b First, find the magnitude of a vector, and then use an extended ratio.
A
a P(0,4)
B
b P(-6,0)
Practice makes perfect
a We are asked to find the coordinates of a point P that lies on AB and partitions the vector into segments in the ratio of 2:1. We will start with drawing AB using the given coordinates of A and B.
As we can see, this vector is vertical, which means we can read its magnitude directly from the graph.
The magnitude of AB is 6. Since we know that point P divides this vector into segments AP and PB that have lengths in the ratio of 2:1, we will use the given ratio to evaluate the exact lengths of these segments. First, we will rewrite the ratio by adding x to each number.
2:1 ⇒ 2 x:1 x
Now, we will use the fact that the magnitude of AB is 6. This means that the sum of the above terms needs to be equal to 6.
As we know the value of x, we can find the lengths of AP and PB.
2 x:1 x
2( 2):1( 2)
4:2
We found that the length of AP is 4. This means that point P is 4 units up from the point A, which is located at the origin, and we know that the x-coordinate of P needs to be equal to 0 as this point lies on a vertical vector. Therefore, P has coordinates of (0, 4).
b Again, we are asked to find the coordinates of a point P that lies on AB, but this time P partitions the vector into segments in the ratio of 2:3. Let's start with drawing AB using the given coordinates of A and B.
As we can see, this vector is horizontal, which means we can read its magnitude directly from the graph.
The magnitude of AB is 15. Since we know that point P divides this vector into segments AP and PB that have lengths in the ratio of 2:3, we will use the given ratio to evaluate the exact lengths of these segments. First, we will rewrite the ratio by adding an x to each number.
2:3 ⇒ 2 x:3 x
Now, we will use the fact that the magnitude of AB is 15. This means that the sum of the above terms needs to be equal to 15.
As we know the value of x, we can find the lengths of AP and PB.
2 x:3 x
2( 3):3( 3)
6:9
We found that the length of AP is 6. This means that point P is 6 units left from the point A, which is located at the origin, and we know that the y-coordinate of P needs to be equal to 0 as this point lies on a horizontal vector. Therefore, P has a coordinates of ( -6,0).
