Question

Find the intersection of those two lines r=(2,9)+s(1,1) and r= (-3,2) + t(1,2) and then find the area of the triangle whose vertices are the intersection point, s = 0 on L1 and t = 3 on L2.

I already found the point of intersection. I just need to know how to go about solving for the area of the triangle whose vertices are the intersection point, s=0 on L1 and t=3 on L2.

Answer 1

When s = 0 , we get the point (2,9)
when t=3, we get the point (0,8)
Using your intersection point that you already found, find the lengths of the 3 line segments using the "distance between two points" formula.
To find the area, use Heron's Formula, which says
Area = √(s(s-a)(s-b)(s-c)) where s = 1/2 the perimeter, and a, b, and c are the lengths of the three sides.

Heron's Formula:
http://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php

Happy calculating !

Answer 2

By "find the length of the 3 line segments" you're referring to the points on each side of the triangle which is (2,9), (0,8) and (-1,6) (intersecting point)?

Answer 3

yes, those three point define three line segments.