# C1 QuestionsWatch

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#1
Heres a few questions I have

1) The points A(-1,-2), B(7,2) and C(k,4), where k is a constant, are the vertices of ∆ABC. Angle ABC is a right angle.

a. Find the gradient of AB - which I have found, it is -1/2

b. Calculate the value of k.

The other question is more general. Lets say I have 3 coordinates that join together to make a triangle. What would be the best way of finding the area of that triangle? The coordinates are A(0,-2), B(0,8) and C(4,4).

Thanks.
0
14 years ago
#2
(Original post by ntrik)
.....

The other question is more general. Lets say I have 3 coordinates that join together to make a triangle. What would be the best way of finding the area of that triangle? The coordinates are A(0,-2), B(0,8) and C(4,4).

Thanks.
Draw a sketch of the three points on a graph.
You will see that the points A and B lie on the y-axis, a distance of 10 apart.
The point C is a perpindicular distance of 4 from the line AB, the y-axis.
So area of triangle is half the base times the perpindicular height.

A = ½.10.4
A = 20
0
14 years ago
#3
(Original post by ntrik)
Heres a few questions I have

1) The points A(-1,-2), B(7,2) and C(k,4), where k is a constant, are the vertices of ∆ABC. Angle ABC is a right angle.

a. Find the gradient of AB - which I have found, it is -1/2

b. Calculate the value of k.

The other question is more general. Lets say I have 3 coordinates that join together to make a triangle. What would be the best way of finding the area of that triangle? The coordinates are A(0,-2), B(0,8) and C(4,4).

Thanks.
b)

For angle ABC to be 90 degrees, line AB must be perpendicular to BC.

gradient of AB = (2-(-2))/(7-(-1) = 4/8 = 1/2 not -1/2

-1/0.5 = -2 = gradient of BC

therefore;
(4-2)/(k-7) = -2
-2(k-7) = 2
-2k + 14 = 2
2k = 12
k = 6

To find area of ABC
A(0,-2), B(0,8) and C(4,4).

Length AB = 8-(-2) = 10 ignore x-coordinates as they are the same. E.g. they both lie on the y=-axis.

Height of ABC = 4

Area of triangle = 0.5*base*height = 0.5*10*4 = 20
0
#4
Thanks again people!
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