Find all lines thru (6,-1) for which the product of the x and y intercepts is 3.?

Question

What is the best way to complete this problem? Im not just looking for an easy answer. I need to know how to work it out on paper. Thanks for any real help.

Answer 1

Sure. I wouldn't give you an answer without a method. I will assume you want straight lines of the form

y = mx + b

We have to find all the values of m and b that will cause the product of the intercepts to be 3.

In

y = mx +b

the y intercept is b and the x intercept is -b/m. So we want

b * -b/m = 3
-(b^2)/m = 3

The straight lines we are looking for have the form

-1 = 6m + b

We thus have two equations in two unknowns.

Let's solve the first one for m:

m = -(b^2)/3

and substitute that value of m into the second equation:

-1 = -(6)(b^2)/3 + b
-1 = -2b^2 + b
2b^2 - b -1 = 0
(2b + 1)(b -1) = 0

b = -1/2; m = -1/12
b = 1; m = -1/3

There are thus two lines that satisfy the requirements:

y = -x/12 - 1/2

[Check: the y intercept is -1/2; the x intercept is -6; product 3. -1 = -6/12 -1/2; line goes through (6, -1).]

and

y = -x/3 + 1

[Check: the y intercept is 1; the x intercept is 3; product 3. -1 = -6/3 + 1; line goes through (6, -1).]


Edit: Louise's two lines (below) are the same as mine. My equations are in standard form, hers are perhaps easier to write and use. Her method, using two equations in two unknowns and substituting one into the other, is the same as mine. Choose whichever explanation is clearer to you for the 10 points.

Answer 2

Let
a = x-intercept
b = y-intercept

This means that (a,0), (0,b) and (6,-1) are all solutions to the equation
y = mx + b
and
ab = 3

Thus,
-1 = 6m + b
0 = am + b
ab = 3

Since 0 = am + b,
a = -b/m

Substitute to ab = 3
(-b/m)(b) = 3
-b² = 3m
m = -b²/3

-1 = 6m + b
-1 = 6(-b²/3) + b
-1 = -2b² + b
2b² - b - 1 = 0
(b - 1)(2b + 1) = 0
b = 1, b = -1/2
m = -(b²)/3
m = -1/3 or m = -1/12

Thus, the lines are
y = -1/3 x + 1 and
y = -1/12 x - 1/2



^_^

Answer 3

All lines through (6, -1) have different slopes:

y - (-1) = m(x - 6)

y + 1 = mx - 6m
y = mx - (6m + 1)

x-intercept is when y = 0
0 = mx - 6m - 1
mx = 6m + 1
x = 6 + (1/m)

y - intercept is when x = 0
y = m(0) - (6m + 1)
y = -6m - 1

Set the product equal to 3 and you can solve for "m"

x*y = 3 = [6 + (1/m)]* [-6m - 1]
3 = -36m - (1/m) - 6 - 6
-36m - (1/m) -15 = 0
m*(-36m - (1/m) -15) = m*0
36m^2 + 15m + 1 = 0
(12m + 1)(3m + 1) = 0
m = (-1/12) or (-1/3)

Use:
y - (-1) = m(x - 6) for m = (-1/12) or (-1/3)

and you have two possibilities.

Answer 4

If an equation of a line is in the form
Ax + By = C,
the x-intercept is C / A, and
the y-intercept is C / B.
The product of these intercepts is
C² / AB, which is supposed to equal 3.

Since the line(s) must go through (6, -1), the equation(s) must hold true for
6A - B = C.
Since C² = 3AB, B = C² / 3A.
Substituting,
6A - C² / 3A = C
Multiplying by 3A,
18A² - C² = 3AC
18A² - 3AC - C² = 0
18A² - 6AC + 3AC - C² = 0
6A(3A - C) + C(3A - C) = 0
(6A + C)(3A - C) = 0
6A + C = 0 or 3A - C = 0
C = -6A or C = 3A.

Substituting into 6A - B = C,
6A - B = -6A or 6A - B = 3A.
If C = -6A, then B = 12A.
If C = 3A, then B = 3A.

Case 1: Ax + By = C, B = 12A, C = -6A.
Ax + 12Ay = -6A
x + 12y = -6
passes through (6, -1), and has x- and y-intercepts of -6 and -1/2, respectively.

Case 2: Ax + By = C, B = 3A, C = 3A.
Ax + 3Ay = 3A
x + 3y = 3
passes through (6, -1), and has x- and y-intercepts of 3 and 1, respectively.

The lines are:
x + 12y = -6 and x + 3y = 3.

Answer 5

So you want the graph of :
There are infinite number of lines passing from a given point.
If you draw the graph of xy = 3 i.e y = 3 / x, you get a rectangular hyperbola. This does not intersect the point (6, -1).

So, according to me no line which passes through (6, -1) has the product of its x & y intercepts 3.

Answer 6

gud ques. i cant answer