### Video Transcript

Given the points π΄ one, π₯; π΅ π¦, one; πΆ three, negative one; and π· two, five, find the values of π₯ and π¦ if the area of the triangle π΄π΅πΆ is two square units and the points π΅, πΆ, and π· are collinear.

In this question, weβre given four points, π΄, π΅, πΆ, and π·, and we can see that points π΄ and π΅ have an unknown coordinate. We need to find the values of these unknown coordinates by using the information given. Weβre told the area of triangle π΄π΅πΆ is two square units, and weβre also told that the points π΅, πΆ, and π· are collinear. And thereβs many different ways we can determine the values of π₯ and π¦; however, weβre only going to go through one of these.

We can start by noticing we can calculate both the area of a triangle from the coordinates of its vertices and whether three points are collinear by using determinants. So, weβll answer this question by using determinants. We might be tempted to start with the first piece of information weβre given that the area of triangle π΄π΅πΆ is two square units. However, two of the vertices of this triangle are π΄ and π΅, and we donβt know the value of π₯ or the value of π¦. So, this would give us an expression involving two unknowns. Instead, by using the fact that π΅, πΆ, and π· are collinear, weβll find an expression only involving π¦.

So, weβll start by recalling if three points π₯ sub one, π¦ sub one; π₯ sub two, π¦ sub two; and π₯ sub three, π¦ sub three are collinear, then the determinant of the three-by-three matrix π₯ sub one, π¦ sub one, one, π₯ sub two, π¦ sub two, one, π₯ sub three, π¦ sub three, one must be equal to zero. Therefore, since weβre told points π΅, πΆ, and π· are collinear, we can substitute their coordinates into this equation. This gives us the determinant of the three-by-three matrix π¦, one, one, three, negative one, one, two, five, one must be equal zero.

Now, we can find the value of π¦ by evaluating the determinant. To do this, weβre going to expand over the first row. Remember, the parity of the sum of the row number and column number will change the sign of the term. In particular, for this case, we multiply the first term by positive one, the second term by negative one, and the third term by positive one. We get π¦ times the determinant of negative one, one, five, one minus the determinant of three, one, two, one plus the determinant of three, negative one, two, five. Now all thatβs left to do is evaluate this expression.

Remember, to find the determinant of a two-by-two matrix, we need to find the difference in the product of the diagonals. We get π¦ times negative one minus five minus three minus two plus 15 plus two. If we distribute and simplify, we get negative six π¦ minus one plus 17. And remember, this is equal to zero. Finally, we can solve this for π¦. Negative one plus 17 is 16. We subtract 16 from both sides of the equation and divide through by negative six. π¦ is negative 16 divided by negative six, which simplifies to give us eight divided by three.

Now that weβve determined the value of π¦, we can use the area of triangle π΄π΅πΆ to determine the value of π₯. To do this, letβs start by clearing some space and keeping in mind π¦ is eight divided by three. We now need to recall how to determine the area of a triangle given the coordinates of its vertices by using determinants. We recall the area of a triangle with vertices π₯ sub one, π¦ sub one; π₯ sub two, π¦ sub two; π₯ sub three, π¦ sub three is one-half the absolute value of the determinant of the three-by-three matrix π₯ sub one, π¦ sub one, one, π₯ sub two, π¦ sub two, one, π₯ sub three, π¦ sub three, one.

Therefore, we can just substitute the coordinates of π΄, π΅, and πΆ into this formula to find the area of the triangle. This gives us the area of the triangle is one-half times the absolute value of the determinant of the three-by-three matrix one, π₯, one, eight over three, one, one, three, negative one, one. And weβre told in the question the area of this triangle is two square units, so this must be equal to two.

We now want to solve this for π¦. Weβll start by multiplying both sides of the equation through by two. Now, we need to evaluate the determinant of the three-by-three matrix. Weβll do this once again by expanding over the first row. This gives us that four is equal to the absolute value of the determinant of the two-by-two matrix one, one, negative one, one minus π₯ times the determinant of the two-by-two matrix eight over three, one, three, one plus the determinant of the two-by-two matrix eight over three, one, three, negative one.

Next, weβll evaluate the determinant of the two-by-two matrices. We take the difference in the product of the diagonals. This gives us that four is equal to the absolute value of one plus one minus π₯ times eight over three minus three plus negative eight over three minus three. Now, weβll distribute over our parentheses and simplify. We get four is equal to the absolute value of π₯ over three minus 11 over three. And now, we just need to solve this absolute value equation. To make it easier, weβll start by taking the factor of one-third outside of the absolute value. This gives us four is equal to one-third times the absolute value of π₯ minus 11. And we can simplify this further by multiplying both sides of our equation through by three. We get 12 is equal to the absolute value of π₯ minus 11.

And finally, for the absolute value of a number to be equal to 12, either itβs equal to 12 or negative 12. In other words, the only solutions to this equation is when π₯ minus 11 is 12 or when π₯ minus 11 is negative 12. And we can solve both of these equations separately. In our first equation, we add 11 to both sides to get π₯ is 23. And in our second equation, we add 11 to both sides to get π₯ is negative one. And this then gives us our final answer: π₯ can be equal to 23 or negative one and π¦ must be equal to eight over three.