First, we must determine the slope. The slope can be found by using the formula: m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))m=y2−y1x2−x1
Where mm is the slope and (color(blue)(x_1, y_1)x1,y1) and (color(red)(x_2, y_2)x2,y2) are the two points on the line.
Substituting the values from the points in the problem gives:
m = (color(red)(-3) - color(blue)(1))/(color(red)(3) - color(blue)(-5)) = (color(red)(-3) - color(blue)(1))/(color(red)(3) + color(blue)(5)) = -4/8 = -1/2m=−3−13−−5=−3−13+5=−48=−12
We can now use the point slope formula to get an equation for the line. The point-slope formula states: (y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))(y−y1)=m(x−x1)
Where color(blue)(m)m is the slope and color(red)(((x_1, y_1))) is a point the line passes through.
Substituting the slope we calculated and the the first point from the problem gives:
(y - color(red)(1)) = color(blue)(-1/2)(x - color(red)(-5))
(y - color(red)(1)) = color(blue)(-1/2)(x + color(red)(5))
We can now transform this equation into the standard form. The standard form of a linear equation is: color(red)(A)x + color(blue)(B)y = color(green)(C)
Where, if at all possible, color(red)(A), color(blue)(B), and color(green)(C)are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
First, we will multiply each side of the equation by 2 to eliminate the fraction and allow each of the coefficients to be an integer.
2(y - color(red)(1)) = 2 xx color(blue)(-1/2)(x + color(red)(5))
2y - 2 = color(blue)(cancel(color(black)(2))) xx color(blue)(-1/cancel(2))(x + color(red)(5))
2y - 2 = -(x + color(red)(5))
2y - 2 = -x - 5
Next, add color(red)(x) and color(blue)(2) to each side of the equation to move both the x and y term to the left side of the equation and the constants to the right side of the equation:
color(red)(x) + 2y - 2 + color(blue)(2) = color(red)(x) - x - 5 + color(blue)(2)
color(red)(x) + 2y - 0 = 0 - 3
color(red)(1)x + color(blue)(2)y = color(green)(-3)
