First, we must determine the slope of the line. The slope can be found by using the formula: m = y 2 − y 1 x 2 − x 1 Where m is the slope and ( x 1 , y 1 ) and ( x 2 , y 2 ) are the two points on the line. Substituting the values from the points in the problem gives: m = − 3 − 5 − 3 − 5 = − 8 − 8 = 1 Next, we can use the point-slope formula to get an equation for the line. The point-slope formula states: ( y − y 1 ) = m ( x − x 1 ) Where m is the slope and ( x 1 y 1 ) is a point the line passes through. Substituting the slope we calculated and the first point from the problem gives: ( y − 5 ) = 1 ( x − 5 ) The standard form of a linear equation is: A x + B y = C Where, if at all possible, A , B , and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1 We can now transform the equation we wrote to standard form as follows: y − 5 = x − 5 − x + y − 5 + 5 = − x + x − 5 + 5 − x + y − 0 = 0 − 0 − x + y = 0 − 1 ( − x + y ) = − 1 ⋅ 0 x − y = 0 1 x + − 1 y = 0
5x-3y=15 first move the 5x to the right side of the equation which becomes 3y= -5x +15 and then divide the whole equation by 3 because u want y to be on its own so then the answer is y= -5/3 + 5