A(0,1) B(-2,3) C(2,-1) are the vertices of a triangle ABC. P and Q are the midpoint of the side AB and AC.
(a) find the equation of PQ,
(b) show that PQ is parallel to BC.
(c) show the length PQ = 1/2 length BC.
find P: P is halfway between a,b, so P=(-1,2) and likewise Q=(0,1)
line PQ:
Y-y1=m(x-x1) first find slope
m=1/-1=-1 (change in y/changeinx)
y-1=-1(x-0)
slope of BC: m=-2/2=-1 same slope, has to be parallel
Length PQ=sqrt( (-1)^2+1^2)=sqrt2
length BC= sqrt( (-4)^2+(4)^2)=2sqrt2 which is twice as long as PQ
To find the equation of the line PQ, we need to first find the coordinates of P and Q, and then use those coordinates to derive the equation of the line.
(a) Finding the coordinates of P and Q:
To find the coordinates of P, we need to find the midpoint of side AB. The midpoint formula is given by:
P = [(x1 + x2)/2, (y1 + y2)/2]
Applying the formula, we get:
P = [(0 + (-2))/2, (1 + 3)/2]
= [-1, 2]
Similarly, to find the coordinates of Q, we need to find the midpoint of side AC. Again using the midpoint formula, we have:
Q = [(x1 + x2)/2, (y1 + y2)/2]
= [(0 + 2)/2, (1 + (-1))/2]
= [1, 0]
So the coordinates of P are (-1, 2) and the coordinates of Q are (1, 0).
Now, we can find the equation of the line PQ using the point-slope formula:
y - y1 = m(x - x1)
Where m is the slope of the line and (x1, y1) is a point on the line.
(b) Showing that PQ is parallel to BC:
To prove that PQ is parallel to BC, we need to show that the slopes of the two lines are equal.
The slope of BC can be found using the formula:
m = (y2 - y1)/(x2 - x1)
Using the coordinates of points B(-2, 3) and C(2, -1), we have:
m_BC = (-1 - 3)/(2 - (-2))
= (-4)/(2 + 2)
= -1
The slope of PQ can be found using the coordinates of points P(-1, 2) and Q(1, 0):
m_PQ = (0 - 2)/(1 - (-1))
= (-2)/(1 + 1)
= -1
Since the slopes of BC and PQ are equal (-1 = -1), we can conclude that PQ is parallel to BC.
(c) Showing the length PQ = 1/2 length BC:
To show that the length PQ is half the length of BC, we can use the distance formula.
The distance between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Using the coordinates of points B(-2, 3) and C(2, -1), we have the length of BC:
Length_BC = √((2 - (-2))^2 + (-1 - 3)^2)
= √((4)^2 + (-4)^2)
= √(16 + 16)
= √32
Using the coordinates of points P(-1, 2) and Q(1, 0), we have the length of PQ:
Length_PQ = √((1 - (-1))^2 + (0 - 2)^2)
= √((2)^2 + (-2)^2)
= √(4 + 4)
= √8
Since Length_PQ = √8 and Length_BC = √32, we need to show that √8 = (1/2)√32.
√8 = √(4 * 2)
= √4 * √2
= 2 * √2
Similarly,
(1/2)√32 = (1/2)√(16 * 2)
= (1/2)√16 * √2
= (1/2) * 4 * √2
= 2 * √2
Therefore, we can see that √8 = (1/2)√32, demonstrating that the length of PQ is indeed half the length of BC.