1. **State the problem:** We are given points $A(-1,2)$, $B(5,-2)$, $C(1,3)$, and $D(0,0)$. We want to express the vector $\overrightarrow{BA}$ as a linear combination of vectors $\overrightarrow{BC}$ and $\overrightarrow{BD}$. 2. **Write the vectors in component form:** $\overrightarrow{BA} = A - B = (-1 - 5, 2 - (-2)) = (-6, 4)$ $\overrightarrow{BC} = C - B = (1 - 5, 3 - (-2)) = (-4, 5)$ $\overrightarrow{BD} = D - B = (0 - 5, 0 - (-2)) = (-5, 2)$ 3. **Set up the linear combination:** We want scalars $x$ and $y$ such that $$\overrightarrow{BA} = x \overrightarrow{BC} + y \overrightarrow{BD}$$ which means $$(-6,4) = x(-4,5) + y(-5,2)$$ 4. **Write the system of equations:** From the $i$-component: $$-6 = -4x - 5y$$ From the $j$-component: $$4 = 5x + 2y$$ 5. **Solve the system:** From the first equation: $$-6 = -4x - 5y \implies 6 = 4x + 5y$$ From the second equation: $$4 = 5x + 2y$$ Multiply the second equation by 2 to align $y$ coefficients: $$8 = 10x + 4y$$ Multiply the first equation by -4: $$-24 = -16x - 20y$$ Multiply the second equation by 5: $$20 = 50x + 20y$$ Add these two equations: $$-24 + 20 = (-16x + 50x) + (-20y + 20y)$$ $$-4 = 34x + 0$$ So, $$x = \frac{-4}{34} = -\frac{2}{17}$$ Substitute $x$ back into the first equation: $$6 = 4\left(-\frac{2}{17}\right) + 5y = -\frac{8}{17} + 5y$$ $$6 + \frac{8}{17} = 5y$$ $$\frac{102}{17} + \frac{8}{17} = 5y$$ $$\frac{110}{17} = 5y$$ Divide both sides by 5: $$\cancel{5}y = \frac{110}{17 \cancel{5}} \implies y = \frac{110}{85} = \frac{22}{17}$$ 6. **Final expression:** $$\overrightarrow{BA} = -\frac{2}{17} \overrightarrow{BC} + \frac{22}{17} \overrightarrow{BD}$$ This means vector $\overrightarrow{BA}$ can be written as a linear combination of $\overrightarrow{BC}$ and $\overrightarrow{BD}$ with coefficients $-\frac{2}{17}$ and $\frac{22}{17}$ respectively.