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March 6, 2015

March 6, 2015

Posted by **sana** on Saturday, March 2, 2013 at 9:27am.

- maths -
**bobpursley**, Saturday, March 2, 2013 at 9:57amI 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:

y=mx+b

-3=1/5 (-4)+b or b=-3+4/5= you do it.

Now, the line perpendicular to DA going thru 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.

- maths -
**Reiny**, Saturday, March 2, 2013 at 10:44amThis 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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