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 · 22 Jan · Shift 1 · Q30
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 · 22 Jan · Shift 1 · Q30

Straight Lines and Pair of Straight Lines question

2025 · 22 Jan · Shift 1 · Q30

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let the triangle PQR be the image of the triangle with vertices (1,3),(3,1)(1,3),(3,1)(1,3),(3,1) and (2,4)(2,4)(2,4) in the line x+2y=2x+2 y=2x+2y=2. If the centroid of △PQR\triangle \mathrm{PQR}△PQR is the point (α,β)(\alpha, \beta)(α,β), then 15(α−β)15(\alpha-\beta)15(α−β) is equal to :
  1. A
    21
  2. B
    19
  3. C
    22
  4. D
    24
View written solutionFree

Correct answer: C

  1. Find the centroid of the original triangle

The given triangle has vertices A(1,3),  B(3,1),  C(2,4).A(1,3),\; B(3,1),\; C(2,4).A(1,3),B(3,1),C(2,4).

The centroid of a triangle 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​​).

So, G=(1+3+23,3+1+43)=(63,83)=(2,83).G=\left(\frac{1+3+2}{3},\frac{3+1+4}{3}\right)=\left(\frac{6}{3},\frac{8}{3}\right)=\left(2,\frac{8}{3}\right).G=(31+3+2​,33+1+4​)=(36​,38​)=(2,38​).

  1. Use the fact that reflection preserves centroid structure

Since △PQR\triangle PQR△PQR is the image of the given triangle in the line x+2y=2,x+2y=2,x+2y=2, the centroid of the image triangle is the reflection of the original centroid across the same line.

So we reflect the point (2,83)\left(2,\frac{8}{3}\right)(2,38​) in the line x+2y−2=0.x+2y-2=0.x+2y−2=0.

  1. Formula for reflection of a point in a line

For reflection of point (x0,y0)(x_0,y_0)(x0​,y0​) in the line ax+by+c=0,ax+by+c=0,ax+by+c=0, the image is (x0−2a(ax0+by0+c)a2+b2,  y0−2b(ax0+by0+c)a2+b2).\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).(x0​−a2+b22a(ax0​+by0​+c)​,y0​−a2+b22b(ax0​+by0​+c)​).

Here, a=1,  b=2,  c=−2,  (x0,y0)=(2,83).a=1,\; b=2,\; c=-2,\; (x_0,y_0)=\left(2,\frac{8}{3}\right).a=1,b=2,c=−2,(x0​,y0​)=(2,38​).

First compute: ax0+by0+c=2+2⋅83−2=163.ax_0+by_0+c=2+2\cdot \frac{8}{3}-2=\frac{16}{3}.ax0​+by0​+c=2+2⋅38​−2=316​.

Also, a2+b2=12+22=5.a^2+b^2=1^2+2^2=5.a2+b2=12+22=5.

Thus, α=2−2⋅1⋅(16/3)5=2−3215=30−3215=−215,\alpha=2-\frac{2\cdot 1\cdot (16/3)}{5}=2-\frac{32}{15}=\frac{30-32}{15}=-\frac{2}{15},α=2−52⋅1⋅(16/3)​=2−1532​=1530−32​=−152​,

β=83−2⋅2⋅(16/3)5=83−6415=40−6415=−2415=−85.\beta=\frac{8}{3}-\frac{2\cdot 2\cdot (16/3)}{5}=\frac{8}{3}-\frac{64}{15}=\frac{40-64}{15}=-\frac{24}{15}=-\frac{8}{5}.β=38​−52⋅2⋅(16/3)​=38​−1564​=1540−64​=−1524​=−58​.

So the centroid of △PQR\triangle PQR△PQR is (α,β)=(−215,−85).\left(\alpha,\beta\right)=\left(-\frac{2}{15},-\frac{8}{5}\right).(α,β)=(−152​,−58​).

  1. Compute 15(α−β)15(\alpha-\beta)15(α−β)

α−β=−215−(−85)=−215+2415=2215.\alpha-\beta=-\frac{2}{15}-\left(-\frac{8}{5}\right)=-\frac{2}{15}+\frac{24}{15}=\frac{22}{15}.α−β=−152​−(−58​)=−152​+1524​=1522​.

Therefore, 15(α−β)=22.15(\alpha-\beta)=22.15(α−β)=22.

  1. Check options

The correct option is: C: 22\boxed{\text{C: }22}C: 22​

PreviousNext

More from Straight Lines and Pair of Straight Lines

  • Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on…2025 · Numerical
  • A rod of length eight units moves such that its ends A and B always lie on the lines x−y+2=0 and y+2=0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is 9(x2+αy2+βxy+γx+28y)−76=0…2025 · MCQ
  • Let the lines 3x−4y−α=0,8x−11y−33=0, and 2x−3y+λ=0 be concurrent. If the image of the point (1,2) in the line 2x−3y+λ=0 is (1357​,13−40​), then ∣αλ∣ is equal to2025 · MCQ
  • Let the points (211​,α) lie on or inside the triangle with sides x+y=11,x+2y=16 and 2x+3y=29. Then the product of the smallest and the largest values of α is equal to :2025 · MCQ
  • Two equal sides of an isosceles triangle are along −x+2y=4 and x+y=4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is:2025 · MCQ
  • If A and B are the points of intersection of the circle x2+y2−8x=0 and the hyperbola 9x2​−4y2​=1 and a point P moves on the line 2x−3y+4=0, then the centroid of ΔPAB lies on the line :2025 · MCQ
  • 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 :2025 · MCQ
  • 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