NARRATOR: Exercise 6. Part a.
Find the equation of the line parallel to the x-axis
which passes through the point where the lines
4x plus 3y minus 6 equals 0 and x minus 2y minus 7 equals 0 meet.
Part b.
Find the gradient of the line
which passes through the point (2, negative 3)
and the point of intersection of the lines
3x plus 2y equals 2 and 4x plus 3y equals 7.
Part a.
Find the equation of the line parallel to the x-axis
which passes through the point of intersection of the two given lines.
If the line is parallel to the x-axis, then it has a gradient of 0.
It passes through the point of intersection between the lines
4x plus 3y minus 6 equals 0 and x minus 2y minus 7 equals 0.
These equations can be solved simultaneously to find that point.
Calling the equations 'one' and 'two' respectively,
and then multiplying equation two by 4
in order to create the same x term in both equations
allows us then to subtract to get the equation 11y plus 22 equals 0,
so 11y equals negative 22 and y equals negative 2.
Substituting y equals negative 2 into the second equation
gives x equals 3.
And so the line passes through the point (3, negative 2).
The line has a gradient of 0
and passes through the point (3, negative 2).
A line with a gradient of 0 is of the form y equals c.
And in this case the required line has equation y equals negative 2.
Part b.
Find the gradient of the line
which passes through the point (2, negative 3)
and the point of intersection of the given lines.
We must first find the point of intersection of the lines
3x plus 2y equals 2 and 4x plus 3y equals 7.
Calling these equations 'one' and 'two' respectively
and then multiplying equation one by 4 and equation two by three
in order to create the same x term in both equations.
Subtracting these new equations gives y equals 13.
And substituting y equals 13 into the first equation
gives x equals negative 8.
So the lines intersect at the point (negative 8, 13).
It's now been established that we wish to find
the gradient of the line joining the points
(2, negative 3) and (negative 8, 13).
Gradient is equal to rise over run,
that is the difference in the y-coordinates
divided by the difference in the x-coordinates.
And so in this case, we have
13 minus negative 3 divided by negative eight minus 2,
giving 16 over negative 10,
which is equivalent
to negative eight-fifths
or negative 1.6.