How do you write the equation of a line given (-6,6),(3,3)?

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Answer 1 · 2016-08-31T22:10:30

$x+3y-12=0$

The equation of a line joining $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)$

Hence line joining $(-6,6)$ and $(3,3)$ is given by

$(y-6)/(3-6)=(x-(-6))/(3-(-6))$ or

$(y-6)/-3=(x+6)/(3+6)$ or

$9×(y-6)=-3×(x+6)$ or

$9y-54=-3x-18$ or

$3x+9y-54+18=0$ or

$3x+9y-36=0$ and dividing by $3$

$x+3y-12=0$