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Question 548115: A line with slope equal to 1 and a line with slope equal to 2 intersect at the point P(4,6) as shown. Points Q and R are both on the x-axis. What is the area of triangle PQR
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! slope = 1 and passes through (4,6)
The equation of this line
m= 1
Plug value of the slope and point ( 4,6) in
Y = m x + b
6.00 = 4 + b
b= 6 - 4
b= 2
So the equation will be
Y = 1 x + 2
The x intercept is Q(-2,0))
slope 2 passing through (4,6)
------------
m= 2
Plug value of the slope and point (4, 6) in
Y = m x + b
6.00 = 8 + b
b= 6 - 8
b= -2
So the equation will be
Y = 2x-2
x intercept = (1,0)
The intersection of these two lines have to be calculated.
-1 x -1 y = 2 .............1
2 x -1 y = 2 .............2
Eliminate y
multiply (1)by -1
Multiply (2) by 1
1 x 1 y = -2
2 x + -1 y = 2
Add the two equations
3 x = 0.00
/ 3
x = 0
plug value of x in (1)
-1 x + -1 y = 2
0 + -1 y = 2
-1 y = 2 + 0
-1 y = 2
y = -2
P(0,-2) Q(-2,0),R(1,0)
Area of triangle with vertices p(x1,y1)Q(x2,y2)&R(x3,y3)=
1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
the points are
1/2[0+0+1(2)]
=1/2*2
=1 sq.unit
m.ananth@hotmail.ca
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