Question:
If \(O\) is the origin and \(H\) is the orthocenter of a triangle formed by the lines
\[
x+y=1,
\]
\[
6x^2-13xy+6y^2=0,
\]
then
\[
OH=
\]
If \(O\) is the origin and \(H\) is the orthocenter of a triangle formed by the lines
\[
x+y=1,
\]
\[
6x^2-13xy+6y^2=0,
\]
then
\[
OH=
\]
Show Hint
When a homogeneous quadratic
\[
ax^2+hxy+by^2=0
\]
appears, first factor it into two lines through the origin. Then find the triangle vertices and obtain the orthocenter by intersecting any two altitudes.
Updated On: Jul 9, 2026
- \(12\sqrt2\)
- \(\dfrac{12\sqrt2}{25}\)
- \(\dfrac{24\sqrt2}{25}\)
- \(24\sqrt2\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The equation
\[
6x^2-13xy+6y^2=0
\]
represents a pair of straight lines through the origin. These two lines together with
\[
x+y=1
\]
form a triangle. The orthocenter is obtained by intersecting two altitudes.
Step 1: Factorize the pair of lines. \[ 6x^2-13xy+6y^2 = (2x-3y)(3x-2y). \] Hence the two lines are \[ 2x-3y=0 \] and \[ 3x-2y=0. \] Let \[ A=(0,0) \] be their point of intersection.
Step 2: Find the other two vertices of the triangle. Intersection of \[ x+y=1 \] and \[ 2x-3y=0 \] gives \[ x=\frac35, \qquad y=\frac25. \] Thus, \[ B\left(\frac35,\frac25\right). \] Intersection of \[ x+y=1 \] and \[ 3x-2y=0 \] gives \[ x=\frac25, \qquad y=\frac35. \] Thus, \[ C\left(\frac25,\frac35\right). \]
Step 3: Find the altitude through \(B\). Slope of \(AC\) is \[ \frac{\frac35-0}{\frac25-0} = \frac32. \] Therefore the altitude through \(B\) has slope \[ -\frac23. \] Its equation is \[ y-\frac25 = -\frac23 \left(x-\frac35\right). \] \[ 10x+15y-12=0. \] \[ \cdots (1) \]
Step 4: Find the altitude through \(C\). Slope of \(AB\) is \[ \frac{\frac25}{\frac35} = \frac23. \] Hence the altitude through \(C\) has slope \[ -\frac32. \] Its equation is \[ y-\frac35 = -\frac32 \left(x-\frac25\right). \] \[ 15x+10y-12=0. \] \[ \cdots (2) \]
Step 5: Find the orthocenter. Solving \[ 10x+15y=12 \] and \[ 15x+10y=12, \] subtracting, \[ 5x-5y=0. \] \[ x=y. \] Substituting into \[ 10x+15x=12, \] \[ 25x=12. \] \[ x=y=\frac{12}{25}. \] Hence \[ H\left(\frac{12}{25},\frac{12}{25}\right). \]
Step 6: Find \(OH\). \[ OH = \sqrt{ \left(\frac{12}{25}\right)^2 + \left(\frac{12}{25}\right)^2 }. \] \[ = \frac{12}{25}\sqrt2. \] \[ OH = \frac{12\sqrt2}{25}. \]
Step 7: Write the final answer. \[ \boxed{\frac{12\sqrt2}{25}} \]
Step 1: Factorize the pair of lines. \[ 6x^2-13xy+6y^2 = (2x-3y)(3x-2y). \] Hence the two lines are \[ 2x-3y=0 \] and \[ 3x-2y=0. \] Let \[ A=(0,0) \] be their point of intersection.
Step 2: Find the other two vertices of the triangle. Intersection of \[ x+y=1 \] and \[ 2x-3y=0 \] gives \[ x=\frac35, \qquad y=\frac25. \] Thus, \[ B\left(\frac35,\frac25\right). \] Intersection of \[ x+y=1 \] and \[ 3x-2y=0 \] gives \[ x=\frac25, \qquad y=\frac35. \] Thus, \[ C\left(\frac25,\frac35\right). \]
Step 3: Find the altitude through \(B\). Slope of \(AC\) is \[ \frac{\frac35-0}{\frac25-0} = \frac32. \] Therefore the altitude through \(B\) has slope \[ -\frac23. \] Its equation is \[ y-\frac25 = -\frac23 \left(x-\frac35\right). \] \[ 10x+15y-12=0. \] \[ \cdots (1) \]
Step 4: Find the altitude through \(C\). Slope of \(AB\) is \[ \frac{\frac25}{\frac35} = \frac23. \] Hence the altitude through \(C\) has slope \[ -\frac32. \] Its equation is \[ y-\frac35 = -\frac32 \left(x-\frac25\right). \] \[ 15x+10y-12=0. \] \[ \cdots (2) \]
Step 5: Find the orthocenter. Solving \[ 10x+15y=12 \] and \[ 15x+10y=12, \] subtracting, \[ 5x-5y=0. \] \[ x=y. \] Substituting into \[ 10x+15x=12, \] \[ 25x=12. \] \[ x=y=\frac{12}{25}. \] Hence \[ H\left(\frac{12}{25},\frac{12}{25}\right). \]
Step 6: Find \(OH\). \[ OH = \sqrt{ \left(\frac{12}{25}\right)^2 + \left(\frac{12}{25}\right)^2 }. \] \[ = \frac{12}{25}\sqrt2. \] \[ OH = \frac{12\sqrt2}{25}. \]
Step 7: Write the final answer. \[ \boxed{\frac{12\sqrt2}{25}} \]
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