Equations for Symmetric Lines in Triangle ABC

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The discussion revolves around finding the equations of the symmetric lines, or angle bisectors, of triangle ABC based on given equations. The symmetric lines are x + 4 = 0 and 4x + 7y + 5 = 0, while one side of the triangle is represented by 3x + 4y = 0. A participant identified point A as (-4, 3) and noted that the lines pass through the origin (0, 0). Clarification was sought on the terminology, confirming that "symmetric line" refers to the angle bisector. The conversation emphasizes the need to derive the remaining angle bisectors for the triangle.
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Homework Statement



The equations of the symmetric lines of two inner angles of triangle ABC are given:

x+4=0 and 4x+7y+5=0, and the equation of the line of one of the sides of the triangle is

given 3x+4y=0 (which goes through the points of the side where the symmetric lines are

going). Find the equations of the lines which are going among the other two sides of the

triangle.

Homework Equations




d=\frac{|Ax + By + C|}{|\sqrt{A^2+B^2}|}


The Attempt at a Solution



I found one of the dots A(-4,3). Also I find that the lines are going through the point (0,0)
 
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By "symmetric line" of an angle, do you mean the angle's bisector?
 
HallsofIvy said:
By "symmetric line" of an angle, do you mean the angle's bisector?

yes, sorry for mistranslation.
 
anybody know?
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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