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 \[ OH= \]

• A. \(12\sqrt2\)
• B. \(\dfrac{12\sqrt2}{25}\)
• C. \(\dfrac{24\sqrt2}{25}\)
• D. \(24\sqrt2\) \bigskip

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.

Answer 1

The Correct Option is B

Concept: Factor the pair of lines, find vertices of the triangle formed with \(x+y=1\). The orthocenter of a right triangle is at the vertex with the right angle. Here, the two lines through the origin are perpendicular? Check slopes: \(2x-3y=0 \Rightarrow m=2/3\); \(3x-2y=0 \Rightarrow m=3/2\). Product = 1, not -1, so not right. Use intersection of altitudes.

Step 1: Factor: \(6x^2-13xy+6y^2=(2x-3y)(3x-2y)=0\). Lines L1: \(2x-3y=0\), L2: \(3x-2y=0\). Origin A(0,0). Intersect with \(x+y=1\): B = L1∩line = (3/5,2/5); C = L2∩line = (2/5,3/5).

Step 2: Find slopes. AC: from (0,0) to (2/5,3/5) has slope 3/2. Altitude from B perpendicular to AC: slope -2/3, equation \(y-2/5 = -2/3(x-3/5) \Rightarrow 10x+15y=12\). AB: slope 2/3. Altitude from C perpendicular to AB: slope -3/2, equation \(y-3/5 = -3/2(x-2/5) \Rightarrow 15x+10y=12\).

Step 3: Solve altitudes: subtract to get \(5x-5y=0 \Rightarrow x=y\). Then \(25x=12 \Rightarrow x=y=12/25\). Orthocenter H(12/25,12/25).

Step 4: Distance from origin O: \(\sqrt{(12/25)^2+(12/25)^2} = \frac{12\sqrt2}{25}\).

Step 5: Write the final answer. \(\boxed{\frac{12\sqrt2}{25}}\)