Let ABC be a triangle formed by the lines 7mathrmx - 6mathrmy + 3 = 0, mathrmx + 2mathrmy - 31 = 0$7\mathrm{x} - 6\mathrm{y} + 3 = 0, \mathrm{x} + 2\mathrm{y} - 31 = 0$ and 9mathrmx - 2mathrmy - 19 = 0$9\mathrm{x} - 2\mathrm{y} - 19 = 0$ . Let the point (mathrmh,mathrmk)$(\mathrm{h},\mathrm{k})$ be the image of the centroid of Delta ABC$\Delta ABC$ in the line 3mathrmx + 6mathrmy - 53 = 0$3\mathrm{x} + 6\mathrm{y} - 53 = 0$ . Then \mathrm{h}^2 + \mathrm{k}^2 + \mathrm{hk} is equal to
A.37
B.47
C.40
D.36
Solution & Explanation
### Related Formula
textCentroid G = left(fracx_1+x_2+x_33, fracy_1+y_2+y_33right)$$\text{Centroid } G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$textImage of (x_1, y_1) text in line ax+by+c=0: fracx-x_1a = fracy-y_1b = -2fracax_1+by_1+ca^2+b^2$$\text{Image of } (x_1, y_1) \text{ in line } ax+by+c=0: \frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$$
### Core Logic
First, find the vertices A, B, C$A, B, C$ by solving the lines pairwise.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and x + 2y - 31 = 0$x + 2y - 31 = 0$ gives A(9,11)$A(9,11)$.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives B(3,4)$B(3,4)$.
Solving x + 2y - 31 = 0$x + 2y - 31 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives C(5,13)$C(5,13)$.
Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 1: Determine the Centroid
G = left(frac9 + 3 + 53, frac11 + 4 + 133right) = left(frac173, frac283right)$$G = \left(\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}\right) = \left(\frac{17}{3}, \frac{28}{3}\right)$$
### Step 2: Find the Image (h, k)
Using the line 3x + 6y - 53 = 0$3x + 6y - 53 = 0$:
frach - frac1733 = frack - frac2836 = -2 frac3left(frac173right) + 6left(frac283right) - 533^2 + 6^2$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{3\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53}{3^2 + 6^2}$$frach - frac1733 = frack - frac2836 = -2 frac17 + 56 - 5345 = -2 frac2045 = -frac89$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{17 + 56 - 53}{45} = -2 \frac{20}{45} = -\frac{8}{9}$$
Solving for h$h$ and k$k$ yields:
h = 3, quad k = 4$$h = 3, \quad k = 4$$Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 3: Compute final algebraic target
h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$
### Pattern Recognition
Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship (k - y_G)/(h - x_G) = -1/m$(k - y_G)/(h - x_G) = -1/m$ to accelerate competitive solving time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Keywords:#image of the centroid of triangle ABC#JEE Main 2025 Morning Q54#Straight Lines JEE Main 2025#Centroid and Image of a Point JEE Main 2025
More Straight Lines Previous-Year Questions — Page 5
Q2jee_main_2024_31_jan_eveningCentroid and Orthocentre
Let A (a, b)$A (a, b)$, B(3, 4)$B(3, 4)$ and (-6, -8)$(-6, -8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5)$P(2a + 3, 7b + 5)$ from the line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$ measured parallel to the line x - 2y - 1 = 0$x - 2y - 1 = 0$ is
A.frac15 sqrt57$\frac{15 \sqrt{5}}{7}$
B.frac17sqrt56$\frac{17\sqrt{5}}{6}$
C.frac17 sqrt57$\frac{17 \sqrt{5}}{7}$
D.fracsqrt517$\frac{\sqrt{5}}{17}$
Solution
### Related Formula
textCentroid divides the line joining Orthocentre and Circumcentre in 2:1$$\text{Centroid divides the line joining Orthocentre and Circumcentre in } 2:1$$
Distance in parametric form: x = x_1 + rcostheta, y = y_1 + rsintheta$x = x_1 + r\cos\theta, y = y_1 + r\sin\theta$
### Core Logic
Centroid and Orthocentre diagram for Q2 - JEE Main 2024 Evening
Let Orthocentre C(-6, -8)$C(-6, -8)$ and Circumcentre B(3, 4)$B(3, 4)$. Centroid A(a, b)$A(a, b)$ divides CB$CB$ in 2:1$2:1$.
a = frac2(3) + 1(-6)2+1 = 0$$a = \frac{2(3) + 1(-6)}{2+1} = 0$$b = frac2(4) + 1(-8)2+1 = 0$$b = \frac{2(4) + 1(-8)}{2+1} = 0$$
So, P(2a+3, 7b+5) = (3, 5)$P(2a+3, 7b+5) = (3, 5)$.
The line along which distance is measured is parallel to x - 2y - 1 = 0$x - 2y - 1 = 0$, giving slope m = tantheta = frac12$m = \tan\theta = \frac{1}{2}$.
Using parametric coordinates from P(3,5)$P(3,5)$:
x = 3 + rcostheta, quad y = 5 + rsintheta$$x = 3 + r\cos\theta, \quad y = 5 + r\sin\theta$$
Substitute into the target line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$:
2(3 + rcostheta) + 3(5 + rsintheta) - 4 = 0$$2(3 + r\cos\theta) + 3(5 + r\sin\theta) - 4 = 0$$r(2costheta + 3sintheta) = -17$$r(2\cos\theta + 3\sin\theta) = -17$$
From tantheta = 1/2$\tan\theta = 1/2$, we get sintheta = frac1sqrt5$\sin\theta = \frac{1}{\sqrt{5}}$ and costheta = frac2sqrt5$\cos\theta = \frac{2}{\sqrt{5}}$.
rleft(2left(frac2sqrt5right) + 3left(frac1sqrt5right)right) = -17$$r\left(2\left(\frac{2}{\sqrt{5}}\right) + 3\left(\frac{1}{\sqrt{5}}\right)\right) = -17$$rleft(frac7sqrt5right) = -17 implies |r| = frac17sqrt57$$r\left(\frac{7}{\sqrt{5}}\right) = -17 \implies |r| = \frac{17\sqrt{5}}{7}$$
### Pattern Recognition
Standard Euler line property: O, G, C$O, G, C$ are collinear and G$G$ divides OC$OC$ in 2:1$2:1$. Use parametric equation to find intersection distance directly without finding the intersection point.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Let A(-2, -1)$A(-2, -1)$, B(1, 0)$B(1, 0)$, C(alpha, beta)$C(\alpha, \beta)$ and D(gamma, delta)$D(\gamma, \delta)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x - y = 5$2x - y = 5$ and the point D lies on 3x - 2y = 6$3x - 2y = 6$, then the value of |alpha + beta + gamma + delta|$|\alpha + \beta + \gamma + \delta|$ is equal to
Q10jee_main_2024_31_jan_morningProperties of Parallelogram
Let alpha, beta, gamma, delta in Z$\alpha, \beta, \gamma, \delta \in Z$ and let A (alpha, beta)$A (\alpha, \beta)$, B (1, 0)$B (1, 0)$, C (gamma, delta)$C (\gamma, \delta)$ and D (1, 2)$D (1, 2)$ be the vertices of a parallelogram ABCD$ABCD$. If AB = sqrt10$AB = \sqrt{10}$ and the points A$A$ and C$C$ lie on the line 3y = 2x + 1$3y = 2x + 1$, then 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$ is equal to
A.10$10$
B.5$5$
C.12$12$
D.8$8$
Solution
### Core Logic
Let E$E$ be the midpoint of the diagonals AC$AC$ and BD$BD$.
Since ABCD$ABCD$ is a parallelogram, the diagonals bisect each other.
Properties of Parallelogram diagram for Q10 - JEE Main 2024 Morning
Midpoint E$E$ from BD$BD$: left( frac1+12, frac0+22 right) = (1, 1)$\left( \frac{1+1}{2}, \frac{0+2}{2} \right) = (1, 1)$.
### Step 1: Apply Midpoint on AC
Midpoint E$E$ from AC$AC$: left( fracalpha+gamma2, fracbeta+delta2 right)$\left( \frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2} \right)$.
Equating both:
fracalpha+gamma2 = 1 implies alpha + gamma = 2$$\frac{\alpha+\gamma}{2} = 1 \implies \alpha + \gamma = 2$$fracbeta+delta2 = 1 implies beta + delta = 2$$\frac{\beta+\delta}{2} = 1 \implies \beta + \delta = 2$$
### Step 2: Final Value
The expression requires 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$.
2(2 + 2) = 2(4) = 8$$2(2 + 2) = 2(4) = 8$$
*(Note: Additional conditions like AB = sqrt10$AB = \sqrt{10}$ and the line equation are extraneous data not needed to find the sum).*
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
More Straight Lines Questions — jee_main_2025_29_jan_morning
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