Equation of Line?

Question

A line passes thru the point (3,3) and forms a triangle in the first quadrant with the x and y coordinate axes. The area of the triangle is 24. What is/are the equation(s) of the line(s)?

I can see the answer just by looking at it, but deriving a solution is elusive. I am looking for a solution that doesn’t just involve stating what the x and y intercepts are without deriving them, because that trivializes the problem.

Answer 1

Given: 1) point on the hypotenuse (3,3);
2) the area is 24
3) Vertex of the trianlge are (0,0) (0, y) and (x, 0)

formula for area of a triangle is A = 1/2 bh
This will be a right triangle with the height being the y value and the base being the x value so the area will be:
24= 1/2xy.
Also, not that because the height is the y axis, the y intercept will be equal to y
Using point point formula for slope and the points (0,y) (x,0), the slope will be:
(y-0)/(0-x) = y/-x
Using the known point on the line, the slope will be:
(y-3)/(0-3) = (y-3)/-3
the two slopes are equal thus:
(y/-x)=(y-3)/-3
Now, consider the area:
24 = 1/2xy
48 = xy
48/y = x
Go back to the slopes--since they are equal
y/-x = (y-3)/-3
substitue -48/y for -x and solve for y
y/(-48/y) = (y-3)/-3
(y^2)/(-48) = (y-3)/(-3)
cross multiply gives
(-3y^2)= -48y + 144
3y^2 -48y + 144 = 0
3(y^2 - 16y + 48) = 0
3(y-4)(y-12)=0
y = 4 or y = 12
Because of the square, there are two potential values for y that will work as the vertex (0,4) and (0,12). So, there are two potential triangles or equations. Each will have to be tested to determine if they are true.
Remember from above, the y intercept is equal to y.
The vertexs of the triangle will be (0,0) (0,4) (12,0) for 1 potential triangle and (0,0) (0,12) (4,0) for the second.
The y intercept will of course be y.
The slope for triangle 1 becomes (4-3)/(0-3) = -1/3 and the slope for triangle 2 is: (12-3)/(0-3)= -3
The equation of the hypotenuse for triangle 1 then becomes:
y=(-1/3)x + 4
the equation for triangle 2 hypotenuse becomes
y = (-3)x + 12
Checking for validity, both triangles and equations are valid.

Answer 2

To form a triangle with the x and y axes, one vertex of the triangle must be at the origin (0,0). Otherwise the resulting figure would be a parallelogram.

Recall triangle area is 1/2*height*base.

Height is 3, from the point (3,3). Therefore, using the formula above, the base is must be 16. (1/2*3*16=24).

The three points are (3,3), (0,0), and (16,0).

Answer 3

The slope we all know is negative.

We want an equation of the form y = mx + h

Where h is the height of the triangle formed

m = (h-3)/(b-3)

y-int = h

y = 3 and x = 3

So we can get two equations

48=b*h

3 = 3(h-3)/(b-3) + h

48/h = b

3 = 3(h-3)/((48/h) - 3)

3 = (3h^2 - 9h)/(48-3h)

3(48-3h) = 3h^2 - 9h

144 -9h = 3h^2 - 9h

144 = 3h^2

12 = h*sqrt(3)

h = 12/sqrt(3) = 12*sqrt(3) / 3 = 4*sqrt(3)

y intercept = 4*sqrt(3)

b = 48/h = 48*sqrt(3)/12 = 4*sqrt(3)

So the slope = -h/b = -1

The equation is then

y = -x + 4*sqrt(3)

Answer 4

Draw the line through point E(3, 3). Let it cut y-axis
at point A(0,y1) and x-axis at point C(x1,0). Let point B(0,0).

Now ABC is a right angled triangle. Draw perpendiculars EF and
ED on x-axis and y-axis repectively.

ED = 3

EF = 3

BC = x1

AB = y1

CF = x1 - 3

AD = y1-3

area of triangle = bh/2 = x1y1/2

so x1y1 = 48 -------------------------eqn(1)

Now area of triangle ABC =

area of square DBFE + area of triangle EFC + area of ADE

9 + (1/2)(FC*EF) + (1/2)(DE*AD)

9 + (1/2)(3(x1-3)) +1/2(3(y1-3)) = 24

18 + 3x1 - 9 + 3y1 - 9 = 48

3x1 + 3y1 = 48

x1 + y1 = 16

x1 = 16 -y1 ------------------------------eqn(2)

substituting x1 in eqn(1)

(16 -y1)y1 = 48

16y1 - y1^2 = 48

y1^2 - 16y1 + 48 = 0

(y1 - 12)(y1-4) = 0

so y1 = 12 or 4

x1 = 4 or 12

equation of AC (in x and y intercept form)

(x/12) + (y/4) = 1

x + 3y = 12

or 3x + y = 12

equations of other two lines are x and y axes

Answer 5

Here are my thoughts...
xy/2=24
xy=48

hypotenuse line y=-x+b
at (3,3)
3=-3+b
b=6 which is y
x is 48/6 or 8

triangle lines are
y axis which is x=0
x axis which is y=0
diagonal which is y=-x+6

????
y=6

Answer 6

Hey you seem like a smart guy, can you tutor me?

Answer 7

Isn't it something like y=mx+b squared???