Question 570882
Find an equation of the line containing the given pair of points.

(1/4, -1/2) and (3/4, 6)
x1		y1	x2	y2				
  1/ 4		  1/ 2	  3/ 4	6				
								
slope m =		(y2-y1)/(x2-x1)						
(	6	-	 1/2	)/(	 3/4	-	 1/4	) 
(	5.5	/	0.5	)  				
m=		11      						 
								
Plug value of  the slope  and point (	 1/4	,	 1/2	) in	
Y 	=	m	x 	+	b			
1/2	=	11/4	+	b				
b=	0.5	-	11/4					
b=	-9/ 4							
So the equation  will be								
Y 	=	11    	x 		-9/4			

What is the equation of the line? y=

second question
Find an equation of the line having the given slope and containing the given point.
m=6/7, (1, -7)

m=		  6/ 7				
						
Plug value of  the slope  and point (	1,	-7
Y 	=	m	x 	+	b	
-7.00	=	 6/7	+	b		
b=	-7	-	 6/7			
b=	-55/ 7					
So the equation  will be						
Y 	=	 6/7	x 		-55/7	

The equation of the line is Y=

Third question
Find an equation of the line containing the given pair of points.
(-1, -2) and (-4, -6)

x1		y1	x2	y2			
-1      		-2      	-4      	-6			
							
slope m =		(y2-y1)/(x2-x1)					
(	-6	-	-2    	)/(	-4    	-	-1    
(	-4	/	-3	)  			
m=		4/ 3					
							
Plug value of  the slope  and point (	-1    	,	-2    	) in
Y 	=	m	x 	+	b		
-2.00	=	-4/3	+	b			
b=	-2	-	-4/3				
b=	-  2/ 3						
So the equation  will be							
Y 	=	4/3	x 		- 2/3		

The equation of the line in slope-intercept from is y=

Fourth Question
Write an equation of the line containing the given point and parallel to the given line. Express your answer in the form y=mx+b
(-4, 7); 4x=3y+2
4	x	+	-3	y  	=	2	
Find the slope of this line 							
make y the subject							
-3	y  	=	-4	x	+	2	
Divide by 	-3						
	y  	=	4/ 3	x		- 2/3	
Compare this equation with y=mx+b							
slope m =	4/3						
The slope of a line parallel to the above line will be the same							
The slope of the required line will be 			4/3				
m=	4/3	,point	(	-4	,	7	)
Find b by plugging the values of m & the point in							
y=mx+b							
7	=	-16/3	+	b			
b=	37/3						
m=	4/3						
Plug value of  the slope  and b		in y = mx +b		
The required equation is y  	=	4/3	x	+	37/3



The equation of the line is y=

Fifth question
Write an equation of the line containing the given point and parallel to the given line. Express your answer in the form y=mx+b
(5, 7); 5x+y=4
5	x	+	1	y  	=	4x+5y=2	
Find the slope of this line 							
make y the subject							
1	y  	=	-5	x	+	4	
Divide by 	1						
	y  	=	-5      	x	+	4    	
Compare this equation with y=mx+b							
slope m =	-5    						
The slope of a line parallel to the above line will be the same							
The slope of the required line will be 			-5    				
m=	-5    	,point	(	5	,	7	)
Find b by plugging the values of m & the point in							
y=mx+b							
7	=	-25.00	+	b			
b=	32						
m=	-5    						
Plug value of  the slope  and b		in y = mx +b		
The required equation is y 	=	-5    	x	+	32    

the equation of the line is y=

Sixth question
Write an equation of the line containing the given point and perpendicular to the given line.
(5, -4); 4x+3y=8
4      	x	+	3	y  	=	8	
Find the slope of this line 							
							
3	y  	=	-4    	x	+	8	
Divide by 	3						
	y  	=	-1  1/ 3	x	+	2.67	
Compare this equation with y=mx+b							
slope m =	-4/ 3						
							
The slope of a line perpendicular to the above line will be the negative reciprocal	 3/4						
m1*m2=-1							
							
							
m=	 3/4	,point	(	5	,	-4	)
Find b by plugging the values of m & the point in							
y=mx+b							
-4	=	15/4	+	b			
b=	-31/4						
m=	 3/4						
The required equation is		y  	=	  3/ 4	x		-31/ 4


The equation of the line is y=

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