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
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 :)
1 answer
1 answer