Make PQ the base. How will the height and the third vertex relate to each other?
B
Practice makes perfect
We have two points, P(- 2, 1) and Q(2,1), which together with a third point will define a triangle with the area 4 square units. Since P and Q will be two of the three points, the segment PQ will be one of the sides. Let's making a diagram and mark the points P and Q and the segment between them.
We calculate the area of a triangle using the formula
A=1/2bh.To find the area we need to know the triangle's base, b, and its height, h. We can pick any side we like as the base. Let's pick PQ. The points P and Q have the same y-coordinate and is therefore parallel to the x-axis. We can can calculate its length using the Ruler Postulate.
We now know that the triangle's base is 4 units. Since we know that our triangle is supposed to have the area 4 square units we can now calculate its height.
The height is perpendicular to the base. Since the base is horizontal the height must be vertical. The point which is the third vertex in the triangle must be 2 units higher up or lower down than the base. Let's mark points which gives us a height of 2 units.
The third vertex must lie on one of these lines to give us a triangle with the area 4 square units. There are four points given as alternatives for the position of the third vertex. They are
A: R(2,0)
B: S(- 2, - 1)
C: T(- 1, 0)
D: U(2,- 2)
Let's now mark the four points in the diagram and see if any of them lie on the lines we have drawn.
We see that one of the four points, S, sits right on one of the lines. The triangle Δ SPQ will therefore have the area 4 square units.
