Find the equation of the line passing through B and perpendicular to ac

Question

Finding the area of a triangle is straightforward if you know the length of the base and the height of the triangle. But is it possible to find the area of a triangle if you know only the coordinates of its vertices? In this task, you’ll find out. Consider ΔABC, whose vertices are A(2, 1), B(3, 3), and C(1, 6); let line segment AC represent the base of the triangle.

 

Answer 1

You could also use the "bounding box" method. Here's how it works:

 

1. Draw your triangle on a piece of graph paper (this makes visualizing it easier).

 

2. Draw a box to enclose your triangle. The box will have vertices at (1,1); (3,1); (1,6); and (3,6).

 

3. Figure out the area of your box: Length is 2, Height is 5. Area = 10.

 

4. look at the three right triangles that are within the box, but outside your triangle. figuring out their areas should be fairly simple. These areas turn out to be: 3, 1, and 3.

 

5. Logically, then, the area of your triangle is equal to the area of the box, MINUS the sum of the areas of these three "extra" triangles (3+1+3).

 

6. The area of your triangle is 3.

 

You can see more about the "bounding box" method here: 

http://www.mathopenref.com/coordtriangleareabox.html

Answer 2

Hi Katie,

Let's call "D" the point where the line passing through B and perpendicular to AC meet. We need to get the (x,y) coordinates of D. Two facts to know are:

1. The slope of a line, "m", can be calculated from m = (y₁ - y₂) / (x₁ - x₂), where 1 & 2 are two points on the line.

2. Two lines are perpendicular if the product of their slopes is -1.

So let's calculate the slope of AC: m_AC =  (y_A - y_B) / (x_A - x_B) = (1 - 6) / (2 - 1) = -5. (Note that if you swap A and B, you still get the same result, -5) Therefore, the slope of the perpendicular line BD is m_BD=1/5 or 0.2 because -5 * 0.2 = -1. We now have the slope, but still we need the (x,y) coordinates of D.

The next piece of the information we can use is that point D is on the line BD, and B is (3,3). The equation of a line is y = m * x + b, where m is the slope and b is the y-intercept (that is, the value of y when x=0). We know the slope is 0.2, so let's get the value of b. We have, 3 = 0.2 * 3 + b ; so, b = 3 - 0.2 * 3 = 3 - 0.6 = 2.4. Thus the equation for BD is y = 0.2 * x + 2.4.

We also know that D is on line AC, and A is (2,1). Or, C is (1,6) -- we can use A or C; I'll use A. We already know the slope of AC is -5, because we calculated m_AC=-5 two paragraphs above. To get the equation of AC, knowing the slope is -5 and A is (2,1), we write y = mx + b ; so, 1 = -5*2 + b ; so, 1 + 5*2 = b ; so b=11. Thus the equation for AC is y = -5 * x + 11.

Wow ... this is a lot of work ... but we go onward ... we are almost to getting the (x,y) coordinate of D ... now we use the fact that BD and AC intersect at D. That means the x-values of D for both lines are equal, and the y-values are equal. In other words, we need to find (x,y) for which

y_D = 0.2 * x_D + 2.4  (line BD)  &  y_D = -5 * x_D + 11  (line AC)

have the same (x_D,y_D). We have two equations and two unknowns; because both equations are in the form of y=..., we can equate the two right-hand sides. That is,

0.2 * x_D + 2.4 = -5 * x_D + 11 ; so, 0.2*x_D + 5*x_D = -2.4 + 11 ; so, 5.2*x_D = 8.6 ; so, x_D = 8.6/5.2

We can insert x_D=8.6/5.2 into either the BD equation of AC equation to get y_D. I'll use the AC equation

y_D = -5 * x_D + 11 ; so, y_D = -5 * (8.6/5.2) + 11 ; so, y_D = (-43/5.2) + 11 

Finally we have the (x,y) coordinates of D: x=8.6/5.2 and y=(-43/5.2)+11 ; or x≈1.654 and y≈2.731

Almost done ... we need to calculate the length of CA, and DB, and the area of the triangle is 0.5 * base * height = 0.5 * length CA * length DB.

We know the (x,y) of all the points C, A, and D. The formula for the length between two points (x₁, y₁) and (x₂, y₂) is length = square root [(x₁ - x₂)² + (y₁ - y₂)²]

By now I'm pretty tired ... so your job is to finish it --

1. calculate the length of CA using C = (1,6) = (x₁,y₁) and A = (2,1) = (x₂,y₂)

2. calculate the length of DB using D = (1.654,2.731) = (x₁,y₁) and B = (3,3) = (x₂,y₂)

3. multiple the two lengths, then divide by 2 to get the area

and we are DONE !