Thread: Assistance sought on how to go about solving graphing problem.

The text asks: Find the possible slopes of a line that passes through (4,3) so that the portion of the line in the first quadrant forms a triangle of area 27 with the positive coordinate axes.

I know that the equation for finding the area of a triangle is . How do i go about solving this algebraically? The answers in the book are . I do not know how they arrived at these answers. I don't know how you could graph it without first determining more than one set of points. Any help would be greatly appreciated. Thanks.

Let (X, 0) and (0, Y) be the points at which such a line intercepts the x and y axes, respectively. Then the area of the triangle formed is (1/2)XY= 27 we we must have XY= 54.

Any line through (4,3) can be written y= m(x- 4)+ 3. That means we must have y= m(0-4)+ 3= -4m+ 3= Y and 0= m(X- 4)+ 3 so that mX- 1= 0, mX= 1, and X= 1/m. XY= (-4m+ 3)(1/m)= 54. Solve that for m.

Hello, KhanDisciple!

Here's another approach . . .

Find the possible slopes of a line that passes through (4,3)
so that the portion of the line in the first quadrant forms a triangle of area 27
with the positive coordinate axes.

Answers:

Code:

      |
     b*
      |  *
      |     *  (4,3)
      |        o
      |           *
      |              *
  - - + - - - - - - - - * - -
      |                 a

A line through with slope
. . has the equation:

It has x-intercept:

It has y-intercept:

The area of the triangle is: .


The area is 27: .

. . . . . . . .

. . . . . . .


Therefore: .

Thanks a ton hallsofivy, and soroban, you guys are the best. Thank you for showing me how to solve this algebraically (the image of the graph helped also), although in hallsofivy's answer I don't see how solving for the x-intercept 0=m(x-4)+3 can be reduced to mx-1=0. I really appreciate it thank you guys.