the coordinates of the endpoints of AB are A (-8, -2) and B(16,6). Point P is on AB what are the coordinates of point P such tha
t AP:PB is 3:5?
1 answer:
We have that
<span>A (-8, -2) and B(16,6)
step 1
find the distance AB in the x coordinates
dABx=(16-(-8))-----> 24 units
step 2
find coordinate x of P (Px)
Px=Ax+(3/5)*dABx------> Px=(-8)+(3/5)*24----> 6.4
step 3
F</span>ind the distance AB in the y coordinates
dABy=(6-(-2))-----> 8 units
step 4
find coordinate y of P (Py)
Py=Ay+(3/5)*dABy------> Py=(-2)+(3/5)*8----> 2.8
the coordinates of P are (6.4,2.8)
see the attached figure
<span>A (-8, -2) and B(16,6)
step 1
find the distance AB in the x coordinates
dABx=(16-(-8))-----> 24 units
step 2
find coordinate x of P (Px)
Px=Ax+(3/5)*dABx------> Px=(-8)+(3/5)*24----> 6.4
step 3
F</span>ind the distance AB in the y coordinates
dABy=(6-(-2))-----> 8 units
step 4
find coordinate y of P (Py)
Py=Ay+(3/5)*dABy------> Py=(-2)+(3/5)*8----> 2.8
the coordinates of P are (6.4,2.8)
see the attached figure
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<h2>Answer:</h2>
We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:
Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:
Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:
Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:
Step 4: solve for z, then for y, then for x:
By substituting into the first equation, we get the
. So:
The range of the inverse has to match the domain of the original function. We want to have map to
, so we restrict the domain of
to this interval.