Question:
Suppose that the sides passing through the vertex \((\alpha,\beta)\) of a triangle are bisected at right angles by the lines
\[
y^2-8xy-9x^2=0.
\]
Then the centroid of the triangle is
Suppose that the sides passing through the vertex \((\alpha,\beta)\) of a triangle are bisected at right angles by the lines
\[
y^2-8xy-9x^2=0.
\]
Then the centroid of the triangle is
Show Hint
If a side is bisected at right angles by a line, then the opposite endpoint is the reflection of the given vertex in that line.
Updated On: Jun 22, 2026
- \(\dfrac{1}{123}(\alpha,\beta)\)
- \(\dfrac{1}{123}(\alpha+32\beta,81\beta+32\alpha)\)
- \(\dfrac{1}{123}(\alpha-32\beta,81\beta+32\alpha)\)
- \(\dfrac{1}{123}(\alpha-32\beta,81\beta-32\alpha)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the slopes of the given pair of lines.
Given, \[ y^2-8xy-9x^2=0 \] Put \[ y=mx \] Then, \[ m^2x^2-8mx^2-9x^2=0 \] \[ m^2-8m-9=0 \] \[ (m-9)(m+1)=0 \] So, \[ m=9,\quad m=-1 \]
Step 2: Use reflection formula.
If a point \((x,y)\) is reflected in the line \(y=mx\), then its image is \[ \left( \frac{(1-m^2)x+2my}{1+m^2}, \frac{2mx+(m^2-1)y}{1+m^2} \right) \] Here, the vertex is \[ A=(\alpha,\beta) \]
Step 3: Reflection of \(A\) in the line \(y=9x\).
For \[ m=9, \] the reflected point is \[ B= \left( \frac{(1-81)\alpha+18\beta}{82}, \frac{18\alpha+(81-1)\beta}{82} \right) \] \[ B= \left( \frac{-40\alpha+9\beta}{41}, \frac{9\alpha+40\beta}{41} \right) \]
Step 4: Reflection of \(A\) in the line \(y=-x\).
For \[ m=-1, \] the reflected point is \[ C= \left( \frac{(1-1)\alpha-2\beta}{2}, \frac{-2\alpha+(1-1)\beta}{2} \right) \] \[ C=(-\beta,-\alpha) \]
Step 5: Find the centroid.
Centroid is \[ G=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right) \] So, \[ G_x= \frac{1}{3}\left(\alpha+\frac{-40\alpha+9\beta}{41}-\beta\right) \] \[ G_x=\frac{\alpha-32\beta}{123} \] Also, \[ G_y= \frac{1}{3}\left(\beta+\frac{9\alpha+40\beta}{41}-\alpha\right) \] \[ G_y=\frac{81\beta-32\alpha}{123} \]
Step 6: Final conclusion.
Therefore, the centroid is \[ \boxed{\frac{1}{123}(\alpha-32\beta,81\beta-32\alpha)} \]
Given, \[ y^2-8xy-9x^2=0 \] Put \[ y=mx \] Then, \[ m^2x^2-8mx^2-9x^2=0 \] \[ m^2-8m-9=0 \] \[ (m-9)(m+1)=0 \] So, \[ m=9,\quad m=-1 \]
Step 2: Use reflection formula.
If a point \((x,y)\) is reflected in the line \(y=mx\), then its image is \[ \left( \frac{(1-m^2)x+2my}{1+m^2}, \frac{2mx+(m^2-1)y}{1+m^2} \right) \] Here, the vertex is \[ A=(\alpha,\beta) \]
Step 3: Reflection of \(A\) in the line \(y=9x\).
For \[ m=9, \] the reflected point is \[ B= \left( \frac{(1-81)\alpha+18\beta}{82}, \frac{18\alpha+(81-1)\beta}{82} \right) \] \[ B= \left( \frac{-40\alpha+9\beta}{41}, \frac{9\alpha+40\beta}{41} \right) \]
Step 4: Reflection of \(A\) in the line \(y=-x\).
For \[ m=-1, \] the reflected point is \[ C= \left( \frac{(1-1)\alpha-2\beta}{2}, \frac{-2\alpha+(1-1)\beta}{2} \right) \] \[ C=(-\beta,-\alpha) \]
Step 5: Find the centroid.
Centroid is \[ G=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right) \] So, \[ G_x= \frac{1}{3}\left(\alpha+\frac{-40\alpha+9\beta}{41}-\beta\right) \] \[ G_x=\frac{\alpha-32\beta}{123} \] Also, \[ G_y= \frac{1}{3}\left(\beta+\frac{9\alpha+40\beta}{41}-\alpha\right) \] \[ G_y=\frac{81\beta-32\alpha}{123} \]
Step 6: Final conclusion.
Therefore, the centroid is \[ \boxed{\frac{1}{123}(\alpha-32\beta,81\beta-32\alpha)} \]
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