Coordinate and Circle Geo Q's (1 Viewer)

[hon]

I'm stuck with the following q's

- The points X (3,1) , Y (9,1) and Z(5,5) are the vertices of a triangle

1. Find the area of the triangle XYZ
2. Find the coordinates of G, the point that divides ZY in the ratio 1:3
3. Write down the equation of the line, m, thgough G parallel to XY.
4. The line m, cuts XZ at H. Find the area of triangle ZGH.

- An arc of length, L cm extends an agnel of a "theta" radians at the centre of a circle radius "r" units. If L = 8 cm and r = 4 cm, find the area of the sector.
Find the length of the chord cut off by the two radii bouding the sector to 3 sig figs.

With this q, the sector appears to be a full circle?? I'm confused about the angle of "theta".

-Find the equation of the tangent to y = 2x^2 - 4x at the point x = 1.

Do you approach this q by finding subbing 1 into the equation (y = -2. Then find y' and use the form y-y1 = m (x-x1)?? I'm just confused if u do that, the equation fo the tangent has a x^2 to it.. Which isn't a tangent?? (If you kinda get what im saying)



Thanks in advance

 

[jet]

Looking at the diagram, the area = 0.5 x base x perpendicular height
= 0.5 x XY x BZ
= 0.5 x 6 x 4
= 12 units²

For the ratio question, you use the formula x = (mx2 + nx1)/(m+n), y = (my2+ny1)/(m+n), assuming that it is an internal division.
Therefore for the point G,
x = (5 + 3(9))/(4) = 8
y = (5 + 3)/(4) = 2
therefore G = (8,2)

Looking at the diagram, we can see that XY is horizontal, hence, the gradient is zero.
Therefore y - 2 = 0(x - 8)
y = 2

The line XZ has gradient (4)(2) = 2
Hence, y - 1 = 2(x-3)
y = 2x-5
Solving the two lines at the same time, 2 = 2x-5, 2x = 7x = 7/2 = 3.5

The length of the arc is given by L = r x theta
hence, theta = 8/4 = 2 radians.
The area of the sector is given by A = 1/2r² x theta
= 0.5 x 16 x 2
= 16 cm²

For the last question, find the derivative first, y' = 4x - 4.
At x = 1, y' = 0, y = -2, hence, the line has eqn y + 2 = 0(x - 1)
y = -2 is the tangent.