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Step-by-Step Solution
Step 1: Identify the Given Sides and Orthocenter
We have two sides of a triangle given by the lines
3x − 2y + 6 = 0 (call this L₁),
4x + 5y − 20 = 0 (call this L₂),
and the orthocenter of the triangle at (1, 1).
We seek the equation of the third side, say L₃.
Step 2: Label the Intersection of L₁ and L₂
• The intersection of L₁: 3x − 2y + 6 = 0 and L₂: 4x + 5y − 20 = 0 gives one vertex of the triangle (call it A).
• Solve simultaneously:
From L₁: 3x − 2y + 6 = 0 ⇒ −2y = −3x − 6 ⇒ y = (3x + 6)/2.
From L₂: 4x + 5y − 20 = 0 ⇒ 5y = 20 − 4x ⇒ y = (20 − 4x)/5.
Equating: (3x + 6)/2 = (20 − 4x)/5.
Cross-multiply: 5(3x + 6) = 2(20 − 4x).
15x + 30 = 40 − 8x.
15x + 8x = 40 − 30.
23x = 10 ⇒ x = 10/23.
Then y = (3×(10/23) + 6)/2 = (30/23 + 6)/2 = (30/23 + 138/23)/2 = 168/23 ÷ 2 = 84/23.
So A = (10/23, 84/23).
Step 3: Recognize Which Side Is Opposite the Found Vertex
The line sought, L₃, is the side opposite the vertex A. In a triangle, the altitude from A must be perpendicular to the side opposite A (i.e., L₃) and must pass through the orthocenter (1, 1).
Step 4: Find the Slope of the Altitude from A to L₃
The altitude from A to L₃ passes through A = (10/23, 84/23) and the orthocenter H = (1, 1).
Slope of AH = (1 − 84/23) / (1 − 10/23) = [ (23/23) − (84/23 ) ] / [ (23/23) − (10/23 ) ] = (−61/23) / (13/23) = −61/13.
Since L₃ is perpendicular to this altitude, the slope of L₃ is the negative reciprocal of −61/13, which is 13/61.
Step 5: Write the General Form of L₃
A line with slope 13/61 can be written as
y = (13/61)x + c,
or equivalently
13x − 61y + (some constant) = 0.
Let us denote this constant by C:
13x − 61y + C = 0.
Step 6: Impose the Perpendicularity Condition from the Other Vertex
To determine C, note that another vertex occurs where L₃ meets L₁ or L₂. Let us pick the intersection of L₃ and L₁ as vertex C. Then the altitude from this vertex C must be perpendicular to L₂ and pass through the orthocenter (1, 1).
1) Intersection of L₁: 3x − 2y + 6 = 0 and L₃: 13x − 61y + C = 0 gives a point (call it C).
2) The slope of L₂ is −4/5, so its perpendicular slope is 5/4. The line from C to (1, 1) must therefore have slope 5/4.
3) This condition fixes C. Solving leads to C = −1675/2 in the equation 13x − 61y + C = 0.
Step 7: Final Equation of the Third Side
Substituting C = −1675/2 and multiplying throughout by 2 to clear the fraction:
2(13x) − 2(61y) + 2 × (−1675/2) = 0 ⇒
26x − 122y − 1675 = 0.
Hence, the required third side is
26x − 122y − 1675 = 0.
Answer
26x − 122y − 1675 = 0
