The desired equation is then y -4 = 12/15 (x -9)
Equations of degree 1 are linear equations. It is the straight line's equation. A linear equation with parameters a and b equal to zero has the conventional form ax+by+c = 0.where a ≠ 0 and b ≠ 0.
We need this line's equation in the point-slope form, y - k = m. (x - h).
The point (h, k) in the point-slope form shown above can be chosen at random from the two points (9, 4) and (-6, -8). In the above point slope form, the other point is thus denoted by (x, y):
Let (x, y) be and (h, k) be (9,4). (-6,-8). Consequently, y - k = m(x - h) becomes
-8-4 = m (-6-9), so that -12 = m(-15), and the slope is thus m = 12/15
The desired equation is then y -4 = 12/15 (x -9) (which is in point-slope form).
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