A point C between two other points A and B, in the line joining A to B, is a linear combination of A and B.
If C is a/n the distance from B to A, then it is (n-a)/n the distance from A to B. We can write that as:
![C=(a)/(n)A+(n-a)/(n)B](https://img.qammunity.org/2023/formulas/mathematics/college/gz1yqn7agyxs6mfnzqqfg9j5myxo4ao9q6.png)
Notice that the farther it is from B (as a increases), the larger is the factor (a/n) by which we multiply A.
Step 1
Identify the points A and B, and the fractions a/n and (n-a)/n.
We have:
![\begin{gathered} A=(3,-4) \\ B=(-4,4) \\ \\ (a)/(n)=(3)/(5) \\ \\ (n-a)/(n)=(5-3)/(5)=(2)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c2fmiwycpi02zim77cgk78i91iazas9ud5.png)
Step 2
Use the previous result in the formula to find C:
![\begin{gathered} C=(3)/(5)(3,-4)+(2)/(5)(-4,4) \\ \\ C=\mleft((9)/(5),-(12)/(5)\mright)+\mleft((-8)/(5),(8)/(5)\mright) \\ \\ C=\mleft((9-8)/(5),(-12+8)/(5)\mright) \\ \\ C=\mleft((1)/(5),-(4)/(5)\mright) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hike8yxhkbf37lv52lk98bqhuq5niecomz.png)
Answer
Therefore, the point that is 3/5 the distance from B to A is
![\mleft((1)/(5),-(4)/(5)\mright)](https://img.qammunity.org/2023/formulas/mathematics/college/xpgsxjhdvtdz6c7y7kzm8049xj8y6pi6j5.png)