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 with an area of 27 with positive coordinate axes.

Answer 1

Hi Lebron D.,

Area=1/2*b*h, in this case the base is x and the height is y, so 27 = 1/2*x*y

The slope (m) at (4,3) and (x,0) is m = (3 - 0)/4 - x) = 3/(4 - x).

The slope (m) at (4,3) and (0,y) is m = (3 - y)/(4 - 0) = (3 - y)/4.

Rearrange both slopes as functions of x(m) and y(m):

m = 3/(4 - x); m = (3 - y)/4

4 -x = 3/m; 3 - y = 4m

x(m) = 4 - 3/m; y(m) = 3 - 4m

You have a choice to write the Area as a function of y(x) or x(y), either way will work.

y(x) = 54/x or x(y) = 54/y,

I'll choose the latter and substitute y(m) into x(y) which will become x(m):

x = 54/y = 54/(3 - 4m) = x(m)

Now I have x(m) = 4 - 3/m, and, x(m) = 54/(3 - 4m), I can set them equal to each other and solve for m:

4 - 3/m = 54/(3 - 4m)

(4m - 3)/m = 54/(3 - 4m)

54m = (4m - 3)*(3 - 4m)

54m = -16m² + 24m -9

0 = 16m² +30m +9

(8m +3)*(2m +3) = 0

m = -3/8 and m = -3/2

We can check our work:

y - 3 = (-3/8)*(x - 4) or y = -3x/8 +9/2 and y - 3 = (-3/2)*(x - 4) or y = -3x/2 + 9

@ (x, 0) x = 12 and x = 6

@ (0,y) y = 9/2 and y = 9

Area = (1/2)*12*(9/2) = 27 and Area = (1/2)*6*9 = 27

I hope this helps, Joe.

Answer 2

Here's what I got... (same set up as above I.e. x -intercept = (x, 0); y - intercept = (0, y) and A = 27; passing through (4,3))

Area = (1/2)(x)(y)

27 = (1/2)xy

xy = 54 ................. (1)

Now, since all three points (x, 0); (0, y); and (4,3) are collinear, we know that their slopes are all equal. So,

m = (0 - y)/(x - 0) = -y/x ......................(i)

or m = (3 - 0)/(4 - x) = 3/(4 - x)............(ii)

or m = (3 - y)/(4 - 0) = (3 - y)/4............(iii)

Any one of these slopes will work. In this case since all three slopes are the same value, I'm going to set

(i) = (ii) (if you want you can try setting (i) = (iii) or (ii) = (iii), the answers will be the same regardless).

(i) = (ii)

-y/x = 3/(4 - x)

-y = (3x)/(4 - x)

y = (-3x)/(4 - x) = (3x)/(x - 4) <------ I distributed the negative from the numerator to the denominator).

y = (3x)/(x - 4) .........(2)

Okay, now we have equations (1) and (2). Notice that I can now take (2) and substitute into eqn (1)

xy = 54

x(3x)/(x - 4) = 54

(3x²)/(x - 4) = 54

3x² = 54(x - 4)

3x² = 54x - 216

3x² - 54x + 216 = 0

3(x² - 18x + 72) = 0

3(x - 6)(x - 12) = 0

x = 6, 12

So, if x = 6, we can see by eqn (1) .... y = 9

Thus the slope between (6, 0); (0,9) is

m = -9/6 = -3/2

If x = 12, by eqn (1)....y = 54/12 = 9/2

Thus, the slope between (12, 0); (0, 9/2) is

m = (9/2)/(-12) = -9/24 = -3/8

Hope this helps

Mr. K

Answer 3

The way I read this, one side of the triangle is on the X-axis, another side is on the Y-axis, and the 3rd side passes through the point (4,3). This gives us a right triangle with the hypotenuse passing through (4,3).

Call the X-intercept A, and the Y-intercept B. The area of the triangle is then (A*B) / 2. Set this equal to 27, and we have A*B = 54.

Next, write the equation for the line that passes through the 2 intercepts: Y = B - (B/A)*X

Next, plug in the X and Y values for the given point:

3 = B - (B/A)*4

Now we have 2 equations for A and B. The next step will be to solve the pair.

Answer 4

This problem is quite interesting, we immediately see the x and y intercepts must be such that

1) x*y=54 for the area to be 27.

Let's look at what we know next.

for slop intercept form:

mx+b=y and we know the line passes through (4,3)

so

2) m*4+b=3

In addition we know the y intercept will be b

To find the x intercept, we use:

m*x+b=0

to obtain x=-b/m

Now we go back to our area form and plug in to 1):

yx=-b/m*b=54

54m=-b²

m=-b²/54

Plugging this back into 2)

-b²/54*4+b=3

Or

-2*b²/27+b-3=0

Using the quadratic formula, we obtain

b=(-1±Sqrt[1-8/9])/(-4/27)

=(27±27/3)/4 Just some simplification going on here since I don't want to type out all the little steps...

=(27±9)/4

Then b can be either 9 or 4.5.

Going back to 2) to find m, we use

in case b=9: m*4+9=3

4m=-6

m=-1.5

In case b=4.5: m*4+4.5=3

4m=-1.5

m=-3/8

Let's double check our areas:

For -1.5x+9=y

We obtain 6 and 9 as our x/y intercepts respectively, which do multiply to 54

For -3x/8+4.5=y

X-intercept: -3/8x=-9/2

x=-9/2*-8/3

x=12

We obtain 12 and 4.5 as our x/y intercepts respectively, which do multiply to 54. Yay!

Answer 5

The line intersects the positive x and y axes at (x,0) and (0,y).

So, we have a right triangle with vertex of the right angle at (0,0) and with legs of lengths x and y.

Area of triangle = (1/2)(x)(y) = 27. So, xy = 54

The slope of the line (using the point (x,0) and (4,3)) is 3 / (4-x). Using the points (0,y) and (4,3), the slope

is (3-y) / 4

3 / (4-x) = (3-y)/4

(3-y)(4-x) = 12

12 - 4y - 3x + xy = 12

So, 3x + 4y = 54

We have the system of equations,

3x + 4y = 54

xy = 54

3x + 4(54/x) = 54

3x² - 54x + 216 = 0

x² - 18x + 72 = 0

(x-12)(x-6) = 0

x = 12 or 6

Possible slopes:

if x = 12, then slope = 3 / (4-12) = -3/8

if x = 6, then slope = 3 / (4-6) = -3/2

Answer 6

Let Θ bet the angle between the line and the x axis . Also let x be the length of the triangle and y be its height. We can develop three equations involving x, y and tan(Θ)

First : y = x tan(Θ) standard definition of tangent

Second: 3 = (x -4) tan(Θ) line must pass through (4,3)

Third: xy = 54 Area of the triangle is 27

Since there are three equations and three unknowns, the solutions can be found by standard methods.

There are two of them : tan(Θ) = 3/2 , x = 6, y = 9

and tan(Θ) = 3/8 , x = 12 , y = 9/2

These solutions are easily checked by substitution into the three equations.

The desired slopes are - 3/2 and - 3/8

Answer 7

the x-intercept is (x,0)

the y-intercept is (y,0)

The slope shall be negative in quadrant #1

Since the area is 27, (1/2)xy = 27 ---> xy = 54

Please label this equation, xy = 54, labels it equation ALPHA.

slope M = (y-3)/x = 3/(4-x)

Cross multiplies: 3x = (y-3)(4-x)

y = (3x)/(4-x) + 3

Substitutes into equation ALPHA above in bold.

x [ (3x)/(4-x) + 3] = 54

(3x^2)/(4-x) + 3x = 54 <-- distributes the x

(3x^2) + 3x(4-x) = 54(4-x) <--- multiplies both sides by (4-x)

3x^2 + 12x - 3x^2 = 216 - 5x <--- distributes 3x on left and 54 on right

12x = 216 - 5x <--- 3x^2 cancels

17x = 216 <--- adds 5x to both sides

x = 12 and 12/17

From equation ALPHA:

(216/17) y = 54

y = 54*17/216 = 4 and 1/4 = 4.25

M = -4.25/ (216/17) = -0.3344907407407407....

y = (-0.3344907407....)x + 4.25

Answer 8

the x-intercept is (x,0)

the y-intercept is (y,0)

The slope shall be negative in quadrant #1

Since the area is 27, (1/2)xy = 27 ---> xy = 54

Please label this equation, xy = 54, labels it equation ALPHA.

slope M = (y-3)/x = 3/(4-x)

Cross multiplies: 3x = (y-3)(4-x)

y = (3x)/(4-x) + 3

Substitutes into equation ALPHA above in bold.

x [ (3x)/(4-x) + 3] = 54

(3x^2)/(4-x) + 3x = 54 <-- distributes the x

(3x^2) + 3x(4-x) = 54(4-x) <--- multiplies both sides by (4-x)

3x^2 + 12x - 3x^2 = 216 - 5x <--- distributes 3x on left and 54 on right

12x = 216 - 5x <--- 3x^2 cancels

17x = 216 <--- adds 5x to both sides

x = 12 and 12/17

From equation ALPHA:

(216/17) y = 54

y = 54*17/216 = 4 and 1/4 = 4.25

M = -4.25/ (216/17) = -0.3344907407407407....

y = (-0.3344907407....)x + 4.25

Answer 9

the x-intercept is (x,0)

the y-intercept is (y,0)

The slope shall be negative in quadrant #1

Since the area is 27, (1/2)xy = 27 ---> xy = 54

Please label this equation, xy = 54, labels it equation ALPHA.

slope M = (y-3)/x = 3/(4-x)

Cross multiplies: 3x = (y-3)(4-x)

y = (3x)/(4-x) + 3

Substitutes into equation ALPHA above in bold.

x [ (3x)/(4-x) + 3] = 54

(3x^2)/(4-x) + 3x = 54 <-- distributes the x

(3x^2) + 3x(4-x) = 54(4-x) <--- multiplies both sides by (4-x)

3x^2 + 12x - 3x^2 = 216 - 5x <--- distributes 3x on left and 54 on right

12x = 216 - 5x <--- 3x^2 cancels

17x = 216 <--- adds 5x to both sides

x = 12 and 12/17

From equation ALPHA:

(216/17) y = 54

y = 54*17/216 = 4 and 1/4 = 4.25

M = -4.25/ (216/17) = -0.3344907407407407....

y = (-0.3344907407....)x + 4.25