## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 947: 75

#### Answer

See the explanation below.

#### Work Step by Step

Express the line equation passing through these 2 distinct points in the form of matrix “A” as follows: $A=\left[ \begin{matrix} x & y & 1 \\ {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ \end{matrix} \right]$ To obtain the equation of the line, put $Det\left( A \right)=0$: \begin{align} & x\left\{ \left( {{y}_{1}}.1 \right)-\left( {{y}_{2}}.1 \right) \right\}-y\left\{ \left( {{x}_{1}}.1 \right)-\left( {{x}_{2}}.1 \right) \right\}+1\left\{ \left( {{x}_{1}}{{y}_{2}} \right)-\left( {{x}_{2}}{{y}_{1}} \right) \right\}=0 \\ & x\left( {{y}_{1}}-{{y}_{2}} \right)-y\left( {{x}_{1}}-{{x}_{2}} \right)+\left( {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}} \right)=0 \\ & x\left( {{y}_{1}}-{{y}_{2}} \right)+\left( {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}} \right)=y\left( {{x}_{1}}-{{x}_{2}} \right) \\ & y=\left( \frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)x+\left( \frac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{{{x}_{1}}-{{x}_{2}}} \right) \end{align} Therefore, the equation now has obtained a form of $y=mx+c$, where $m=\left( \frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)\text{ and }c=\left( \frac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{{{x}_{1}}-{{x}_{2}}} \right)$ Hence, the determinant equation can be used to give the equation if the line is passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$.

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