The straight line L1 passes through the points (–1, 3) and (11, 12). Find an equation for L1 in the form ax + by + c = 0, where a, b and c are integers

When finding the equation of a straight line there are two important figures to calculate. The first being the gradient (the slope of the line) and the second being the y intercept (where the line crosses the y axis). Firstly we will find the gradient. The equation for a gradient is the chage in y coordinates divided by the change in x coordinates so in the example it equals: (12-3)/(11-(-1)) =9/12 which we can simplify to 3/4 by dividing top and bottom by 3.  We now need to find the y intercept. A good way to do this is to use the formula y=mx+c where m is the gradient an c is the intercept. We then substitute one of the points for x and y into the equation as well as the gradient we just worked out. 12=(11x3/4)+c we rearrage this equation to get c=15/4. We can now write the full equation of the line using the y=mx+c format so the answer becomes y=3/4x+15/4. However this is not how the question wants us to present our answer. To simplify the whole equation we need to multiply everthing by 4 so it becomes: 4y=3x+15 and then rearrage it so it becomes 4y-3x-15=0