Math Problem. Cant figure it out but need the answer in like 2 hours

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  • #1
Ok, i have a question. I have an answer but i don't really think it sright or i am doing it right. Can anybody explain to me how to go about doing this question, and give me the answer so i can know if i am doing these kinds of questions right?
Here it is:
A line passes through the point (1,4) as shown. The line and the two axis make a triangle of area 9 in the first quadrant. Determine the two possible points where the line could meet the x-axis.
Hope i can get some help, Brandon
and if the pic don't work, here's the ink:
Oh yeah and by the way, I am in quite the hurry to figure this out so any help would be awesome!
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Answers and Replies

  • #2
You should tell us what you've tried first.
  • #3
i think that knowing the segmental equation of a line would be helpful. that is

x/a +y/b =1, where a is the x-intercept, b-is the y intercept. look for other stuff as well. And pls post your work so far, so we might know where are you stuck,ane somebody will point to the right direction.
  • #4
honestly, i don't have a clue how to do it, i though tit was a lot simpler than it is, all that i tried was if it has an area of 9, then b*h=18, and i tryed to just guess and check, but i realized my answer was wrong so i need some help.
  • #5
A=(1/2)a*b so a*b=18, why did u think you were wrong. But look for what else from the figure youc can use to figure it out? Try to somehow use the fact that the line passes through (1,4) and then use the equation of the line i postedon post #3, and you will be just fine! And also use the fact that it crosses y, and x axes to determine the values of a, and b.
  • #6
k man ill give that some work, ill let you know if i gets it but more help wouldn't hurt lol. i don't really get how i am suposed to get 2 points that x can pass throgh from that equation.
  • #7
yeah but remember you are on homework forum, you are supposed to show your work first, before anyone here can help you!
  • #8
There are only 2 equations that you need.

[tex]A=\frac 1 2 x y[/tex]


[tex]9=\frac 1 2 xy \rightarrow 18=xy[/tex]

Now, find the slope of your line, use the point-slope equation and plug it in your slope. You're next step should be obvious as the answer has fallen out!
  • #9
okay i think i have an answer, let me know if its right. i have that the line can intercept x at 2.5 and 3.6. why i have this is because 2.5*7.2=18, and 3.6*5=18, and 3.6/7.2+2.5/5= 1. Let me no if that looks right.
  • #10
This response seems now no longer necessary, but here:

Your three points indicated are all on the same line, and so they have the same slope, m.
\frac{{y - 4}}{{0 - 1}} = m\quad \frac{{0 - 4}}{{x - 1}} = m

The formula for area of a right triangle gives you [tex] \[
{\textstyle{1 \over 2}}xy = 9

First, develop your slope equations into one equation of two variables; then use the triangle area
equation to transform the slope equation into a quadratic equation in one variable. The resulting
quadratic equation is factorable. The algebra is uncomplicated. Finishing the solution requires finding
the other unsolved variable. You will have two answers.
  • #11
k thanks for the help guys, think i have the right answer now. 1.5 and 3.
  • #12
k thanks for the help guys, think i have the right answer now. 1.5 and 3.
Wrong, use those values to compute the area of the triangle and tell me if it equals 9.

[tex]P(1,4) \rightarrow y-4=(y-4)(x-1)[/tex]

Simplify and you will get xy, but you know what xy is. Plug in and you have one of your solutions!
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  • #13
ok but the way i look at it is if the x value is 3, then the y value must be 6./ That will give you a slope of -2, with a y-intercept of 6. Go over 1, down 2, and you hit (1,4) so it works. This is how i figured it out for 1.5 too, checki it out, i think it makes sense
  • #14
[tex]A_{Triangle}=\frac 1 2 \cdot \frac 3 2 \cdot 3[/tex]

I don't think it equals 9 better check your work again and take our advice.

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