How do I approach finding the slope of the third line?

Question

Line l₁ has equation 3x - 2y = 1 and goes through A (-1, -2). Line l₂ has equation y = 1 and meets line l₁ at point B. Line l₃ has positive slope, goes through point A, and meets l₂ at point C. The area of triangle ABC is 3. What is the slope of l₃?

(I do not know how I should start this problem).

Answer 1

Line l₁ : 3x - 2y = 1 : we observe that A(-1, -2) is on the line.

( This is a good start, always good to check ! There are misleading problems out there )

Line l₂ : y = 1 : a horizontal line (slope 0) that intersects Line l₁ at y=1

==> When y=1 occurs on Line l₁ , there is only 1 answer for x: 3x - 2y = 1 ==> 3x - 2(1) = 1 ==> 3x = 3 ==> x = 1

==> B(1, 1)

Line l₃ : “positive" slope through A, meets Line l₂ at C. Everywhere on Line l₂, y = 1, therefore C(Cₓ, 1)

We have 2 choices for C(Cₓ, 1)… to the “left” of B (case a), or to the right of B (case b)...

Case (a) - C to the “left” of B along Line l₂ : y = 1

Consider the ΔABC with BC as the “base” (along the line y = 1), whose height is the “altitude” from BC to A. This altitude runs (upside down) from BC. Since BC runs along y=1, we only need to compare y=1 with the y-value of A, which is -2

Therefore, the altitude for base BC is (1 - (-2)) = 3

Area of ΔABC = 3 = (1/2)base*height = (1/2)BC*3 = 3 ==> (1/2)BC = 1 ==> BC = 2

==> Cₓ = Bₓ - 2 = 1 - 2 = -1 ==> C(-1, 1)

Given that C(-1, 1) has the same x-value as A(-1, -2), Line l₃ would be the “vertical” line through A and C.

Conclusion: Line l₃ : x = -1, with infinite slope (undefined)

Problem: We are told that Line l₃ has a “positive" slope through A… but for Case (a), the slope is “undefined”.

So we try Case (b)...

Case (b) - C to the “right” of B along Line l₂ : y = 1

Same story with the area of ΔABC = 3, altitude = 3, base BC = 2, along y = 1. Starting from B(1, 1), we have C(1+2, 1) = C(3, 1)

The line through A(-1, -2) and C(3, 1) is:

(y - (-2)) / (x - (-1)) = m = (1 - (-2)) / (3 - (-1)) = 3/4

==> “positive" slope of Line l₃ : m = 3/4 [1/2 the slope of Line l₂ ]

Answer 2

The intersection of the line 3x - 2y = 1 with the line y = 1 is point B(1,1), found by substituting 1 for y in the 1st equation. The altitude from vertex A to side BC is 3, because side BC is a horizontal segment.

A = (1/2)(Base)(Height)

3 = (1/2)(distance from B to C)(3)

distance from B to C = 2

If C has coordinates (x, 1) then how to find the distance from (1,1) to (x, 1)?

That distance is really just the change in x.

BC = |x - 1|

|x - 1| = 2

x - 1 = 2 or x - 1 = -2

x = 3 or x = -1

So the candidates for C are (3,1) and (-1, 1). Since the target line has positive slope and passes through A(-1,-2) calculate the slope between A and C to determine if either gives a positive slope:

Between C(3,1) and A(-1,-2):

m₁ = [(1-(-2)]/[3-(-1)] = 3/4.

Between C(-1,1) and A(-1,-2):

m₂ is undefined since the x-coordinates are equal (vertical line) which is actually the altitude from vertex A to side BC. In this case the area of triangle ABC is indeed 3, but the slope of the line through A and C is not positive.

desmos.com/calculator/8qud5oxoaw

Answer 3

Great question Elizabeth — this kind of problem can feel a little tricky at first, but once you break it down step by step, it becomes much more manageable.

To get started, let’s look at what you’re given. Line l1 has the equation 3x - 2y = 1, and you’re told it goes through point A, which is at (−1, −2). Line l2 is the horizontal line y = 1, and it intersects l1 at a point B. There’s also a third line, l3, that has a positive slope, goes through point A, and intersects line l2 at point C. We’re told the area of triangle ABC is 3, and we’re trying to find the slope of line l3.

Let’s begin by finding the coordinates of point B, the intersection of lines l1 and l2. Since line l2 is y = 1, we can plug that into the equation for l1: 3x - 2(1) = 1, which simplifies to 3x = 3, so x = 1. That means point B is at (1, 1).

Next, we want to find the coordinates of point C, which lies on line l2, so its y-coordinate is 1. Let’s call C = (x, 1). Since line l3 goes through A = (−1, −2) and C = (x, 1), we can later compute its slope once we know x. For now, let’s use the fact that the area of triangle ABC is 3 to find the value of x that makes this true.

We can use the coordinate formula for the area of a triangle given three points:

Area = (1/2) | x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

Plugging in A = (−1, −2), B = (1, 1), and C = (x, 1), we get:

(1/2) | (−1)(1−1) + 1(1−(−2)) + x((−2)−1) | = 3

This simplifies to (1/2) | 0+3−3x | = 3, or | 3 - 3x | = 6.

Solving this, we get two possible values: x = -1 or x = 3.

Now we check which x-value makes sense. If C = (−1, 1), then line l3 goes from A = (−1, −2) to (−1, 1), which would be a vertical line—and that has an undefined slope, not a positive one. If C = (3, 1), then we can compute the slope of l3 as rise over run: (1 - (−2)) / (3 - (−1)) = 3/4, which is positive. That’s exactly what we want.

So the final answer is that the slope of line l3 is 3/4.

Let me know if you'd like a visual or more clarification—this was a great question to ask!

Answer 4

Because Line  has the equation y=1, point B must have a y-value of 1 (since that is where intersects  ). Substituting that into the equation, we have:

 

 simplifying

 Adding 2 to both sides

 Dividing by 3 on both sides

Giving us point B at (1, 1)

We know that the area of the triangle is 3. And we know that the triangle has to be obtuse (because the two diagonal lines both have positive slopes), so we know that altitude to the horizontal line (y=1) is on the exterior of the triangle.

The altitude will go from point (-1, -2) to the line y=1, giving it a height of 3. This means the length along that line to point C is 2 units, putting it at (3,1).

Using the two points we have for line   we can find the slope: .

I’ve graphed it out on Desmos to make it more understandable: https://www.desmos.com/calculator/heah0frlnj

Answer 5

Point B is (1,1). Let C = (x, 1). Then 3(x - 1) = 6; x = 3. Clope of AC(l₃) = (1 + 2)/(3 +1 )= 3/4