Question 1180511

Find the equation of the line which passes through point (1,1) and is perpendicular to line AB which passes through points A(1,-2) and B(5,6). 

Equation of AB

x1		y1	x2	y2				
1		-2	5	6				
								
slope m =		(y2-y1)/(x2-x1)						
(	6	-	-2	)/(	5	-	1	)
(	8	/	4	)  				
m=		2.00						 
								
Plug value of  the slope  and point			(	1	,	-2	) in	
Y	=	m	x	+	b			
-2.00	=	2    	+	b				
b=	-2.00	-	2    					
b=	-4      							
So the equation  of AB will be								
Y 	=	2      	x	-4    	

slope of the line = 2

The slope of a line perpendicular to the above line will be the negative reciprocal							
m1*m2=-1							
The slope of the required line will be			0.50				
							
m=	 1/2	,point	(	1	,	1	)
Find b by plugging the values of m & the point in	-0.875						
y=mx+b							
2	=	1.75	+	b			
b=	0.25						
m=	-0.875						
The required equation is		y  	=	-  7/8 	x+	  1/4