SOLUTION: 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

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Question 1103837: 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
(b) the coordinates of C
(c) the area of triangle ABC

Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
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
===================
(b) the coordinates of C
Find the x-intercept of line BC
BC is 3y + 4x = 9
4x = 9
x = 9/4 --> C(9/4,0)
===========================
(c) the area of triangle ABC
---
 A    B    C    A
 4    0   9/4   4
 0    3    0    0

---
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

Answer by ikleyn(52877)   (Show Source): You can put this solution on YOUR website!
.
From the other tutor response, you just know that the coordinates of the point C are C = (9/4,0).


Now notice that the triangle ABC has the side AC lying at the x-axis.


Its length is   -  =  -  = .


The altitude drawn from the point B to the side AC is 3 units long.


So, the area of the triangle ABC is   =   square units.


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