maths
posted by sana .
If the point A(1,2), B(2,3), C(3,2), and D(4,3) are vertices parallelogram ABCD, then taking AB as the base , find the height of this parallelogram

I would take the line DA, write it as y=mx+b, then find the vertical line from it to B. The distance of that is the height.
sketch it so you understand.
a. slope DA=(ydya)/(xdxa)=(3+2)/(41)= 1/5
Now using that, and the point D, the equation of DA can be found:
y=mx+b
3=1/5 (4)+b or b=3+4/5= you do it.
Now, the line perpendicular to DA going through the point C is what we are going after. Slope of that line is 5.
y=5x+b but you know point C is in it
so, 2=5(2)+b so b=12
and then this means the equation of the altitude is y=5x+12
Now we need the distance of the altitude. Where does it intersect DA?
y=5x+12 and
y=1/5 x 3+4/5
solve for x and y, that is the point of intersection.
Now use the distance formula to find the distance between C and the intersection point, that is the altitude. 
This question was posted earlier, and I had suggested that there might be a typo, since none of the line segments are parallel
http://www.jiskha.com/display.cgi?id=1362226321#1362226321.1362230628
I see it posted exactly the same way, but ....
slope AB = 1/3
slope BC = 1/5
slope CD = 5/1 = 5
slope DA = 5/3
Once you have found your typo and corrected it, follow bobpursley's method.
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