Section Check In – 1.03 Coordinate GeometryQuestions

1. Determine the equation of the straight line through the points and . Give your

answer in the form .

2. Determine the centre and radius of the circle with equation .

3.* The parametric equations of a curve are . Determine the

coordinates of the points where the straight line meets the curve.

4.* The parametric equations of a curve are . Use an appropriate trigonometric identity to determine the cartesian equation of the curve.

5. The coordinates of three points are and .

(a) Show that is perpendicular to .

(b) Explain why the mid-point of is the centre of the circle through , and .

(c) Determine the equation of the circle through , and .

6.* The parametric equations of a curve are .

(a) Find a cartesian equation of the curve.

(b) Explain why the curve exists only for , and sketch the curve.

7. The point has coordinates and the point has coordinates . The gradient

of is and the distance is . Find the possible values of and .

8. A circle with centre at the origin has radius . A straight line has equation and meets the circle at two distinct points. Find the possible values of .

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9. Alan is conducting an experiment involving the variables and . The measurements are given in the following table.

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Alan suspects that the variables are linked by an equation of the form . By plotting

values of against values of , comment fully on Alan’s suspicions and, if appropriate, suggest values for the constants and .

10.* A point is moving on a screen and its position, referred to and axes at time (where ) is given by the parametric equations

.

(a) Find the value of when and show that, at this instant, the point is moving parallel to the -axis.

(b) Show that there is no instant when the point is moving parallel to the line .

Extension

Points and have coordinates and respectively.

(a) The point moves in the plane so that its distance from is twice its distance from . Find the equation of the path followed by and determine what sort of curve this equation represents.

(b) More generally, investigate the situation if the distance of from is times its distance from .

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Worked solutions

1. Gradient of line is

Equation of line is or

Multiplying both sides by 2 and rearranging to required form, equation is

2. Equation is

Completing the square twice,

Simplifying, equation is

Hence centre is and radius is

3. Substituting expressions for and into equation , curve and line meet where

Simplifying gives equation

Factorising and solving, and so or

Substituting in the parametric equations: when and

when and

Curve and line meet at the points and

4. Rearranging the parametric equations, and

Substituting into the identity gives

Multiplying both sides by , cartesian equation is

5. (a) Gradient of is ; gradient of is

Product of gradients = and so lines are perpendicular.

(b) Since angle , the angle in a semicircle property means that is diameter of circle through . Mid-point of the diameter is the centre of the circle.

(c) Mid-point of is

Radius

Equation of circle is

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6. (a) From the first equation,

Substituting in the second equation,

Simplifying,

Cartesian equation is or, in completed square form,

(b) For all values of the parameter, and so , i.e.

From cartesian equation, curve has minimum point at but note that the sketch does not involve the whole parabola, just the part for which

7. Gradient of , giving the equation and so Simplifying,

Distance , giving the equation

Expanding and simplifying, From first equation,

Substituting in second equation,

Simplifying, and, factorising, Solving, or , giving or

Possible values are and

8. Equation of circle is

Line meets circle where

Expanding and simplifying,

For two distinct roots, discriminant , giving

Simplifying, and so Possible values of are those for which

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9. Forming table showing values of and ,

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If Alan is correct, plotting these points should show the points lying on a straight line (more or less, allowing for experimental inaccuracies)

Checking the graph, the points (except for one) do lie close to a straight line

Ignoring the point with the assumption that there was an error made with this reading

For , value of is gradient of line in graph

Line through and has gradient Substituting values gives Ignoring the one apparently wrong reading, Alan’s suspicions are justified and is given

approximately by , rounding the two constants to 2 significant figures

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10. (a) Putting Solving, (ignoring the root since )

When

To consider direction, need to consider

Differentiating, and and therefore

When , , i.e. the gradient of path taken by point is ‘infinite’ and so the point is moving parallel to the axis at this instant

(b) Gradient of line is

If point is moving parallel to this line, gradient of curve is , i.e. Equation is and there is no possible value of satisfying this, i.e. at no instant is the point travelling parallel to the line

Extension

(a) gives

This leads to the equation

Expanding and simplifying, and hence

Equation represents a circle with centre and radius 4

(b) Now leading to

Expanding, simplifying and dividing by leads to

Completing the square and simplifying,

Equation represents a circle with centre and radius

Check that substituting agrees with your answers to part (a)What happens if ?

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