Text description - screencast for exercise 6

[The narrator reads out the onscreen text.]

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...

[Equation 2 times 4: 4x minus 8y minus 28 equals 0. Equation 1: 4x plus 3y minus 6 equals 0. Equation 1 minus equation 2: 11y plus 22 equals 0.]

NARRATOR: ..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.

[Equation: x plus 4 minus 7 equals 0. x minus 3 equals 0. x equals 3.]

NARRATOR: 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.

[Equation 1 times 4. 12x plus 8y equals 8. Equation 2 times 3. 12x plus 9y equals 21. Equation 1 minus equation 2. y equals 13.]

NARRATOR: Subtracting these new equations gives y equals 13. And substituting y equals 13 into the first equation gives x equals negative 8.

[Equation: 3x plus 26 equals 2. 3x equals negative 24. x equals negative 8.]

NARRATOR:  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.