Respuesta :
Given that the line passes through the points [tex](-4,\ -1)[/tex] and [tex](1 \frac{1}{2}, \ 2)[/tex]
The equation of a line passing through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by:
[tex] \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}[/tex]
Thus, the required equation is obtained as follows:
[tex]\frac{y-(-1)}{x-(-4)} = \frac{2-(-1)}{( \frac{3}{2}) -(-4)} \\ \\ \Rightarrow \frac{y+1}{x+4} = \frac{2+1}{ \frac{3}{2}+4}= \frac{3}{ \frac{11}{2} }= \frac{6}{11} \\ \\ \Rightarrow 11(y+1)=6(x+4)\\ \\ \Rightarrow 11y+11=6x+24\\ \\ \Rightarrow \bold{-6x+11y=13}[/tex]
The equation of a line passing through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by:
[tex] \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}[/tex]
Thus, the required equation is obtained as follows:
[tex]\frac{y-(-1)}{x-(-4)} = \frac{2-(-1)}{( \frac{3}{2}) -(-4)} \\ \\ \Rightarrow \frac{y+1}{x+4} = \frac{2+1}{ \frac{3}{2}+4}= \frac{3}{ \frac{11}{2} }= \frac{6}{11} \\ \\ \Rightarrow 11(y+1)=6(x+4)\\ \\ \Rightarrow 11y+11=6x+24\\ \\ \Rightarrow \bold{-6x+11y=13}[/tex]
Answer:
What is the equation of this line in standard form?
6x−11y=−13
is the answer