Straight line L1 has a gradient of -1/2 and passes through the point T (-1,-3). Find the equation of the line in the form of y=mx+c were m and c are constant
The equation of a straight line can be expressed in the form y = mx + c, where m is the gradient of the line and c is the y-intercept.
Given that the gradient of the line L1 is -1/2 and it passes through the point T(-1, -3), we can substitute the values into the equation:
-3 = (-1/2)(-1) + c
-3 = 1/2 + c
c = -3 - 1/2
c = -3.5
Therefore, the equation of the line L1 in the form of y = mx + c is:
y = -1/2x - 3.5