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Co-ordinate methods in geometry, some questions. (1 Viewer)
- Thread starter Shoom
- Start date
jet
Banned
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- Jan 4, 2007
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- HSC
- 2009
1) For the first one (cant be bothered) you can just find the length of each side using the distance formula, then, by finding the ratio of corresponding sides, the triangles should be similar, as all the ratios are equal.
2) I'm guessing that C is the centre of a circle. To find the radius, know that the tangent is perpendicular to the radius. Hence, using the line 3x - 4y - 36 = 0, and the point C(14, 4) we have,
r = |3(14) - 4(4) - 36|/sqrt(9 + 16)
=|42 - 16 - 36|/sqrt(25)
=10/5
=2
Hence the circle has raduis 2.
Area=πr^2
=π(2^2)
=4π
3) A median joins the centre of a side of a triangle to the opposite vertex.
So, M(AB)=(10/2,14/2)=E(5,7)
Hence, Gradient(CE)=(11)/(-5)
y - 7 = (-11/5)(x - 5)
5y - 35 = -11x + 55
11x + 5y -90 = 0
2) I'm guessing that C is the centre of a circle. To find the radius, know that the tangent is perpendicular to the radius. Hence, using the line 3x - 4y - 36 = 0, and the point C(14, 4) we have,
r = |3(14) - 4(4) - 36|/sqrt(9 + 16)
=|42 - 16 - 36|/sqrt(25)
=10/5
=2
Hence the circle has raduis 2.
Area=πr^2
=π(2^2)
=4π
3) A median joins the centre of a side of a triangle to the opposite vertex.
So, M(AB)=(10/2,14/2)=E(5,7)
Hence, Gradient(CE)=(11)/(-5)
y - 7 = (-11/5)(x - 5)
5y - 35 = -11x + 55
11x + 5y -90 = 0