Assistance sought on how to go about solving graphing problem.

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January 10th 2013, 01:54 PM
KhanDisciple

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 $\frac{1}{2}bh$ . How do i go about solving this algebraically? The answers in the book are $\frac{-3}{2}, \frac{-3}{8}$ . 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.
January 10th 2013, 02:03 PM
HallsofIvy

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

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.
January 10th 2013, 04:10 PM
Soroban

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

Hello, KhanDisciple!

Here's another approach . . .

Quote:

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: $\text{-}\tfrac{3}{2},\;\text{-}\tfrac{3}{8}$

Code:

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

A line through $(4,3)$ with slope $m$
. . has the equation: $y - 3\:=\:m(x-4) \quad\Rightarrow\quad y \:=\:mx + 3 - 4m$

It has x-intercept: $\left(\frac{4m-3}{m},\:0\right)$

It has y-intercept: $\big(0,\:-[4m-3]\big)$

The area of the triangle is: . $A \:=\:\tfrac{1}{2}bh \:=\:\tfrac{1}{2}\left(\frac{4m-3}{m}\right)\big(-[4m-3]\big)$

The area is 27: . $-\tfrac{1}{2}\frac{(4m-3)^2}{m} \;=\;27$

. . . . . . . . $(4m-3)^2 \:=\:-54m \quad\Rightarrow\quad 16m^2 - 24m + 9 \:=\:-54m$

. . . . . . . $16m^2 + 30m + 9 \:=\:0 \quad\Rightarrow\quad (2m+3)(8m+3) \:=\:0$

Therefore: . $\begin{Bmatrix}2m+3 \:=\:0 & \Rightarrow & m \:=\:\text{-}\frac{3}{2} \\ \\[-4mm] 8m+3 \:=\:0 & \Rightarrow & m \:=\:\text{-}\frac{3}{8} \end{Bmatrix}$
January 10th 2013, 10:33 PM
KhanDisciple

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

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.