<PRE><B>Question 1103837</B>
A,B and C are vertices of a triangle. Given that A(4,0),B(0,3),C lies on the x-asis and that the equation of BC is 3y + 4x = 9, find

(a) the equation of AB in the form y =mx+c
3x + 4y = 3*4 = 12
y = (-3/4)x + 3
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(b) the coordinates of C
Find the x-intercept of line BC
BC is 3y + 4x = 9
4x = 9
x = 9/4 --&gt; C(9/4,0)
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(c) the area of triangle ABC 
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&lt;pre&gt;
 A    B    C    A
 4    0   9/4   4
 0    3    0    0
&lt;/PRE&gt;
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Add the diagonal products starting upper left:
4*3 + 0 + 0 = 12
Add the diagonal products starting lower left:
0 = 27/4 = 0 = 27/4
the difference 12 - 27/4 = 21/4
Area is 1/2 that
Area = 27/8 sq units

</PRE>