SOLUTION: I need help: I have the formulas I just do not know how to apply them correctly. Here is the problem it is 9 ? out of : given points p(-1,2) and n(5,-7)...find (1) the midpoint.

Question 14818: I need help: I have the formulas I just do not know how to apply them correctly. Here is the problem it is 9 ? out of : given points p(-1,2) and n(5,-7)...find (1) the midpoint. (2) the distance (3) the slope (4) slope of a line parallel to segment pn (5) slope of a line perpendicular to segment pn (6) slope-intercept form of the equation the line (7) point-slope form of the equeation of the line (8) standard form of the equation of the line (9) graph the points....Thanks
(1) midpoint --I can do...-1+5/2= 4/2 =2 2+-7/2= -5/2= -5/2 (2, -5/2)
(2) distance-
(3) the slope y2-y1/x2-x1= -7-2/5- (-1)= -9/6=-3/2
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
I need help: I have the formulas I just do not know how to apply them correctly. Here is the problem it is 9 ? out of : given points p(-1,2) and n(5,-7)...find (1) the midpoint. (2) the distance (3) the slope (4) slope of a line parallel to segment pn (5) slope of a line perpendicular to segment pn (6) slope-intercept form of the equation the line (7) point-slope form of the equeation of the line (8) standard form of the equation of the line (9) graph the points....Thanks
(1) midpoint --I can do...-1+5/2= 4/2 =2 2+-7/2= -5/2= -5/2 (2, -5/2)
(2) distance-
(3) the slope y2-y1/x2-x1= -7-2/5- (-1)= -9/6=-3/2
Good You know the formulae and did fairly well..let us standardise and recount all the formulae for the above questions
Let the 2 points as you called P and Q say be represented by their coordinates
(x1,y1) and (x2,y2) respectively . So in your example we have x1=-1 ; y1=2 ;
x2=5 ; and y2 = -7...you did substitutions in the formula correctly for the midpoint and slope . hence ,knowing the fomulae for the other questions , you should be able to do the problem your self by substituting for x1,y1,x2 and y2.
mid point is ...(x1+x2)/2 ; and (y1+y2)/2..you did this ok.=(2,-5/2)
distance is .... =square root of 117
slope is ........(y2-y1)/(x2-x1)...you did this ok .=-3/2
slope of a line parallel to the above segment ....same as above since parallel lines have identical slopes =-3/2
slope of a line perpendicular to the above line ....-(x2-x1)/(y2-y1) since the rule is that product of slopes of 2 perpendicular lines should be equal to -1. answer is 2/3
slope intercept form equation of the line
Y=X*((y2-y1)/(x2-x1))+[y1-x1*((y2-y1)/(x2-x1)))answer is Y=(-3/2)x+(1/2)
where ((y2-y1)/(x2-x1))is the slope and (y1-x1*((y2-y1)/(x2-x1)))is the intercept.
point slope form is Y-y1=(X-x1)*((y2-y1)/(x2-x1))answer is Y-2=(-3/2)*(X+1)
standard form of eqn. is X*((y2-y1)/(x2-x1))-Y+(y1-x1*((y2-y1)/(x2-x1)))=0
answer is (-3/2)*X - Y +(1/2)=0 ..or multiplying with -2 throughout to remove 2 in the denominator we get 3X+Y-1=0
to plot the graph take the last equation 3X+Y-1=0 , give different values for X , say 0,2 and 4 (2 or 3 would do as it is a straight line),find corresponding values of Y from this equation and plot to a suitable scale.

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