### Video Transcript

For what value of π do the lines π₯ over negative five is equal to π¦ minus two over negative one is equal to π§ divided by negative two and π₯ minus one divided by π is equal to π¦ plus two over four is equal to π§ plus one over four intersect?

In this question, weβre given the equations of two lines. And one of the lines is an unknown value of π. We need to determine the value of π that will make the two lines intersect. To do this, we can start by noticing both of the lines weβre given are written in Cartesian form. And itβs difficult to find the point of intersection between two lines by using the Cartesian form. So, instead, weβre going to start by converting both lines into vector form.

And we can recall the vector form for the equation of the line is the form π« is equal to π« sub zero plus π‘ multiplied by π, where the vector π« sub zero is the position vector of a point on the line and the vector π is a direction vector of the line. Itβs a nonzero parallel vector to the line. And we can find both of these vectors directly from the Cartesian form of the line.

Letβs start with finding the position vector of the first line. Weβll do this by solving each part of the equation equal to zero. And if we do this, we see the point with position vector zero, two, zero lies on this line. Next, we want to find the direction vector of this line. And we can do this by recalling in the Cartesian form of a line, the denominators represent the direction vector. So our direction vector is the vector negative five, negative one, negative two. And this tells us the position vector of any point which lies on this line satisfies this equation for some scalar value of π‘.

We can do the exact same for our other line. First, we find a point on the line by solving each part of the equation equal to zero. This gives us the point with position vector one, negative two, negative one. And we could also find the direction vector of this line by looking at the denominators. The direction vector is the vector π, four, four. This gives us the following vector equation of the second line. And itβs worth noting weβve called our scalar π in the second example so we donβt confuse it with the scalar weβve used above, π‘.

And now we can note if the two lines intersect, this means thereβs a position vector of a point which lies on both lines. In other words, their vector equations must be equal. So weβll set the two equations to be equal to each other and see if we can solve for the values of π‘, π, and π. First, letβs start by simplifying our vector equations. Weβll start by distributing both π‘ and π over our direction vectors. And remember, we do this by multiplying all of the components of the vector by the scalar. This gives us the following equation.

Now, letβs evaluate the vectorsβ sum on both sides of our equation. This gives us that the vector negative five π‘, negative π‘ plus two, negative two π‘ must be equal to the vector ππ plus one, four π minus two, four π minus one. And now weβre almost ready to solve for our unknowns. We just need to recall for two vectors of the same dimension to be equal, all of their corresponding components must be equal. This gives us three equations, one for equating each component.

Weβre going to start by only looking at the second and third components. This is because this gives us two equations in two unknowns. So we need to solve the simultaneous equations negative π‘ plus two is equal to four π minus two and negative two π‘ is equal to four π minus one. And thereβs many different ways we can solve this equation. For example, we can note the coefficient of π is equal in both equations. So we can find the difference in the left- and right-hand sides of the equations to eliminate our variable π.

Subtracting the left-hand side of the second equation from the first equation gives us negative π‘ plus two minus negative two π‘. And we can evaluate this. It gives us π‘ plus two. And subtracting the right-hand side of the second equation from the first equation gives us four π minus two. And then we subtract four π minus one. This is equal to negative one.

So we have π‘ plus two is equal to negative one. And we can solve this equation for π‘. We subtract two from both sides of the equation. We get π‘ is equal to negative three. We can then use this value of π‘ to find the value of π. We just need to substitute π‘ is negative three into either of our equations. Weβll substitute π‘ is negative three into our first equation. We get negative one times negative three plus two is equal to four π minus two.

And now we can solve this equation for π. First, the left-hand side of this equation simplifies to give us five. Then, we add two to both sides of the equation. We get seven is equal to four π. Finally, we divide both sides of the equation through by four. We get π is equal to seven over four.

And now we can go back and look at equating the first component of our two vectors. This was the equation negative five π‘ is equal to ππ plus one. Initially, we didnβt want to use this equation because it contains three unknowns. However, we now know the value of π‘ and the value of π. So we can substitute these into our equation and solve for π. This gives us negative five times negative three is equal to π multiplied by seven over four plus one.

We can now solve this equation for π. First, we simplify. We get 15 is equal to seven π over four plus one. Then, we subtract one from both sides of the equation. 14 is equal to seven π over four. And then finally we multiply both sides of the equation through by four and divide both sides of the equation through by seven. We then see that 14 divided by seven is equal to two. So π is equal to two times four, which is equal to eight, which is our final answer.

Therefore, we were able to show if the lines π₯ over negative five is equal to π¦ minus two over negative one is equal to π§ over negative two and π₯ minus one over π is equal to π¦ plus two over four is equal to π§ plus one over four intersect, then the value of π must be eight.