JEE Mathematics Straight Lines MCQs Set 25

Answer: (b) 2x + y = 4

Explanation:
1. E is the mid-point of AC and F is the mid-point of AB. Let the vertices be A, B, C. Then A + C = (6, 8), A + B = (2, 4) and A + B + C = 3G = (9, 6).
2. Subtracting: B = (9, 6) − (6, 8) = (3, −2); C = (9, 6) − (2, 4) = (7, 2); A = (9, 6) − (3, −2) − (7, 2) = (−1, 6).
3. Slope of AB = \( \frac{-2-6}{3+1}=-2 \), so \( y-6=-2(x+1) \), i.e. 2x + y = 4.

Teacher's Note:
1. The centroid equals the average of the three vertices, so the three mid-point conditions fix all vertices.
2. The triangle with A(−1, 6), B(3, −2), C(7, 2) is acute (all three dot products at the vertices are positive), as the question states.

Answer: (d) (5, 0)

Explanation:
1. From the data, B = (3, −2) and C = (7, 2).
2. D is the mid-point of BC: \( \left(\frac{3+7}{2},\frac{-2+2}{2}\right)=(5,0) \).
3. Check: G divides AD in the ratio 2 : 1, and \( \frac{2\cdot5+(-1)}{3}=3,\ \frac{2\cdot0+6}{3}=2 \) gives G(3, 2).

Teacher's Note:
1. D is the foot of the median from A on BC, so it is the mid-point of BC.
2. Always verify with the centroid ratio 2 : 1.

Answer: (d) 6\( \sqrt2 \)

Explanation:
1. A = (−1, 6), B = (3, −2), C = (7, 2).
2. Line BC: slope \( \frac{2+2}{7-3}=1 \), so \( y+2=x-3 \), i.e. x − y − 5 = 0.
3. Distance of A from BC: \( \frac{|-1-6-5|}{\sqrt2}=\frac{12}{\sqrt2}=6\sqrt2 \).

Teacher's Note:
1. The altitude from A is the perpendicular distance of A from the line BC.
2. Find the line BC first, then apply the point-to-line distance formula.

Answer: (d) Δ(A, B) + Δ(A, C) = 0

Explanation:
1. The median passes through A and the mid-point M of BC.
2. Δ(A, M) is linear in the co-ordinates of M, so Δ(A, M) = \( \frac12 \)[Δ(A, B) + Δ(A, C)].
3. Hence the median is Δ(A, B) + Δ(A, C) = 0.

Teacher's Note:
1. Because the determinant is linear in the last row, the mid-point turns into an average of the two determinants.
2. Δ(B, C) = 0 is the line BC itself and Δ(A, B) = 0 is the line AB.

Answer: (c) Δ(A, B) − Δ(A, C) = 0

Explanation:
1. A point P(x, y) lies on the line through A parallel to BC if the vector AP is parallel to BC: \( (x-x_1)(y_3-y_2)-(y-y_1)(x_3-x_2)=0 \).
2. Expanding Δ(A, B) − Δ(A, C) gives exactly this expression: the x, y terms are x(y1 − y2) − x(y1 − y3) + ... = x(y3 − y2) − y(x3 − x2) + constant, with constant fixed by A.
3. It vanishes at A, so the line is Δ(A, B) − Δ(A, C) = 0. (Checked with A(1, 2), B(5, 1), C(3, 7): both reduce to 3x + y − 5 = 0.)

Teacher's Note:
1. Subtracting the two determinants cancels the point A and leaves a direction condition parallel to BC.
2. Adding them instead would give the median.

Answer: (d) b Δ(A, B) + c Δ(A, C) = 0

Explanation:
1. The internal bisector from A meets BC at D with BD : DC = AB : AC = c : b.
2. So D = \( \frac{bB+cC}{b+c} \).
3. Since the determinant is linear in its last row, Δ(A, D) = \( \frac{b\,\Delta(A,B)+c\,\Delta(A,C)}{b+c} \).
4. The bisector is b Δ(A, B) + c Δ(A, C) = 0.

Teacher's Note:
1. The angle bisector theorem gives the ratio in which D divides BC: the weight of B is the length AC = b.
2. A numerical check with A(1, 2), B(5, 1), C(3, 7) confirms that only option (c) is a constant multiple of Δ(A, D).

Answer: (c) 5x² − 5y² − 50xy + 92x + 8y + 2 = 0

Explanation:
1. Factorise: \( 2x^2-3y^2+xy-x+11y-10=(x-y+2)(2x+3y-5) \), so the lines meet where x − y + 2 = 0 and 2x + 3y − 5 = 0, i.e. at \( \left(-\frac15,\frac95\right) \).
2. For the homogeneous part a = 2, b = −3, 2h = 1: \( \frac{X^2-Y^2}{5}=\frac{XY}{1/2} \), i.e. \( X^2-Y^2-10XY=0 \), where \( X=x+\frac15,\ Y=y-\frac95 \).
3. Expanding and multiplying by 5: 5x² − 5y² − 50xy + 92x + 8y + 2 = 0.

Teacher's Note:
1. Shift the origin to the point of intersection before applying the bisector formula.
2. Expanding the shifted equation carefully (including the linear terms) is where mistakes usually occur.

Answer: (c) (−9, 5)

Explanation:
1. X = x − h = −7 − 2 = −9.
2. Y = y − k = 9 − 4 = 5.
3. New co-ordinates: (−9, 5).

Teacher's Note:
1. Shifting the origin subtracts the new origin from every point.
2. Do not add: the point moves in the opposite direction to the origin.

Answer: (b) \( \left(3,\frac43\right) \)

Explanation:
1. A point inside a triangle that splits it into three triangles of equal area is the centroid.
2. Centroid of O(0, 0), P(3, 4), Q(6, 0): \( \left(\frac{0+3+6}{3},\frac{0+4+0}{3}\right)=\left(3,\frac43\right) \).

Teacher's Note:
1. The three medians divide a triangle into six equal-area pieces, and the centroid joins each pair into three equal parts.
2. Only the centroid has this property.

Answer: (b) P, Q, R are non-collinear

Explanation:
1. Compute the determinant \( \begin{vmatrix}-\sin(\beta-\alpha)&-\cos\beta&1\\\cos(\beta-\alpha)&\sin\beta&1\\\cos(\beta-\alpha+\theta)&\sin(\beta-\theta)&1\end{vmatrix} \).
2. Evaluating it for several admissible values (for example \( \alpha=0.3,\beta=0.5,\theta=0.2 \)) gives a non-zero value (about −0.137), and the sign does not vanish in the allowed range.
3. A non-zero determinant means the points are not collinear.

Teacher's Note:
1. Collinearity of three points is tested by the area determinant.
2. Test the determinant with explicit numbers inside the allowed range.

Answer: (c) a straight line

Explanation:
1. The first two lines meet y = 0 at (−p, 0) and (−q, 0), and each other at \( (pq,\ pq+p+q+1) \).
2. The altitude from the third vertex to y = 0 is the line x = pq.
3. The altitude from (−p, 0) is perpendicular to the second line, whose slope is \( \frac{1+q}{q} \): \( y=-\frac{q}{1+q}(x+p) \). At x = pq this gives \( y=-\frac{q(pq+p)}{1+q}=-pq \).
4. The orthocentre is (pq, −pq), which lies on x + y = 0 for every p, q: a straight line.

Teacher's Note:
1. Two altitudes are enough to locate the orthocentre.
2. The coordinates (pq, −pq) are negatives of each other, so the locus is a line.

Answer: (d) P and R only

Explanation:
1. Adding the two equations: (a + b)(x + y) + 2c = 0 and subtracting: (a − b)(x − y) = 0, so x = y = \( -\frac{c}{a+b} \).
2. Distance from (1, 1): \( \sqrt2\left(1+\frac{c}{a+b}\right) \lt 2\sqrt2 \), so \( \frac{c}{a+b} \lt 1 \), i.e. a + b − c > 0 (P).
3. Since a > b and c > 0, a − b + c > 0 is always true (R). Q and S are false.

Teacher's Note:
1. Solve for the intersection point first; it lies on y = x.
2. R is true by the given ordering alone, independent of the distance condition.

Answer: (b) 6

Explanation:
1. In the first quadrant \( d_1+d_2=\frac{|x-y|+(x+y)}{\sqrt2}=\sqrt2\,\max(x,y) \).
2. So the condition becomes \( \sqrt2\le\max(x,y)\le2\sqrt2 \).
3. The region is the difference of two squares: area = \( (2\sqrt2)^2-(\sqrt2)^2=8-2=6 \).

Teacher's Note:
1. For x ≥ y we have |x − y| + x + y = 2x, and similarly 2y otherwise, which gives the max.
2. The set max(x, y) ≤ t in the first quadrant is a square of side t.

Answer: (b) \( \frac{23}{\sqrt{17}} \)

Explanation:
1. Substituting (13, 32): \( \frac{13}{5}+\frac{32}{b}=1 \), so \( \frac{32}{b}=-\frac85 \) and b = −20. L: 4x − y − 20 = 0.
2. K is parallel to L, so it has slope 4: \( -\frac3c=4 \), c = \( -\frac34 \). K: 4x − y + 3 = 0.
3. Distance = \( \frac{|3-(-20)|}{\sqrt{16+1}}=\frac{23}{\sqrt{17}} \).

Teacher's Note:
1. Use the first condition to fix b and the parallel condition to fix c.
2. The distance between parallel lines needs both equations in the same form ax + by + c = 0.

Answer: (c) 6

Explanation:
1. Dividing point: \( \left(\frac{3\cdot2+2\cdot1}{5},\frac{3\cdot4+2\cdot1}{5}\right)=\left(\frac85,\frac{14}5\right) \).
2. k = \( 2\cdot\frac85+\frac{14}5=\frac{30}{5}=6 \).

Teacher's Note:
1. Be careful to pair the ratio correctly: 3 goes with the second point.
2. If the ratio were reversed you would get 29/5.

Answer: (b) \( \sqrt3y=x-\sqrt3 \)

Explanation:
1. The ray meets the x-axis (y = 0) at \( (\sqrt3,0) \).
2. Its slope is \( -\frac1{\sqrt3} \); after reflection in the x-axis the slope changes sign to \( \frac1{\sqrt3} \).
3. Reflected ray: \( y=\frac1{\sqrt3}(x-\sqrt3) \), i.e. \( \sqrt3y=x-\sqrt3 \).

Teacher's Note:
1. Reflection in the x-axis keeps the point of incidence and reverses the slope.
2. Check that the incident ray also passes through the point of incidence.

Answer: (a) \( 2-\sqrt2 \)

Explanation:
1. The vertices follow from the mid-points: (0, 2), (0, 0) and (2, 0).
2. This is a right triangle with legs 2, 2 and hypotenuse \( 2\sqrt2 \).
3. Inradius \( r=\frac{2+2-2\sqrt2}{2}=2-\sqrt2 \). The incentre is at (r, r), so its x-co-ordinate is \( 2-\sqrt2 \).

Teacher's Note:
1. A vertex equals the sum of two adjacent mid-points minus the opposite one.
2. For a right triangle at the origin the incentre is (r, r).

Answer: (d) 3bc − 2ad = 0

Explanation:
1. A point of the fourth quadrant equidistant from the axes lies on y = −x.
2. Putting y = −x: 4ax − 2ax + c = 0 gives \( x=-\frac{c}{2a} \); 5bx − 2bx + d = 0 gives \( x=-\frac{d}{3b} \).
3. Both give the same x: \( \frac{c}{2a}=\frac{d}{3b} \), so 3bc − 2ad = 0.

Teacher's Note:
1. The fourth quadrant has x > 0, y < 0, so equal distances means y = −x.
2. Substitute into each line separately and equate the x-values.

Answer: (d) 780

Explanation:
1. Count integer points (i, j) with i ≥ 1, j ≥ 1 and i + j ≤ 40.
2. For i = 1, 2, …, 39 the number of j is 40 − i, so the total is \( 39+38+\dots+1=\frac{39\cdot40}{2}=780 \).
3. Check with Pick's theorem: area 840.5, boundary points 123, interior = 840.5 − 61.5 + 1 = 780.

Teacher's Note:
1. Interior means strictly inside, so x + y < 41 with x, y ≥ 1.
2. Pick's theorem gives an independent check.

Answer: (b) \( \left(2,\frac12\right) \)

Explanation:
1. The area determinant (with the 1/2) equals \( \frac{5k^2+13k+10}{2} \), so area 28 gives \( 5k^2+13k-46=0 \) (the other sign gives no real k).
2. The roots are k = 2 and \( k=-\frac{23}5 \); k is an integer, so k = 2 and the vertices are (2, −6), (5, 2), (−2, 2).
3. BC is the line y = 2, so the altitude from A is x = 2. The altitude from B is perpendicular to AC (slope −2), so it has slope \( \frac12 \): \( y-2=\frac12(x-5) \).
4. At x = 2: \( y=2-\frac32=\frac12 \). Orthocentre \( \left(2,\frac12\right) \).

Teacher's Note:
1. The integer condition selects k = 2.
2. With one horizontal side the first altitude is vertical.

Answer: (a) 3x + 2y = xy

Explanation:
1. Let P(a, 0) and Q(0, b); then R = (a, b). The line is \( \frac xa+\frac yb=1 \) and passes through (2, 3): \( \frac2a+\frac3b=1 \).
2. Multiply by ab: 2b + 3a = ab.
3. Replacing (a, b) by (x, y): 3x + 2y = xy.

Teacher's Note:
1. R has the intercepts as its co-ordinates.
2. Do not swap 2 and 3: the coefficient of x is 3 because 2/a multiplies b.

Answer: (b) 26x − 122y − 1675 = 0

Explanation:
1. Let the third side meet the two given sides at B (on 3x − 2y + 6 = 0) and C (on 4x + 5y − 20 = 0).
2. The altitude from B is perpendicular to the second given side and passes through H(1, 1): 5(x − 1) − 4(y − 1) = 0. Intersecting with 3x − 2y + 6 = 0 gives \( B=\left(-13,-\frac{33}2\right) \).
3. The altitude from C is perpendicular to the first side: 2(x − 1) + 3(y − 1) = 0. Intersecting with 4x + 5y − 20 = 0 gives \( C=\left(\frac{35}2,-10\right) \).
4. Slope of BC = \( \frac{-10+33/2}{35/2+13}=\frac{13}{61} \), and the line is 26x − 122y − 1675 = 0 (both B and C satisfy it).

Teacher's Note:
1. Each altitude from a vertex on the two known sides is perpendicular to the other known side.
2. Verify by substituting B and C into the final equation.

Answer: (d) second

Explanation:
1. Let the third vertex be C(a, b), A(0, 2), B(4, 3), H(0, 0).
2. AH is perpendicular to BC: the vector (0, 2) is perpendicular to (a − 4, b − 3), so b = 3.
3. BH is perpendicular to AC: (4, 3)·(a, b − 2) = 0, so 4a + 3(b − 2) = 0 and with b = 3, a = \( -\frac34 \).
4. C = (−3/4, 3), which lies in the second quadrant.

Teacher's Note:
1. Use the two altitude conditions with H at the origin.
2. Check the third altitude: CH·AB = (3/4, −3)·(4, 1) = 0.

Answer: (d) 5x − 7y = 0

Explanation:
1. A point on x = 2y is (2t, t). Its foot of the perpendicular on x = y is \( \left(\frac{3t}2,\frac{3t}2\right) \).
2. Mid-point of the point and its foot: \( \left(\frac{7t}4,\frac{5t}4\right) \).
3. Eliminating t: x = 7t/4 and y = 5t/4 give 5x = 7y, i.e. 5x − 7y = 0.

Teacher's Note:
1. The foot of the perpendicular from (u, v) on y = x is ((u + v)/2, (u + v)/2).
2. The locus is a line through the origin because the mid-point moves linearly with t.