Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Straight Lines and Pair of Straight Lines question

2025 · 29 Jan · Shift 1 · Q32
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Straight Lines and Pair of Straight Lines
  5. /2025 · 29 Jan · Shift 1 · Q32

Straight Lines and Pair of Straight Lines question

2025 · 29 Jan · Shift 1 · Q32

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let ΔABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of ΔABC in the line 3x + 6y – 53 = 0. Then h2 + k2 + hk is equal to :
  1. A
    47
  2. B
    37
  3. C
    40
  4. D
    36
View written solutionFree

Correct answer: B

  1. Find the vertices of the triangle formed by the lines

Let L1:7x−6y+3=0,L_1: 7x-6y+3=0,L1​:7x−6y+3=0, L2:x+2y−31=0,L_2: x+2y-31=0,L2​:x+2y−31=0, L3:9x−2y−19=0.L_3: 9x-2y-19=0.L3​:9x−2y−19=0.

The triangle vertices are the pairwise intersections.

  1. Intersection of L1L_1L1​ and L2L_2L2​

Solve 7x−6y=−3,7x-6y=-3,7x−6y=−3, x+2y=31.x+2y=31.x+2y=31.

From the second equation, x=31−2y.x=31-2y.x=31−2y. Substitute into the first: 7(31−2y)−6y=−37(31-2y)-6y=-37(31−2y)−6y=−3 217−14y−6y=−3217-14y-6y=-3217−14y−6y=−3 217−20y=−3217-20y=-3217−20y=−3 −20y=−220-20y=-220−20y=−220 y=11.y=11.y=11. Then x=31−2(11)=9.x=31-2(11)=9.x=31−2(11)=9. So one vertex is A=(9,11).A=(9,11).A=(9,11).

  1. Intersection of L2L_2L2​ and L3L_3L3​

Solve x+2y=31,x+2y=31,x+2y=31, 9x−2y=19.9x-2y=19.9x−2y=19.

Add the equations: 10x=50 ⇒ x=5.10x=50 \,\Rightarrow\, x=5.10x=50⇒x=5. Then 5+2y=31 ⇒ 2y=26 ⇒ y=13.5+2y=31 \,\Rightarrow\, 2y=26 \,\Rightarrow\, y=13.5+2y=31⇒2y=26⇒y=13. So B=(5,13).B=(5,13).B=(5,13).

  1. Intersection of L3L_3L3​ and L1L_1L1​

Solve 9x−2y=19,9x-2y=19,9x−2y=19, 7x−6y=−3.7x-6y=-3.7x−6y=−3.

Multiply the first by 333: 27x−6y=57.27x-6y=57.27x−6y=57. Subtract the second: (27x−6y)−(7x−6y)=57−(−3)(27x-6y)-(7x-6y)=57-(-3)(27x−6y)−(7x−6y)=57−(−3) 20x=60 ⇒ x=3.20x=60 \,\Rightarrow\, x=3.20x=60⇒x=3. Then 9(3)−2y=199(3)-2y=199(3)−2y=19 27−2y=1927-2y=1927−2y=19 −2y=−8 ⇒ y=4.-2y=-8 \,\Rightarrow\, y=4.−2y=−8⇒y=4. So C=(3,4).C=(3,4).C=(3,4).

  1. Find the centroid of △ABC\triangle ABC△ABC

The centroid is G(x1+x2+x33,y1+y2+y33).G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).G(3x1​+x2​+x3​​,3y1​+y2​+y3​​).

Thus,

=\left(\frac{17}{3},\frac{28}{3}\right).$$ --- 6. **Reflect the centroid in the line $3x+6y-53=0$** Given line: $$3x+6y-53=0.$$ Here, $$a=3,\quad b=6,\quad c=-53.$$ For reflection of point $(x_0,y_0)$ in line $ax+by+c=0$: $$x' = x_0 - \frac{2a(ax_0+by_0+c)}{a^2+b^2},$$ $$y' = y_0 - \frac{2b(ax_0+by_0+c)}{a^2+b^2}.$$ Now for $$\left(x_0,y_0\right)=\left(\frac{17}{3},\frac{28}{3}\right),$$ we get $$ax_0+by_0+c = 3\cdot\frac{17}{3}+6\cdot\frac{28}{3}-53$$ $$=17+56-53=20.$$ Also, $$a^2+b^2=3^2+6^2=9+36=45.$$ So, $$h=x' = \frac{17}{3}-\frac{2\cdot 3\cdot 20}{45} =\frac{17}{3}-\frac{120}{45} =\frac{17}{3}-\frac{8}{3}=3.$$ $$k=y' = \frac{28}{3}-\frac{2\cdot 6\cdot 20}{45} =\frac{28}{3}-\frac{240}{45} =\frac{28}{3}-\frac{16}{3}=4.$$ Hence the reflected point is $$(h,k)=(3,4).$$ --- 7. **Compute $h^2+k^2+hk$** $$h^2+k^2+hk = 3^2+4^2+(3)(4)$$ $$=9+16+12=37.$$ --- 8. **Compare with options** The value is $$\boxed{37}.$$ So the correct option is **B**. --- 9. **Compare with stored correct answer** Stored correct answer: **B** Our derived answer: **B** So they agree.
PreviousNext

More from Straight Lines and Pair of Straight Lines

  • Let the line x + y = 1 meet the axes of x and y at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of…2025 · MCQ
  • Let ABC be an isosceles triangle in which A is at (−1,0),∠A=32π​,AB=AC and B is on the positve x-axis. If BC=43​ and the line BC intersects the line y=x+3 at $(\alpha,…2024 · Numerical
  • The lines L1​, L2​,…,L20​ are distinct. For n=1,2,3,…,10 all the lines L2n−1​ are parallel to each other and all the lines L2n​ pass through a given point…2024 · Numerical
  • The vertices of a triangle are A(−1,3),B(−2,2) and C(3,−1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to…2024 · MCQ
  • Let two straight lines drawn from the origin O intersect the line 3x+4y=12 at the points P and Q such that △OPQ is an isosceles triangle and ∠POQ=90∘. If l=OP2+PQ2+QO2…2024 · MCQ
  • Let A(−1,1) and B(2,3) be two points and P be a variable point above the line AB such that the area of △PAB is 10. If the locus of P is $\mathrm{a}…2024 · MCQ
  • Let a variable line of slope m>0 passing through the point (4,−9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is2024 · MCQ
  • If the locus of the point, whose distances from the point (2,1) and (1,3) are in the ratio 5:4, is ax2+by2+cxy+dx+ey+170=0, then the value of a2+2b+3c+4d+e is equal to :2024 · MCQ