1. ## More geometry

A and B are points of intersection of y=x^2-2x and y=2-3x
simultaneous equation solved = x^2 + x -2 = 0

a) Find the coordinates of A and B, given that the x coordinate of B is greater than that of A.

The line L is perpendicular to y=2-3x and passes through B.
b) Find an equation of L

The line M is parallel to y=5-3x and passes through A
c) Find an equation of M

The lime M crosses the y-axis at point P. The line L crosses the x-axis at point Q.
d) Find the distance of PQ, giving your answer in square root form where a and b are integers and b is prime. (Hint - use pythagoras' theorem)

2. Originally Posted by x-disturbed-x
A and B are points of intersection of y=x^2-2x and y=2-3x
simultaneous equation solved = x^2 + x -2 = 0

a) Find the coordinates of A and B, given that the x coordinate of B is greater than that of A.
You have identified correctly that the x co-ordinatea of the
points of intersection of A and B are the solutions of

$\displaystyle x^2+x-2\ =\ 0$

so now find those solutions, let's call them $\displaystyle x_1$ and $\displaystyle x_2$, and we
will assume that $\displaystyle x_1\ <\ x_2$, (they are not equal as the LHS
of the equation is not the square of a single linear factor).

Then from what we are told we know that the
x co-ordinate of A is $\displaystyle x_1$, and the x co-ordinate of B
is $\displaystyle x_2$.

To find the y co-ordinates of A and B respectivly we now substitute
$\displaystyle x_1$ and $\displaystyle x_2$ into

$\displaystyle y\ =\ 2-3x$

RonL

3. Originally Posted by x-disturbed-x
The line L is perpendicular to y=2-3x and passes through B.
b) Find an equation of L
The gradient of any line perpendicular to the line:

$\displaystyle y\ =\ m.x\ +\ c$

is $\displaystyle -1/m$.

So you have sufficient information to determine the slope
of L. Also you now know the co-ordinates of B (as you
will have found them in part a), you can find in intercept
for L by plugging the co-ordinates of B into into the
general equation of a line with the required slope.

RonL

4. [QUOTE=x-disturbed-x]The line M is parallel to y=5-3x and passes through A
c) Find an equation of M

This is like part b, except that the slope of M can be
found using the result that the slopes of parallel lines are equal.

Then proceed as in part b.

RonL

5. Originally Posted by x-disturbed-x

The lime M crosses the y-axis at point P. The line L crosses the x-axis at point Q.
d) Find the distance of PQ, giving your answer in square root form where a and b are integers and b is prime. (Hint - use pythagoras' theorem)
You should by now have the equations for M and L, let these be:

$\displaystyle \mbox{L: }\ \ y\ =\ m_L.x\ +\ c_L$

$\displaystyle \mbox{M: }\ y\ =\ m_M.x\ +\ c_M$

M crosses the y-axis when $\displaystyle x\ =\ 0$, which is at

$\displaystyle y\ =\ c_M$

so $\displaystyle P$ is the point $\displaystyle (0,c_M)$.

Similarly L crosses the x-axis when $\displaystyle y\ =\ 0$, which is at

$\displaystyle x\ =\ -c_L/m_M$

so $\displaystyle Q$ is the point $\displaystyle (-c_L/m_L,0)$.

So now you should know where $\displaystyle P$ and $\displaystyle Q$ are
and so you can find the distance between them in the usual manner.

RonL