) \( =\frac{4}{10}=\frac25 \)
2. Slope between \( (5,5) \) and \( (10,K) \) \( =\frac{K-5}{5} \)
3. \( \frac{K-5}5=\frac25\Rightarrow K=7 \)

Teacher's Note:
1. Equal slopes give a linear equation in K.
2. Check by using the third pair as well.

Answer: (d) collinear

Explanation:
1. The line through \( (3a,0) \) and \( (0,3b) \) is \( \frac x{3a}+\frac y{3b}=1 \)
2. At \( (a,2b) \): \( \frac13+\frac23=1 \), so the point lies on it
3. The points are collinear

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

Answer: (d) collinear

Explanation:
1. The third point is the section formula applied to the first two points
2. It divides the segment joining them in the ratio \( k:l \), so it lies on that segment
3. The three points are collinear

Teacher's Note:
1. Recognise the section formula.
2. A dividing point always lies on the line through the end points.

Answer: (a) \( \left(5,\frac{11}2\right) \)

Explanation:
1. \( AC=\sqrt{64+36}=10 \)
2. \( AB=2.5=\frac14AC \), so B (on segment AC) divides AC in the ratio \( 1:3 \)
3. \( B=\left(3+\frac{8}4,4+\frac64\right)=\left(5,\frac{11}2\right) \)

Teacher's Note:
1. Find AC first, then the fraction AB over AC.
2. Move that fraction of the way from A towards C.

Answer: (d) 2 is the H.M. of x, y

Explanation:
1. The area determinant gives \( x(y-3)-2x(2y-3)+(6y-3y)=0 \)
2. That simplifies to \( -3xy+3x+3y=0 \), so \( xy=x+y \)
3. Then \( \frac{2xy}{x+y}=2 \), so 2 is the harmonic mean of x and y

Teacher's Note:
1. Expand the determinant carefully.
2. The relation xy = x + y is the harmonic mean condition.

Answer: (a) collinear

Explanation:
1. The first three points lie on the line \( y=\frac bax \)
2. The point \( (a^2,ab) \) satisfies \( ab=\frac ba\cdot a^2 \)
3. All four points are collinear

Teacher's Note:
1. Find the line through two points and test the others.
2. All four satisfy the same equation.

Answer: (c) (3, 7)

Explanation:
1. \( x'=x-h=4-1=3 \)
2. \( y'=y-k=5-(-2)=7 \)

Teacher's Note:
1. New co-ordinates = old co-ordinates minus the new origin.
2. Be careful with the minus sign in y.