Question

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

Answer 1

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
\[3x+y=-3\]

Answer 2

option B

Answer 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