Question 548115
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