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Bailey.Nae
Find the equation of the line that passes through the points (-2, 3) and (1, -6). Write your answer in Standard Form. Answer -3x – y = 3 3x + y = -3 3x + y = 3 -3x – y = -3
Standard equation of a line passing through points (x1, y1) and (x2, y2) \[ \frac{y-y2}{x-x2}=\frac{y2-y1}{x2-x1}\] here (x1, y1)= (-2, 3) and (x2, y2)= (1,-6) so we have the equation of line passing through \[ \frac {y-(-6)}{x-1}= \frac{-6-3}{1-(-2)}\] we get \[ \frac {y+6}{x-1}= \frac{-9}{3}\] now we have \[ \frac {y+6}{x-1}= \frac{-3}{1}\] cross multiplying we get \[y+6=-3x+3\] so we get \[3x+y=-3\]
the eqn of line is y=mx+c;m=(y2-y1)/(x2-x1);hence m=(3+6)/(-2-1)=9/-3=-3. hence y=-3x+c; put pt(-2,3) as it lies on line and satisfies the eqn hence 3=3*2+c c=3-6=-3; thus eqn of line is 3x+y=-3 optn b