NARRATOR: Exercise 1.
If a point, P, with coordinates (x, y) lies on the line AB
where A has coordinates (x1, y1) and B has coordinates (x2, y2)
and the ratio of AP to PB is a to b with a and b both positive,
show that x, that is the x-coordinate of the point P,
is equal to b times x1 plus a times x2 all divided by a plus b,
and y, that is the y-coordinate of the point P,
is equal to b times y1 plus a times y2 all divided by a plus b.
Let's first consider the situation geometrically.
And so we have the line AB with the point P on that line.
And the point P divides the line in the ratio A to B.
So for example, we could say that the distance from A to P is a
and the distance from P to B is b.
And so that means the total distance from A to B is a plus b.
Now if we consider the two triangles that we have here,
we note first that they are in fact similar triangles,
so APP-dash is similar to ABB-dash
due to the fact that the two triangles have the same angles.
And we also note that we now have a triangle with a hypotenuse of A
that's similar to another triangle with a hypotenuse a plus b.
And so we can work out some scale factors here.
So if we were to look at the scale factor
going from the smaller triangle to the larger triangle,
we would be multiplying the side lengths by a plus b all divided by a.
And if we were to go from the larger triangle to the smaller one,
we would be multiplying the side lengths by a divided by a plus b.
And so using that idea, then,
we see that if we look at
the horizontal distances,
so dealing with the x-coordinates,
we see that A to P-dash would be
equal to a divided by a plus b
multiplied by A to B-dash.
And we can use this now to create our required expressions.
So AP-dash is equal to a divided by a plus b times AB-dash.
And we know that AB-dash is the horizontal distance
from the point A to the point B
so that's just the difference between
the x-coordinates of those two points,
that is x2 minus x1.
And tidying up, we get a single fraction - ax squared minus ax...
Sorry, ax2 minus ax1 all divided by a plus b.
Now, we're not just looking for that distance,
we're in fact looking for the x-coordinate of the point P.
And we know that the x-coordinate of the point P
is just the x-coordinate of the point A
plus this horizontal distance between A and P
that we've just worked out above.
And so we get this expression and we can now
create common denominators,
put the fraction together to find
that the x-coordinate of the point P
is b times x1 plus a times x2
all divided by a plus b
as was required in the question.
We can now use similar logic for the y-coordinate of the point P.
And we note that PP-dash, that is the vertical distance from A to P,
is equal to a divided by a plus b times BB-dash,
which is the vertical distance from A to B.
And we know that this vertical distance from A to B
is simply the difference in the y-coordinates
between those two points,
so that is y2 minus y1,
which simplifies as a single fraction
to a times y2 minus a times y1 all divided by a plus b.
Again, though, the y-coordinate of the point P
does not equal this expression in itself.
The y-coordinate of the point P would be equivalent to
the y-coordinate of the point A
plus this new vertical distance that we've established above.
And again, creating
a common denominator and putting
those two fractions together, we find
that the y-coordinate of the point P,
that is y, is equal to
b times y1 plus a times y2
all divided by a plus b,
as was stated in the question.
And so we have shown that
the coordinates of the point P
are as stated.