The line y = -2x + 8 meets the y-axis at the point B. Find the equation
of the line with gradient 2 that passes through the point B
The vertical y-axis goes through the horizontal x-axis at the point x = 0
So you substitute x = 0 into the above eq'n.
y = -2(0) + 8
y = 0 + 8
y = 8 ( This is the point B).
The coefficient of 'x' is the gradient. So in the given eq'n y = -2x + 8 the gradient is '-2'.
So for a gradient of '2' through B we write
y = 2x + 8
The line y = 1/4x + 2 meets the y-axis at the point B. The point C has
coordinates (-5,3). Find the gradient of the line joining the points B
Similarly to above if the point B is on the y-axis , then x = 0
y = (1/4)(0) + 2
y = 2
So the coordinates of B are ( 0 ,2). and C is ( -5,3).
To find any gradient from a given pair of points(coordinates).
m = (y(1) - y(2)) / (x(1) - x(2)).
So in our case we can write
m(BC) = (y(B) - y(C)) / ( x(B) - x(C)).
1. The y-coordinates go on the top and the x-coordinates go on the bottom.
2. The B coordinates go to the left and the C coordinates go to the right.
Hence m(BC) = (2 - 3) / ( 0 - - 5) (NB Note the 'double' minus).
m(BC) = (-1) / (5)
m(BC) = -1/5 ( The gradient joining B & C).