If the lines 4x-y+2=0 and x+2ky+1=0 are perpendicular to each other, determine the value of k

For a line in the general form,
,

so

and

we require that

therefore...

so

Find the slope of the following line. 4y=x+4

The slope is the coefficient of the x term when the equation is in slope-intrcept form. Therefore our slope is 1/4

The line thought the two points (x,-8) and (3,5) has a slope of -7/2. Find the missing coordinate.

so...


and then we write




x=47/7

Find the slope of a line perpendicular to each given line. y=-5/3x+3

the perpendicular slope is the negative reciprocal of the original so it is 3/5 

:)

Find the rate of change between the two points: (8, 1) and (4, 2) where x is in months and y is in inches.

Rate of change is computed using slope: 
m=(y2-y1)/(x2-x1)
so 
m=(2-1)/(4-8)=-1/4 
that means the rate is falling at 0.25 inches every month

Find the slope

Find the slope of the following line. 4y=x+4

can be rewritten as


so we see that the slope is m=1/4 (the slope is the coefficient of the e-term when the equation is in slope-intercept form)

:)

Find the value of x so that the line passing through (-2, 7) and (x, 2) has a slope of -1/3

 
and 
 

so 


 


 

(4,11) and (-2,1) perpendicular of the slope

(4,11) and (-2,1) 
leads to 

 
which is perpendicular to 

 


:)

Determine the value(s) of t if P(-3;7), Q(5;-9) and R (-2;t ) are co-linear.

collinear lines have the same slope so... 

 

which gives us t=5 



graphing is actually much easier, but relies on precision and accuracy 
:)

Write an equation of the line containing the specified point and parallel to the indicated line (-3, 6), y=7

horizontal lines have a slope of 0, so the y-coordinate determines the equation of the line containing (-3, 6): 

y=6 


:)

Calculate the slope of the line between the points (3, -4) and (3, 5) Please help. Thank you.

the x-coordinate does not change so slope is undefined for this pair of points 

:)

Equation of a line

Find the equation of the line passing through the points (3, -1) and (1, 2)

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