yellowcoco123
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26.) A line passes through (3, –2) and (8, 2).
a. Write an equation for the line in point-slope form.
b. Rewrite the equation in standard form using integers.

1.) y – 2 = 4/5(x – 3); –4x + 5y = 22
2.) y + 2 = 4/5(x - 3); -4x + 5y = -22
3.) y + 2 = 4/5(x + 3); –4x + 5y = –22
4.) y – 3 = 4/5(x + 2); –4x + 5y = 23

Answer :

[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 3}} &,&{{ -2}}~) % (c,d) &&(~{{ 8}} &,&{{ 2}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{2-(-2)}{8-3}\implies \cfrac{2+2}{8-3}\implies \cfrac{4}{5} \\\\\\[/tex]

[tex]\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-2)=\cfrac{4}{5}(x-3)\implies y+2=\cfrac{4}{5}(x-3)\\\\ -------------------------------\\\\ y+2=\cfrac{4}{5}x-\cfrac{12}{5}\impliedby \begin{array}{llll} \textit{now let's multiply both sides by }\stackrel{LCD}{5}\\ \textit{to do away with the denominators} \end{array} \\\\\\ 5(y+2)=5\left( \cfrac{4}{5}x-\cfrac{12}{5} \right)\implies 5y+10=4x-12 \\\\\\ -4x+5y=-22[/tex]

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