1. ## geometry question

Question: A triangle is bounded by the three lines

x +y +1 =0, y=2x-1, y=k
where k is a positive integer. For what values of k is the area of the triangle less than 2008?
I found out the point of intersections which are
(0, -1)
(-(1+k). k)
$(\frac{1+k}{2} ,k)$

what do I do from here
any help will be appreciate

2. ## Re: geometry question

note there are two possible values for $k$ (a sketch shows an upper triangle and a congruent lower triangle)

$x+y+1=0$ intersects $y=2x-1$ at $(0,-1)$

height of the upper triangle is $(k+1)$

base of the upper triangle is $b = \dfrac{k+1}{2} - [-(k+1)] = \dfrac{3(k+1)}{2}$

$\dfrac{1}{2} \cdot \dfrac{3(k+1)}{2} \cdot k < 2800$

3. ## Re: geometry question

The lines x+ y+ 1= 0, and y= k intersect where x+ k+ 1= 0 or x= -k- 1. That is the point (-k- 1, k).
The lines y= 2x- 1 and y= k intersect where 2x- 1= k or x= (k+1)/2. That is the point ((k+1)/2, k).
The lines x+ y+ 1= 0 and y= 2x- 1 intersect where x+ 2x-1+ 1= 3x= 0. That is the point (0, -1)

Choose the horizontal line y= k as base. The distance from (1- k, k) to (k+1, k) is k+ 1- (1- k)= 2k. That is the length of the base.
The altitude is the distance from the third vertex, (0, -1), to the base, measured along the line perpendicular to the base. Since the base is y= k, the perpendicular is x= 0. That crosses y= k at (0, k) which has distance from (0, -1) k- (-1)= k+ 2 and that is the "height" of the triangle. I presume you know that the area of a trangle is "1/2 base times height".

4. ## Re: geometry question

Originally Posted by HallsofIvy
The lines x+ y+ 1= 0, and y= k intersect where x+ k+ 1= 0 or x= -k- 1. That is the point (-k- 1, k).
The lines y= 2x- 1 and y= k intersect where 2x- 1= k or x= (k+1)/2. That is the point ((k+1)/2, k).
The lines x+ y+ 1= 0 and y= 2x- 1 intersect where x+ 2x-1+ 1= 3x= 0. That is the point (0, -1)

Choose the horizontal line y= k as base. The distance from (1- k, k) to (k+1, k) is k+ 1- (1- k)= 2k. That is the length of the base.
The altitude is the distance from the third vertex, (0, -1), to the base, measured along the line perpendicular to the base. Since the base is y= k, the perpendicular is x= 0. That crosses y= k at (0, k) which has distance from (0, -1) k- (-1)= k+ 2 and that is the "height" of the triangle. I presume you know that the area of a trangle is "1/2 base times height".
Originally Posted by skeeter
note there are two possible values for $k$ (a sketch shows an upper triangle and a congruent lower triangle)

$x+y+1=0$ intersects $y=2x-1$ at $(0,-1)$

height of the upper triangle is $(k+1)$

base of the upper triangle is $b = \dfrac{k+1}{2} - [-(k+1)] = \dfrac{3(k+1)}{2}$

$\dfrac{1}{2} \cdot \dfrac{3(k+1)}{2} \cdot k < 2800$

Thank you both mathematicians a lot
I raise a problem with this question

the answer is $1\leqslant x\leqslant 50$
but applying the advice you gave me
I did it this

$\frac{1}{2}\cdot \frac{3)(k+1)}{2}\cdot (k+1)<2008$
$\frac{3}{4}(k+1)^2 <2008$
how do I get from the above to the answer $1 \leqslant x\leqslant 50$
I believe that the answer may be wrong but I wanted a second opinion.

I uploaded the worksheet it is question number 19C1 Coordinate Geometry Hard.pdf

5. ## Re: geometry question

I would set:

$3\le\frac{3}{4}(k+1)^2<2008$

Multiply through $\frac{4}{3}$:

$4\le (k+1)^2<\frac{8032}{3}$

This implies (keeping in mind that $k$ is a natural number):

$2\le k+1\le\left\lfloor\sqrt{\frac{8032}{3}} \right\rfloor<\sqrt{\frac{8032}{3}}$

$2\le k+1\le51$

Subtract through by 1:

$1\le k\le50$

6. ## Re: geometry question

Originally Posted by MarkFL
I would set:

$3\le\frac{3}{4}(k+1)^2<2008$

Multiply through $\frac{4}{3}$:

$4\le (k+1)^2<\frac{8032}{3}$

This implies (keeping in mind that $k$ is a natural number):

$2\le k+1\le\left\lfloor\sqrt{\frac{8032}{3}} \right\rfloor<\sqrt{\frac{8032}{3}}$

$2\le k+1\le51$

Subtract through by 1:

$1\le k\le50$
Dear Mr Mark thank you but i want to ask some questions
firstly where did you get the number 3 from $3\le\frac{3}{4}(k+1)^2<2008$

secondly, the answer uses a less than or equal to sign. I wanted to know is there a reason for this. This is because i dont want to make a mistake by leaving it out since i can lose marks

7. ## Re: geometry question

The problem, as you stated in your first post, required that k be a positive integer. The smallest positive integer is 1 and, if k= 1, $\frac{3}{4}(1+ 1)^2= 3$. That says that $3\le \frac{3}{4}(k+ 1)^2$. The other inequality, that $\frac{3}{4}(k+ 1)^2< 2008$ follows from the condition that the area is less than 2008. Once you have $3\le k+ 1<\sqrt{\frac{8032}{3}}$, knowing that $\sqrt{\frac{8032}{3}}= 51.7425$, and that k is an integer, it follows that $3\le k+ 1\le 51$.

8. ## Re: geometry question

Originally Posted by HallsofIvy
The problem, as you stated in your first post, required that k be a positive integer. The smallest positive integer is 1 and, if k= 1, $\frac{3}{4}(1+ 1)^2= 3$. That says that $3\le \frac{3}{4}(k+ 1)^2$. The other inequality, that $\frac{3}{4}(k+ 1)^2< 2008$ follows from the condition that the area is less than 2008. Once you have $3\le k+ 1<\sqrt{\frac{8032}{3}}$, knowing that $\sqrt{\frac{8032}{3}}= 51.7425$, and that k is an integer, it follows that $3\le k+ 1\le 51$.
thank you very much