CHAPTER 7 SOLUTIONS TO REINFORCEMENT EXERCISES IN COORDINATE GEOMETRY
7.3.1 Coordinate systems in a plane
7.3.1A.
Plot the points
i) (– 1, – 1)ii)(3, 2)iii)(– 2, 3)iv) (0, 4)v) (4, 0)
vi) (1, 1) vii) (3, – 1) viii) (0, – 2)
Solution
The figure shows the points plotted on a two-dimensional cartesian coordinate system with axes x horizontal and y vertical. So, for example, the point (– 1, – 1) is placed at one unit in each of the negative x and y directions
7.3.1B.
Plot the points with polar coordinates
i)(, )ii)(1, /3)iii)(1,120o) iv)(2, – 60o)
v) (2, /2)vi)(3, ) vii) (2, ) viii) (3, – 150)
Solution
The figure shows the points plotted on a polar coordinate system. For example (, ) is plotted at a distance from the origin or pole along the direction making angle in an anti-clockwise direction from the x-axis or initial line.
7.3.2Distance between two points
Find the distance between each pair of points
i)(– 1, 1), (2, 0)ii)(1, 0), (1, – 2)
iii)(2, 2) , (3, 3)iv)(0, 1), (0, 3)
v)(2, 1), (3, 1)vi)(– 2, – 1), (– 1, – 2)
Solution
The distance between the two points (x1, y1) , (x2, y2) is
d =
i)d((– 1, – 1), (2, 0)) = =
ii)d((1, 0), (1, – 1)) = = = 2
In this last example we could use
d((a, y1), (a, y2)) = |y2 – y1 | = |0 – (– 1) | = 1
iii)d((2, 2), (3, 3)) = =
iv) d((0, 1), (0, 3)) = |3 – 1| = 2
v)d((2, 1), (3, 1)) = |2 – 3| = 1
vi)d((– 2, – 1), (– 1, – 2)) = =
7.3.3 Midpoint and gradient of a line
7.3.3A.
Find the midpoint of the line segments joining the pairs of points in exercise 7.3.2
Solution
Given two points (x1, y1) , (x2, y2) the midpoint of the line segment between them is at the point
i)Applying this to the pair of points (– 1, 1), (2, 0), gives for the midpoint
=
ii)For (1, 0), (1, – 2) we get
= (1, – 1)
iii)For (2, 2) , (3, 3) we get
=
iv)For (0, 1), (0, 3) we get
= (0 , 2)
v)For (2, 1), (3, 1) we get
=
vi)For (– 2, – 1), (– 1, – 2) we get
=
7.3.4B
Ditto for the gradients.
Solution
The gradient of the line segment joining the points (x1, y1), (x2, y2) is
m =
i) For the points (– 1, 1), (2, 0) we get m = = –
ii) If we strictly apply the formula in the case of the points (1, 0), (1, – 2) we get a 'gradient'
m = =
which of course is not defined. This is the case where the gradient is 'infinite', and the line joining the two points is in fact vertical.
iii) For the points (2, 2) , (3, 3) we get m = = 1
Another, picturesque, way of thinking about this one is that it corresponds to a line that increases in x and y values by equal amounts.
iv) For the points (0, 1), (0, 3) we again get a vertical line with 'infinite' gradient.
v) For the points (2, 1), (3, 1) we get m = = 0
In this case the line through the two points is horizontal - they have the same y coordinate. So the gradient of the line is zero.
vi) For the points (– 2, – 1), (– 1, – 2) we get m = = – 1, a line sloping down from left to right (a negative gradient) at 45 to x-axis.
7.3.4C
Ditto for the points dividing each line segment in the ratio 2:3.
Solution
The point that divides the line joining (x1, y1) to (x2, y2) in the ratio µ : is
For µ : = 2 : 3 the coordinates will be
=
i) For (– 1, 1), (2, 0) the coordinates will be
= =
ii) For (1, 0), (1, – 2) the coordinates will be
=
iii) For (2, 2) , (3, 3) the coordinates will be
=
iv) For (0, 1), (0, 3) the coordinates will be
=
v) For (2, 1), (3, 1) the coordinates will be
=
vi) For (– 2, – 1), (– 1, – 2) the coordinates will be
=
7.3.4Equation of a straight line
7.3.4A.
Find the equations of the straight lines through each pair of points in Q7.3.2
Solution
The equation of the straight line through the points (x1, y1), (x2, y2) is
y – y1 = = m(x – x1 )
where m is the gradient. We can choose either of the given points for (x1, y1). We could use the results of Q7.3.3B for the gradient, but we may as well simply use the form
y – y1 =
directly.
i) For the points (– 1, 1), (2, 0) we have
y – 1 = = – 3(x + 1) = – 3x – 3
which on rearrangement gives
3x + y + 2 = 0
ii) For the points (1, 0), (1, – 2) using the formula above will give problems because in this case x2 – x1 = 0. The line is of course vertical and the gradient 'infinite' - x is always 1. Thus in this case the equation of the line is
x = 1
iii) In the case of (2, 2) , (3, 3) you can either use the general formula given above, or you can simply note that the line through these points is characterized by the equation
y = x
iv) For (0, 1), (0, 3) we again have a vertical line given by
x = 0
v) In the case of (2, 1), (3, 1) it is y that is always constant, the line is horizontal and the gradient is zero. The equation of this line is therefore
y = 1
vi) For (– 2, – 1), (– 1, – 2) we use the formula given above, taking due care with the signs. We get
y – (– 1) = = – 1(x + 2) = – x – 2
which rearranges to
x + y + 3 = 0
7.3.4B.
Find the gradients and intercepts of the following lines
i)2x – 3y = 4ii)4x + 2y = 7iii)x – y + 1 = 0
Solution
The easiest way is to rewrite the lines in the standard form y = mx + c and simply read off the gradient m and (y) intercept c directly.
i) Rewriting 2x – 3y = 4 as
y = x –
the gradient is and the y -intercept is – . The x-intercept is found by putting y = 0 and so is x = 2.
ii) 4x + 2y = 7 can be rewritten y = – 2x, so m = – 2 and c =
iii) x – y + 1 = 0 is equivalent to y = x + 1, revealing that the gradient and intercept are both 1
7.3.4C.
Show that the the equations
= = m
may be rearranged to give the equation of the straight line as
y – y1 =
or
y – y1 = m(x – x1)
Also show that it may be arranged into the 'symmetric form'
y = y2 + y1
Solution
The first two results are almost immediate from the first equation. The last result is less obvious. First notice that when we write the equation of a line in the form
=
we can equally well use x2, y2 on the LHS, and we can write the line as
=
Cross multiplying gives
(y – y1)(x – x2) = (y – y2)(x – x1)
or
y (x – x2) – y1(x – x2) = y (x – x1) – y2(x – x1)
Rearranging:
y (x1 – x2) = y1(x – x2) – y2(x – x1)
Now dividing through and changing some signs round gives the required form
y = y2 + y1
7.3.5Parallel and perpendicular lines
For the lines in 7.3.4 B determine
a) lines parallel to each of them through the origin.
b) lines perpendicular to each of them through the point (– 1, 1).
Solution
a) Parallel lines have the same gradient. In the form that the lines are given all we need to do to obtain a parallel line through the origin is to put the RHS equal to zero.
i) 2x – 3y = 0 has the same gradient as 2x – 3y = 4 but it passes through the origin. The answer in the book was intended for another question and is incorrect.
ii)For 4x + 2y = 7 the required line is 4x + 2y = 0
iii)x – y = 0 is the required equation in this case
b) If the gradient of a given line is m, then any line perpendicular to it has a gradient – , and such a line will have an equation of the form
y = – x + c
where c can be found given a point that the line passes through.
i) 2x – 3y = 4 has gradient m = and so any line perpendicular to it has the form
y = – x + c
If the line passes through the point (– 1, 1) then
1 = – + c
from which c = – and so the equation is
y = – x –
which can be rewritten as
3x + 2y + 1 = 0
(again the answer given in the book is incorrect)
ii) 4x + 2y = 7 has gradient – 2 and so the required perpendicular line as the form
y = x + c
As this passes through (– 1, 1) we find c = and so the line is y = x + or
x – 2y + 3 = 0
iii) Perhaps you can see directly now that x + y = 0 is perpendicular to x – y = 1 and also passes through (– 1, 1)?
7.3.6Intersecting lines
Find all points where the following lines intersect
i)x + y = 1ii)2x + 2y = 3iii)x – y = 1
Solution
i) and ii) are parallel and therefore do not intersect anywhere except, as they say, 'at infinity'). i) and iii) intersect where
x + y = 1
x – y = 1
which gives the point (1,0).
ii) and iii) intersect where
2x + 2y = 3
x – y = 1
which have the solution x = and y = , so the point of intersection is ( , ).
7.3.7Equation of a circle
7.3.7A.
Write down the equations of the circles with centres and radii :-
i)(– 1, 1), 4ii)(2, – 1), 1iii)(4, 1), 2
Solution
The equation of the circle centre (a, b) with radius r is
(x – a)2 + (y – b)2 = r2
i) For centre (– 1, 1) and radius 4 we have the equation
(x – (– 1))2 + (y – 1)2 = 42
which simplifies to
x2 + y2 + 2x – 2y – 14 = 0
ii) For centre (2, – 1) radius 1 we get
(x – 2)2 + (y + 1)2 = 1
or
x2 + y2 – 4x + 2y + 4 = 0
iii) For centre (4, 1) radius 2 we get
(x – 4)2 + (y – 1)2 = 22 = 4
or
x2 + y2 – 8x – 2y + 13 = 0
7.3.7B.
Find the centre and radius of each of the circles
i)x2 + y2 – 2x – y = 4ii)x2 + y2 + 3x – 2y – 7 = 0
iii)x2 + y2 + y = 3
Solution
The best approach is to complete the square to obtain the standard form
(x – a)2 + (y – b)2 = r2
This is equivalent to remembering the 'f-g' formulae, but avoids having to remember this (hopefully, completing the square should now be straightforward to you).
i)For x2 + y2 – 2x – y = 4 completing the square gives
(x – 1)2 – 1 + 2 – 2 = 4
or
(x – 1)2 + 2 = 4 + 1 + =
So the centre is (1, ) and the radius is
ii)Completing the square for x2 + y2 + 3x – 2y – 7 = 0 gives
2 – + 2 – 1 – 7 = 0
or
2 + 2 = 8 + =
So the centre is (– , 1 ) and the radius is
iii)x2 + y2 + y = 3 becomes
x2 + 2 – = 3
or
x2 + 2 = 3 + =
So the centre is (0, – ) and the radius is
7.3.7C.
A circle has the equation x2 + y2 – 4y = 0. Find its centre and radius, and the equation of the tangents at the points (± , 2 + ), using only geometry and trig. Also find the point where the two tangents intersect.
Solution
x2 + y2 – 4y = 0 is equivalent to
x2 + (y – 2)2 = 4
so its radius is 2 and the centre is at (0, 2)
The points (± , 2 + ) are as shown in the figure
If we consider the radius from the point (0, 2) to the tangent point at (, 2 + ) we see that it is the hypotenuse of a right-angled 45 triangle (from the centre of the circle, as x increases by to the tangent, so y increases by the same amount). The gradient of this radius is therefore 1. Since the tangent at this point is perpendicular to the radius, its gradient must therefore be –1. So the gradient of the tangent through (, 2 + ) is –1 and, by a similar argument (or symmetry) that through (– , 2 + ) is 1. So the equations of these lines are as follows.
Through (, 2 + ):
y – 2 – = – (x – )
or
x + y – 2 – 2 = 0
Through (– , 2 + ):
y – 2 – = + 1 (x + )
or
x – y + 2 + 2= 0
From the gradients the tangent lines meet where the y-coordinate on the y-axis is equal to the positive value of x where the line cuts the x-axis, ie where y = 0, and therefore x = 2 + 2. So the lines intersect at
(0, 2 + 2 )
The same result can be obtained more laboriously by solving the two equations simultaneously. Note that the answer given in the book is incorrect, a 2 has gone missing!
7.3.7D.
A circle has the equation x2 + y2 – 2x = 0. Determine its centre and radius, and find the equation of the tangent at the point . Determine where this tangent cuts the axes.
Solution
x2 + y2 – 2x = (x – 1)2 + y2 – 1 = 0 shows that the centre of the circle is at (1, 0), ie on the x-axis, and the radius is 1. The figure illustrates the point , denoted P. With C the centre of the circle, and N the point (, 0) the triangle NPC is a 60-30 triangle, because = (see UEM176). If the tangent intersects the x-axis at M then, because MPN = 60, PMN = 30. Thus, the gradient of the tangent is tan 30 = . So, the equation of the tangent through is
y – =
which simplifies to
x – y + 1 = 0
This crosses the x-axis at y = 0 and x = – 1, ie at (– 1, 0) and the y-axis at x = 0 and y = , ie at ) .
7.3.8Parametric representation of curves
Eliminate the parameters in the following pairs of equations
i)x = 3 cos t , y = 3 sin t
ii)x = 1 + 2 cos , y = 3 – sin
iii)x = 2u2 , y = u – 2
iv) x = , y = 3t
v) x = cos2t, y = sin t
Solution
i) By squaring and adding x = 3 cos t , y = 3 sin t we obtain
x2 + y2 = 9 cos2 t + 9 sin2 t
= 9
which is a circle centre the origin and radius 3.
ii)For x = 1 + 2 cos , y = 3 – sin we simply solve for cos and sin and get
cos = , and sin = 3 – y
Now square and add to give:
cos2 + sin2 =2 + (3 – y)2 = 1
iii)For x = 2u2 and y = u – 2 we obtain from the second equation u = y + 2, which we can substitute in the first to get
x = 2(y + 2)2
iv) From x = and y = 3t we have
y = 3t = 3=
or
xy = 6
Or we can get this immediately by multiplying x and y in parametric form - the t cancels out.
v) We note that x = cos2t = 1 – 2 sin2 t (by the double angle formula for cos 2t) = 1 – 2y2, so the implicit form of the equation is
x = 1 – 2y2
- 1 -