Question 872791
Write the equation of the line which passes through the point of intersection of the lines x-2y-4 = 0 and 4x-y-4 = 0  and satisfying the additional conditions
a) parallel to the line 16x -11y +3 =0
b) perpendicular to the line 9x + 22y - 8 =0 



Solve x-2y-4 = 0 and 4x-y-4 = 0

to get (x,y)

1	x		-2	y	=	4	.............1	
								
4	x		-1	y	=	4	.............2	
Eliminate	y							
multiply (1)by		1						
Multiply (2) by		-2						
1	x		-2	y	=	4		
-8	x	+	2	y	=	-8		
Add the two equations								
-7	x				=	-4		
/	-7							
x	=	1						
plug value of			x	in (1)				
1	x		-2	y	=	4		
1			-2	y	=	4		
			-2	y	=	4		-1
			-2	y	=	3		
				y	=	-2		


(1,-2) is the point of intersection

16    	x	 	-11	y  	=	-3	
Find the slope of this line							
make y the subject							
-11	y  	=	-16    	x		-3	
Divide by	-11						
	y  	=	1 4/9	x	+	 2/7	
Compare this equation with y=mx+b							
slope m =	1 4/9						
The slope of a line parallel to the above line will be the same							
The slope of the required line will be			1 4/9				
m=	1 4/9	,point	(	1      	,	-2	)
Find b by plugging the values of m & the point in							
y=mx+b							
-2	=	16/11	+	b			
b=	-3.45						
m=	13/9						
Plug value of  the slope  and b					in y = mx +b		
The required equation is		y  	=	13/9	x		-38/11


for perpendicular take the negative reciprocal of the slope of given line 9x + 22y - 8 =0
and point is (1,-2)
frame the equation