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:
The slope of a line, "m", can be calculated from m = (y_{1} - y_{2}) / (x_{1} - x_{2}), where 1 & 2 are two points on the line.
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 (x_{1}, y_{1}) and (x_{2}, y_{2}) is length = square root [(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}]
By now I'm pretty tired ... so your job is to finish it --
calculate the length of CA using C = (1,6) = (x_{1},y_{1}) and A = (2,1) = (x_{2},y_{2})
calculate the length of DB using D = (1.654,2.731) = (x_{1},y_{1}) and B = (3,3) = (x_{2},y_{2})
multiple the two lengths, then divide by 2 to get the area
and we are DONE !
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