Question

Find the equation of a line that passes through the point (4,2) that is perpendicular to the line y = x. Show your work.

Answer 1

@AnonymousHD

Answer 2

@mww

Answer 3

@math&ing001

Answer 4

Well first of all you need to know every line can be written in the form

y - y1 = m(x-x1)

where (x1,y1) is a point that lies on the line and m is the gradient of that line. The above equation is known as the point gradient form of a line.

Thus to find the equation of the line we just need to know the gradient and any point that lies on it.

The point has already been given. Now you just need a gradient.
The gradient is not provided directly but recall that if two lines are parallel, their gradients are equal. So the line must have the same gradient as y = x since they are parallel.

The gradient of the line y = x can be found by comparing the equation to the gradient-intercept form of a line

y = mx + b.

So first, tell me what the gradient of the line y = x is.

Answer 5

it doesn't say

Answer 6

@mww

Answer 7

What is the value of m for y = x?

Answer 8

i dont know i gave you all that its said on the page.

Answer 9

remember that m is the gradient of the line, which is the coefficient of x if they write a line as y = mx + b (known as gradient intercept form). So all you need to do is find the coefficient of x in the equation y = x.

Answer 10

The gradient of y = x is simply 1, because the coefficient of x is 1.

I hope you can see the pattern. Further examples.
If the line was y = 5x - 2, the gradient would be 5.
For y = -3x + 2, the gradient would be -3.
For y = 5 - 4x, the gradient would be -4
For y = -x, the gradient would be -1.

The other way to calculate gradient of y = x is to find any two points of the line y = x (by subbing in say x = 0 and x = 1) then using the gradient formula:

\[m = \frac{ y_2-y_2 }{ x_2-x_1 }\]
you should get m = 1 as well.

Now sub in this gradient into the first formula I showed you with the given point and after simplifying you're done.

Answer 11

Oops I made one mistake. The question asks for a line PERPENDICULAR to y =x , not parallel to it. So the gradient you need to use will be the negative reciprocal of the gradient for y = x (since m1 x m2 = -1 if the lines are perpendicular) so m2 = -1/m1

Answer 12

thank you so much you're a life saver!!!