Perpendicular line gradient (higher gcse) Watch
The question is:
Line A and B are perpendicular and intersect at (3, 3). If line A has the equation 3y - x = 6, what is the equation of line B ?
Don't say anything about defferentiation please as we down learn that at gcse and I just don't understand that.
(Although if you could explain defferentiation to me also it would be helpful)
then read off the gradient m.
from this you can predict the gradient of the other line...
for instance if m = 2/3 the other m = -3/2
if m = -8 the other m = 1/8
Rearrange the equation for line A to get your gradient.
The using your gradient from line A find the negative reciprocal.
Substitue this into the equation:
Y-y1=m(x-x1) to get line B
Differentiating for linear graphs isn't entirely useful, although it may be useful for normals to curves, for example line A and B being normals to a quadratic curve. In this situation you would differentiate the quadratic curve, and then subsitute the x coordinates of the two given points, through this you would obtain the gradient of the tangents to which the normals touch. Using the negative reciprocal it is then possible to calculate the gradient of the two normals, A and B, and hence using the above equation, as stated previously, can be used to work out the equations of the two normals.
For this question, rearrange the equation of line A in the form y = mx + c to find its gradient.
From that, you can work out the gradient of line B (do you know what the link is between the gradient of perpendicular lines?).
Once you have found the gradient, you can find the equation of line B by using y1 = mx1 + c to find what c is.
Alternatively, you could use the formula y - y1 = m(x - x1) to work out the equation.
(x1,y1) - coordinates through which the line passes
m - gradient of the line.
Hope that helps!