Question

find the equation of a straight line passing through the points (3,2)and (-3,6)giving the answer in the form x/a+y/b ,where a and b are constants

Answer 1

Slope of line = $\frac{6-2}{-3-3}= \frac{4}{-6} =- \frac{2}{3}$

the equation : $(y-2)= \frac{-2}{3}(x-3)$

3y - 6 = -2x + 6,     2x+3y =12  (you can divide by 12 to get the equation straight away)

to get the intercept of both the x-axis and y-axis:

the intercept of x-axis (y=0) , 2x=12 , x = 6

the intercept of y-axis (x=0) , 3y=12 , y = 4

the equation is : $\frac{x}{6} + \frac{y}{4} =1$