9.8

Hi, and welcome back to a step in practice. Today, we are dealing with lines in the plane. In exercise 1, we are asked to find the slope of the line k minus 2 times x plus 6 minus k y equal to 3. Now k is a real parameter. And first of all, in order to speak about slope, we have to be sure that it is not a vertical line. Now, the line is vertical just when its equation is x equal something. So let us consider the case where the line is vertical. Well, the question is x equals something if the coefficient of y is equal to 0.

71

So if 6 minus k is equal to 0, so if k is equal to 6– OK, I know to speak about slope, we will avoid the k is k is equal to 6. So we assume k different from 6. In that case, we have to express y in terms of x. And the equation becomes 6 minus k y equal to 3 minus k minus 2x. That is y equals to 2 minus k divided by 6 minus k x plus 3 over 6 minus k. And so the slope of the line is a 2 minus k divided by 6 minus k. We are asked for which k the slope is less or equal than 0.

133.7

So the slope depends on k equals 2 minus k divided by 6 minus k for k different from 6. And it will be less or equal than 0 if and only if– well, 2 minus k divided by 6 minus k is less or equal than 0. Now the sign of a quotient is the sign of a product. So the sign of 2 minus k over 6 minus k equals the sign of 2 minus k times 6 minus k, which is also the sign of k minus 2 times k minus 6. Now this is a second-degree polynomial in k with two roots, 2 and 6. The coefficient of k squared is 1.

195.2

So this polynomial is positive out of the roots and negative inside the roots, between the two roots. And so it is less or equal than 0 when k varies from 2 to 6. So it is less or equal than 0 if and only if k belongs to the interval 2, 6.

217.8

Now, this is the sign of the product. But we have to exclude k plus 6. So the solution is 2, 6 without taking 6.

233.5

In exercise two, we’ve got two lines on the plane.

243.7

And the first line is 3x plus ky equal to 7. And the other line is 2 minus k x minus 5y equal to 2. And the question is for which values of k the two lines are parallel? Well, Francis showed us that the two lines are parallel if and only if the determinant of the coefficients is 0. Now the determinant is the product of 3 with minus 5 minus the product of 2 minus k with k. So it is minus 5 times 3 minus 2 minus k times k, which gives minus 15 minus 2k plus k squared.

307.6

And the two lines are parallel if and only if this determinant is 0, which means this is equivalent to solve the second-degree equation, k squared minus 2k minus 15 equal to 0. Now, the discriminant of this equation is 4 plus 60– so 64, which is eight to the square. And therefore, it has two roots, which are 2 plus or minus 8 over 2, which gives in the case with plus we have got 5. In the case with the minus, we’ve got minus 3.

374.9

So the lines are parallel for k equal to 5 or equal to minus 3. Now, let us check for k equal to 5. The two lines are 3x plus 5y equal to 7. And the second is minus 3x, minus 5y equal to 2. So these lines are parallel to the line 3x plus 5y equal to 0.

414.5

For k equal to minus 3, the first line is 3x minus 3y equal to 7, whereas the second line for k equal to minus 3 is 5x minus 5y equal to 2. The first is equivalent to x minus y equal to 7/3, and the second is equivalent to x minus y equal to 2/5. So these lines are parallel to the line x minus y equal to 0. And this ends exercise two and also this step in practice. See you in the next and last step in practice of this week.