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=(yd-ya)/(xd-xa)=(-3+2)/(-4-1)= 1/5
Now using that, and the point D, the equation of DA can be found:
-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=-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
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.