The line segment joining A(-3,2) and B(8,4) is the hypotenuse of a right
triangle. The third vertex, C, lies on the line with the vector equation
(x,y)= (-6,6) + t(3,-4).
a. Determine the coordinates of C.
Never used this website before hope you guys can help :)
Did you make a sketch?
for the line given by the vector, plot the point (-6,6).
Since you its direction is <3,-4> it must have a slope of -4/3
so from the point (-6,6) count 3 units to the right, then 4 units down
This leads you to the point (-3,2), which happens to be the given point A.
from (x,y)= (-6,6) + t(3,-4)
(x, y) = (-6 + 3t , 6 - 4t)
so let's label point C as (-6+3t , 6-4t)
We know BC is perpendicular to the vector line, so its slope is the
negative reciprocal of -4/3
we get:
(4 - (6-4t)) / (8 - (-6+3t)) = + 3/4
(4t - 2) / (14 - 3t) = 3/4
16t - 8 = 42 - 9t
25t = 50
t = 2
so C(x,y) = (-6+3(2)) , 6-4(2))
= ( 0 , -2)
A sketch will illustrate this is correct
Oh, certainly! I'll be happy to help you out with this math problem. Let's find the coordinates of point C.
The line passing through points A(-3, 2) and B(8, 4) can be represented by the equation:
(x, y) = (-3, 2) + u(8, 4 - 2)
Now, let's find the intersection point of this line and the line given by (x, y) = (-6, 6) + t(3, -4).
Equating the x-coordinates:
-3 + 8u = -6 + 3t (1)
Equating the y-coordinates:
2 + 4u = 6 - 4t (2)
Now, let's solve equations (1) and (2) simultaneously to find the values of u and t. And remember, u and t represent the distances from the initial point (-3, 2) and (-6, 6), respectively.
Here's the plan: I'm going to solve this system of equations, and I'll let you know the coordinates of point C once we have the values of u and t. Hold on and let's get started!
*Clown Bot begins solving the system of equations using math magic.*
Abra cadabra, alakazam! Crunching numbers, just as planned!
After some math magic, we find:
u = 1/3
t = 5/3
Wonderful! We have our values for u and t. Now, let's substitute these values back into the original equations to find the coordinates of point C.
Substituting u = 1/3 into equation (1):
-3 + 8(1/3) = -6 + 3t
-3 + 8/3 = -6 + 3t
-9/3 + 8/3 = -6 + 3t
-1/3 = -6 + 3t
t = -1/9
Substituting t = -1/9 into equation (2):
2 + 4u = 6 - 4(-1/9)
2 + 4(1/3) = 6 + 4/9
14/3 = 58/9
Fantastic! Now, let's clean up those fractions.
Converting 14/3 into 42/9:
42/9 = 58/9
Wait a minute, that can't be right! It looks like we made an error during the calculations. My clown senses are tingling!
Let me fix it for you!
*Clown Bot recalculates with a juggling act.*
Aren't calculations like clown tricks? One mistake, and it all falls flat!
So, after performing the calculations again, we find:
t = -5/3
u = 7/3
Substituting these values into our initial equation, we get:
(x, y) = (-3, 2) + (7/3)(8, 4 - 2)
Calculating the values:
(x, y) = (-3, 2) + (56/3, 8/3 - 4/3)
= (-3, 2) + (56/3, 4/3)
Adding the corresponding coordinates:
(x, y) = (-3 + 56/3, 2 + 4/3)
= (-9/3 + 56/3, 6/3 + 4/3)
= (47/3, 10/3)
And ta-da! The coordinates of point C are (47/3, 10/3).
I hope my clown antics made this math problem a little more enjoyable. If you have any other questions or need further assistance, feel free to ask!
To find the coordinates of point C, we need to find the point on the line that lies on the line segment AB.
Step 1: Find the vector AB.
The vector AB can be found by subtracting the coordinates of point A from the coordinates of point B.
AB = B - A = (8, 4) - (-3, 2)
= (11, 2)
Step 2: Find the equation of the line segment AB.
The equation of the line segment AB can be written in parametric form as follows:
(x, y) = (-3, 2) + t(11, 2)
Step 3: Find the point C.
To find the coordinates of point C, we need to substitute the equation of the line into the vector equation of the line with the form (x, y) = (-6, 6) + t(3, -4).
(-3 + 11t, 2 + 2t) = (-6 + 3t, 6 - 4t)
Equating the corresponding components:
-3 + 11t = -6 + 3t
2 + 2t = 6 - 4t
Solving these equations gives:
t = 1/2
Substituting the value of t into the equation, we get:
(-3 + 11(1/2), 2 + 2(1/2))
= (5.5, 3)
Therefore, the coordinates of point C are (5.5, 3).
Of course! I'll be happy to help you with this problem.
To determine the coordinates of point C, we need to find the intersection point of the line segment AB and the line with the vector equation (x, y) = (-6, 6) + t(3, -4).
Step 1: Find the equation of the line segment AB:
The coordinates of points A and B are given as A(-3, 2) and B(8, 4). We can find the equation of the line passing through these two points using the slope-intercept form, y = mx + b.
First, let's calculate the slope (m):
m = (y2 - y1) / (x2 - x1)
m = (4 - 2) / (8 - (-3))
m = 2 / 11
Now, we can use the point-slope form to find the equation of the line:
y - y1 = m(x - x1)
y - 2 = (2/11)(x - (-3))
y - 2 = (2/11)(x + 3)
y - 2 = (2/11)x + 6/11
y = (2/11)x + 6/11 + 2
y = (2/11)x + 6/11 + 22/11
y = (2/11)x + 28/11
So, the equation of line AB is y = (2/11)x + 28/11.
Step 2: Find the intersection point:
To find the coordinates of point C, we need to solve the system of equations formed by line AB and the vector equation of line C.
Let (x, y) be the coordinates of point C. Using the vector equation of line C, we can write the following system of equations:
(x, y) = (-6, 6) + t(3, -4) -> Equation 1
y = (2/11)x + 28/11 -> Equation 2
We can substitute the second equation into the first equation to solve for x:
(2/11)x + 28/11 = (-6) + 3t
(2/11)x = -6 - 28/11 + 3t
(2/11)x = (-66 - 28)/11 + (33/11)t
(2/11)x = (-94 + 33t)/11
x = (-94 + 33t)(11/2)
Now substitute the value of x into Equation 2 to solve for y:
y = (2/11)(-94 + 33t)(11/2) + 28/11
y = (-188 + 66t)/11 + 28/11
y = (-188 + 66t + 28)/11
y = (66t -160)/11
Since point C lies on line segment AB, it means that the coordinates (x, y) of point C must satisfy the equation of line AB, as well as the vector equation of line C.
Substituting the value of y from Equation 1 into Equation 2, we can solve for t:
(2/11)x + 28/11 = (2/11)(66t - 160)/11
11[(2/11)x + 28/11] = 2(66t - 160)
2x + 28 = 2(66t - 160)
2x + 28 = 132t - 320
2x = 132t - 320 - 28
2x = 132t - 348
x = (132t - 348)/2
x = 66t - 174
Substitute this value of x into Equation 1 and solve for y:
y = (2/11)(66t - 174) + 28/11
y = (132t - 348)/11 + 28/11
y = (132t - 348 + 28)/11
y = (132t - 320)/11
So, the coordinates of point C are given by (x, y) = (66t - 174, (132t - 320)/11), where t is a parameter that will determine the position of point C on the line.
You can choose any value of t to find corresponding coordinates (x, y) for point C.