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Sagot :
The required equation of line BD is option d.) [tex]y=\frac{2b}{2a-c} x-\frac{2b}{2a-c} c[/tex]
Equation of the line is given be in standard form is
[tex]y-y_{1} =\frac{y_2-y_1}{x_2-x_1} (x-x_1)[/tex]
What is a Line?
Line, curve of the shortest distance between two points.
The questions seem to be incomplete the question, the complete question could be
a)[tex]y=\frac{2b}{2a-c} (x-c)\\[/tex]
b)[tex]y=\frac{b}{-c} x-\frac{2b}{2a-c}[/tex]
c)[tex]y=\frac{2b}{2a-c} x-\frac{bc}{2a-2c}[/tex]
d)[tex]y=\frac{2b}{2a-c} x-\frac{2b}{2a-c} c[/tex]
e)[tex]y=\frac{2b}{a-c} x-\frac{2b}{a-c}[/tex]
Now all the Equation in option represents the triangle with coordinates A(0, 0), B (2a, 2b), C(2c,0) and D(c, 0).
For equation of line BD we have coordinates, (2a, 2b) and (c, 0).
putting these coordinates in the standard form of equation of line i.e.
[tex]y-y_{1} =\frac{y_2-y_1}{x_2-x_1} (x-x_1)[/tex]
⇒ [tex]y=\frac{2b}{2a-c} x-\frac{2b}{2a-c} c[/tex]
Thus the required equation of the line is [tex]y=\frac{2b}{2a-c} x-\frac{2b}{2a-c} c[/tex].
learn more about line here:
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