If the line passes through the points (3,-5) and (-2, 7), then x is the intercept is

Solution

$$y\ -y1\ =\ \frac{\left(y2-y1\right)}{\left(x2-x1\right)}\left(x-x1\right)$$

Putting the values, 

y2 = 7, x2 = -2

y1=-5, x1= 3

$$y\ -\left(-5\right)\ =\ \frac{\left(7-\left(-5\right)\right)}{\left(-2-\left(3\right)\right)}\left(x-3\right)$$

$$y\ +5\ =\ \frac{\left(12\right)}{\left(-5\right)}\left(x-3\right)$$

$$-5\left(y\ +5\right)\ =\ \left(12\right)\left(x-3\right)=>-5y-25=12x-36$$

on reaarranging,

12x + 5y =11

For ,x intercept

putting y=0

$$x=\frac{11}{12}$$ Answer