show that R is vertically bellow the midpoint of line PQ

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
I've been looking at this question for a while and I still can't get my head around it. Can anyone help to explain the answer?

question: the curve y= ( (x^2)/3 ) - 4x+ 7 crosses the y-axis at the point P

Q is the point ( 6, -5 )

a) find the gradient of the line PQ

I managed to do this:

point P is when the curve crosses the y-axis, so this must be when x=0

to find the y-coordinate: y= ( 0/3) - (4 x 0 ) + 7
y = 7

to find the gradient on PQ: ( -5 - 7 ) / ( 6 - 0 ) = -2 which is correct

b) for the curve, find dy/dx

I found this to be dy/dx = ( 2x / 3 ) - 4

c) the gradient of the curve at point R is equal to the gradient of the line PQ. Show that R is vertically below the midpoint of the line PQ

Can anyone help to explain what to do? thanks :)
 

The Tan Gent

New member
Joined
Apr 15, 2019
Messages
4
a.) First of all, you are correct in calculating point P by setting X=0, so the ordered pair of point P is (0, 7). Using the given point Q, the gradient is basically the slope of the line which you found to be -2.
b.) Your derivative is correct.
c.) In order to show point R is vertically below the midpoint of line PQ, you need to find the midpoint of PQ using the midpoint formula, which you should get (3, 1). Here, X=3 so plug 3 into your dy/dx. You should end up with -2. Coincidentally, this is equal to your gradient but it has nothing to do with that. This is simply the Y value of point R. Plug the Y value back into the original function. You will get X=3 for one of those solutions, which matches with the X value of the midpoint of segment PQ, thus proving point R is vertically below the midpoint of segment PQ.
 

Harry_the_cat

Senior Member
Joined
Mar 16, 2016
Messages
1,385
For (c), remember that \(\displaystyle \frac{dy}{dx}\) gives you a formula for calculating the gradient at any point on the curve.

At point R the gradient is -2 (using your result from (a))

So to find R let \(\displaystyle \frac{dy}{dx}=-2\)
ie \(\displaystyle \frac{2x}{3}-4=-2\) which yields \(\displaystyle x = 3\).
So the x-cord of R is 3. Sub in \(\displaystyle y =\frac{x^2}{3} - 4x+ 7 = \frac{3^2}{3}-4*3+7=-2\) since R lies on that curve
So R is the point (3, -2).

Now find the midpoint (M) of PQ using the midpoint formula.

For R to lie vertically below M, the x-coords must be the same, and the y-coord of R must be less than the y-coord of M.
You can finish it off.
 

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
For (c), remember that \(\displaystyle \frac{dy}{dx}\) gives you a formula for calculating the gradient at any point on the curve.

At point R the gradient is -2 (using your result from (a))

So to find R let \(\displaystyle \frac{dy}{dx}=-2\)
ie \(\displaystyle \frac{2x}{3}-4=-2\) which yields \(\displaystyle x = 3\).
So the x-cord of R is 3. Sub in \(\displaystyle y =\frac{x^2}{3} - 4x+ 7 = \frac{3^2}{3}-4*3+7=-2\) since R lies on that curve
So R is the point (3, -2).

Now find the midpoint (M) of PQ using the midpoint formula.

For R to lie vertically below M, the x-coords must be the same, and the y-coord of R must be less than the y-coord of M.
You can finish it off.
ah okay! I understand this method.

PQ midpoint: ( 3, 1 ) so R must be vertically below the midpoint of the line PQ

Thank you!
 
Top