[3 marks)
9. find the value of k, if the points a(2,3) b(4, k) & c(6,3) are collinear.
10. find the perimeter of a triangle whose vertices have the coordinates
i. (3, 10) (5,2) & (4, 12)
ii. (-2, 1) (4,6) & (6,3)
11. find the area of a triangle whose vertices are
i. (1, 1), (-4,6) & (-3,-5)
ii. a (2,3), b(4,4) & c(2,6)
iii. (2,-2), (-2,1) & (5,2)

9. We would normally use point-point form to determine three points on a line.  Here A and C have the same y coordinate, so we skip all that and just write K=3.

10. Perimeters given coordinates can get messy because of the square roots

i. A(3,10), B(5,2), C(4,12)

AB² = (5-3)² + (2-10)² = 68 = 2²·17

BC²= (4-5)² + (12-2)² = 101

AC² = (4-3)² + (12-10)² = 5

Perimeter: 2√17 + √101 + √5

ii. A(-2,1), B(4,6), C(6,3)

AB² = (4 - - 2)² + (6-1)² = 61

BC²= (6-4)² + (3-6)² = 13

AC² = (6 - -2)² + (3-1)² = 68

Perimeter: √61 + √13 + 2√17

I'll leave any calculator approximating to you; I prefer exact answers.

11.

When the triangles have a common x or y coordinate we can just do half base times height. The first and the third question here don't work that way.  We can use the Shoelace Formula which says the signed area is half the sum of the cross products of the sides, true for any polygon.

Actually let's translate the first vertex to the origin because the area of triangle with vertices (0,0), (a,b), (c,d) is (1/2)(ad - bc)

i. Translating (1,1), (-4,6), (-3,-5) becomes (0,0), (-5,5), (-4,-6) so area

(1/2)( -5(-6) - 5(-4) ) = (1/2)(50) = 25

25

ii.  We have common coordinate, one side parallel to an axis, which makes things easier. We have a base along x=2 of length b=6-3=3.  There's an altitude h=4 - 2 = 2, so area is (1/2)(3)(2) = 3.

iii. (2,-2), (-2,1) & (5,2)

No common coordinate.  Let's try the full shoelace:

Δ = (1/2)( 2(1) - -2(-2) + (-2)(2) - 1(5) + 5(-2) - 2(2) ) = -25/2

The Shoelace Formula gives signed area, which may be negative.  The area is the absolute value:  25/2

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.