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What is the equation of line b d, simplified? y − y1 = m(x − x1) y − 0 = (startfraction 2 b over 2 a minus c endfraction)(x − c) y = (startfraction b over a minus c endfraction)x − (startfraction 2 b c over 2 a minus c endfraction) y = (startfraction 2 b over 2 a minus c endfraction)x − (startfraction b c over 2 a minus 2 c endfraction) y = (startfraction 2 b over 2 a minus c endfraction)x − (startfraction 2 b c over 2 a minus c endfraction) y = (startfraction 2 b over a minus c endfraction)x − (startfraction 2 b c over a minus c endfraction)

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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