JEE Mathematics Coordinates and Straight Lines MCQs Set 12

Answer: (c) \( (7,-2) \)

Explanation:
1. C lies on AB produced beyond B with \( AC=2BC \), so B is the mid-point of AC
2. \( C=2B-A=(4+3,\ 2-4)=(7,-2) \)

Teacher's Note:
1. Since C is outside the segment and AC is twice BC, B is the mid-point of AC.
2. Use the mid-point formula backwards.

Answer: (d) 4 : 9

Explanation:
1. Ratio \( =-\frac{3(1)+4(2)-7}{3(-2)+4(1)-7}=-\frac{4}{-9}=\frac49 \)
2. Positive, so the division is internal in the ratio \( 4:9 \)

Teacher's Note:
1. Substitute each end point in the line equation and divide.
2. A positive ratio means the line cuts the segment internally.

Answer: (a) \( \left(\frac83,\frac{10}3\right) \)

Explanation:
1. M lies on AB with \( AM=\frac13AB \), so M divides AB in the ratio \( 1:2 \)
2. \( x=\frac{1(4)+2(2)}{3}=\frac83 \)
3. \( y=\frac{1(2)+2(4)}{3}=\frac{10}3 \)

Teacher's Note:
1. AM = AB/3 means M is a trisection point nearer A.
2. Use the section formula with ratio 1 : 2.

Answer: (a) \( (7,9),\ (-3,-7),\ (1,1) \)

Explanation:
1. Each vertex equals the sum of two mid-points minus the third
2. \( (2,1)+(4,5)-(-1,-3)=(7,9) \)
3. \( (2,1)+(-1,-3)-(4,5)=(-3,-7) \)
4. \( (-1,-3)+(4,5)-(2,1)=(1,1) \)

Teacher's Note:
1. The medial triangle is similar to the original with ratio 1 : 2.
2. A vertex is the sum of the neighbouring mid-points minus the opposite one.

Answer: (c) cannot lie on the y-axis

Explanation:
1. Centroid x co-ordinate \( =\frac{1+2+c^2}{3}=\frac{3+c^2}{3}\ge1>0 \)
2. A point on the y-axis needs \( x=0 \), which is impossible
3. The y co-ordinate \( \frac{a+b+3}{3} \) can be zero for suitable a, b, so the x-axis is possible

Teacher's Note:
1. Test whether each condition can ever be met.
2. The x co-ordinate is always positive here.

Answer: (c) \( (3,1) \)

Explanation:
1. \( x=\frac{4+3+2}{3}=3 \)
2. \( y=\frac{-3-2+8}{3}=1 \)

Teacher's Note:
1. Average the x co-ordinates and the y co-ordinates separately.
2. The centroid is always inside the triangle.

Answer: (b) \( \left(\frac{10}3,\frac73\right) \)

Explanation:
1. \( x=\frac{2+5+3}{3}=\frac{10}3 \)
2. \( y=\frac{1+2+4}{3}=\frac73 \)

Teacher's Note:
1. Add the three x values and divide by 3.
2. Do the same for y.

Answer: (c) \( \left(2,\frac53\right) \)

Explanation:
1. The mid-point triangle has the same centroid as the original triangle
2. \( x=\frac{-2+4+4}{3}=2 \)
3. \( y=\frac{3-3+5}{3}=\frac53 \)

Teacher's Note:
1. The centroid of the medial triangle coincides with the centroid of the triangle.
2. So average the three mid-points directly.

Answer: (c) at least one co-ordinate irrational

Explanation:
1. Let the base vector be \( (p,q) \) with integers p, q not both zero
2. The third vertex is the mid-point plus \( \frac{\sqrt3}{2}(-q,p) \)
3. At least one of \( q\sqrt3 \) and \( p\sqrt3 \) is irrational, so at least one co-ordinate is irrational

Teacher's Note:
1. Root 3 enters through the height of the triangle.
2. Both co-ordinates are irrational only when p and q are both non-zero, so the safe statement is at least one.

Answer: (a) \( (6,2) \)

Explanation:
1. The centroid divides the line from orthocentre H to circumcentre O in the ratio \( 2:1 \), so \( G=\frac{H+2O}{3} \)
2. \( 3=\frac{-3+2x}{3}\Rightarrow x=6 \)
3. \( 3=\frac{5+2y}{3}\Rightarrow y=2 \)

Teacher's Note:
1. H, G, O are collinear with HG : GO = 2 : 1.
2. Use the section formula to find O.

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

Explanation:
1. The vertices are \( (0,0) \), \( (4,0) \) and \( (0,3) \), a right triangle with sides 3, 4, 5
2. Inradius \( r=\frac{3+4-5}{2}=1 \)
3. The incentre is at distance r from both axes: \( (1,1) \)

Teacher's Note:
1. For a right triangle r = (sum of legs - hypotenuse)/2.
2. The incentre is r away from both legs.

Answer: (c) \( \left(3,\frac34\right) \)

Explanation:
1. The altitude from \( (3,4) \) is perpendicular to the x-axis side, so it is \( x=3 \)
2. The side from \( (3,4) \) to \( (4,0) \) has slope \( -4 \), so the altitude from the origin has slope \( \frac14 \): \( y=\frac x4 \)
3. At \( x=3 \), \( y=\frac34 \)

Teacher's Note:
1. Two altitudes are enough to locate the orthocentre.
2. A side lying along an axis gives a vertical or horizontal altitude.

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

Explanation:
1. The lines \( 4x-7y+10=0 \) and \( 7x+4y=15 \) have slopes \( \frac47 \) and \( -\frac74 \), whose product is \( -1 \)
2. They are perpendicular, so the triangle is right angled at their intersection
3. Solving the two lines: \( 4x-7y=-10 \), \( 7x+4y=15 \) gives \( (1,2) \), which is the orthocentre

Teacher's Note:
1. If two sides are perpendicular, the orthocentre is their common vertex.
2. Always check slopes before finding altitudes.

Answer: (d) \( (3,4) \)

Explanation:
1. The lines \( x=3 \) and \( y=4 \) are perpendicular and meet at \( (3,4) \)
2. So the triangle has a right angle there
3. The orthocentre is \( (3,4) \)

Teacher's Note:
1. Vertical and horizontal lines are perpendicular.
2. The orthocentre of a right triangle is the right-angle vertex.

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

Explanation:
1. \( A=(8,0) \), \( B=(0,6) \), so the sides of the right triangle are 6, 8, 10
2. \( r=\frac{6+8-10}{2}=2 \)
3. The incentre is at distance 2 from both axes: \( (2,2) \)

Teacher's Note:
1. The triangle is right angled at the origin.
2. The incentre is (r, r) here.

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

Explanation:
1. The triangle is right angled at the origin, so the orthocentre is \( (0,0) \)
2. The circumcentre is the mid-point of the hypotenuse, \( \left(\frac b2,\frac a2\right) \)
3. Distance \( =\sqrt{\frac{b^2}{4}+\frac{a^2}{4}}=\frac{\sqrt{a^2+b^2}}{2} \)

Teacher's Note:
1. In a right triangle the circumcentre is the mid-point of the hypotenuse.
2. The distance equals half the hypotenuse.

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

Explanation:
1. \( 3(4)=6+2+x\Rightarrow x=4 \)
2. \( 3(6)=4+6+y\Rightarrow y=8 \)

Teacher's Note:
1. The sum of the three x values is three times the centroid's x.
2. Same for y.

Answer: (c) \( c=-4 \)

Explanation:
1. y co-ordinate of the centroid \( =\frac{1+3+c}{3} \)
2. It is zero when \( c=-4 \)

Teacher's Note:
1. The centroid lies on the x-axis when its y co-ordinate is zero.
2. Only the y co-ordinates of the vertices matter.

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

Explanation:
1. The vertices are \( (2,3) \), \( (0,1) \) and \( (4,1) \)
2. Slopes of the first two sides are \( -1 \) and \( 1 \), so the angle at \( (2,3) \) is \( 90^\circ \)
3. The circumcentre is the mid-point of the hypotenuse from \( (0,1) \) to \( (4,1) \), which is \( (2,1) \)

Teacher's Note:
1. A right angle makes the hypotenuse a diameter of the circumcircle.
2. The circumcentre is its mid-point.

Answer: (a) \( (0,0) \)

Explanation:
1. The vertices are \( (10,0) \), \( (0,0) \) and \( (0,24) \)
2. The sides along the axes meet at the origin at a right angle
3. The orthocentre is \( (0,0) \)

Teacher's Note:
1. Recover the vertices from the mid-points first.
2. The triangle turns out to be right angled.

Answer: (c) centroid, circumcentre and orthocentre

Explanation:
1. The centroid is an average of the vertices, so it is rational
2. The circumcentre and the orthocentre are obtained by solving linear equations with rational coefficients, so they are rational
3. The incentre uses side lengths, which are square roots and in general irrational

Teacher's Note:
1. Centres defined by linear equations in the vertex co-ordinates stay rational.
2. The incentre needs distances, which bring in square roots.

Answer: (b) \( \left(\frac53,\frac1{\sqrt3}\right) \)

Explanation:
1. The mid-point of BC is \( (2,0) \), so \( C=(4,0) \)
2. \( A=(2\cos60^\circ,2\sin60^\circ)=(1,\sqrt3) \)
3. Centroid \( =\left(\frac{0+4+1}3,\frac{0+0+\sqrt3}3\right)=\left(\frac53,\frac1{\sqrt3}\right) \)

Teacher's Note:
1. Place C on the x-axis from the mid-point information.
2. Use polar form to place A.

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

Explanation:
1. Let the third vertex be \( C(a,b) \) and \( A(5,-1) \), \( B(-2,3) \), \( H=O \)
2. \( OA\perp BC \): \( 5(a+2)-(b-3)=0 \)
3. \( OB\perp AC \): \( -2(a-5)+3(b+1)=0 \)
4. Solving: \( a=-4 \), \( b=-7 \)

Teacher's Note:
1. Each vertex joined to the orthocentre is perpendicular to the opposite side.
2. Two such conditions give two equations.

Answer: (a) first

Explanation:
1. The vertices are \( A(-3,4) \), \( B\left(-\frac37,\frac{16}7\right) \) and \( C\left(-\frac35,\frac85\right) \)
2. Altitude from A is perpendicular to \( 4x-y+4=0 \) (slope 4): \( y-4=-\frac14(x+3) \)
3. Altitude from B is perpendicular to \( x+y=1 \) (slope \( -1 \)): \( y=x+\frac{19}7 \)
4. Solving: \( x=\frac37 \), \( y=\frac{22}7 \), both positive, so the first quadrant

Teacher's Note:
1. Find the vertices first, then two altitudes.
2. The signs of the co-ordinates fix the quadrant.