Find the coordinates of point P
P
along the directed line segment AB
AB
, from A(1, 6)
A(1, 6)
to B(−2,−3)
B(−2,−3)
, so that the ratio of AP
AP
to PB
PB
is 5
5
to 1
1
.

The calculated coordinates of the point P is P (-3/2, -3/2)

From the question, we have the following parameters that can be used in our computation:

A = (1, 6)

Also, we have

B = (-2, -3)

And we have the partition to be

m : n = 5 : 1

The coordinate of the partition is calculated as

P = 1/(m + n) * (mx₂ + nx₁, my₂ + ny₁)

Substitute the known values in the above equation, so, we have the following representation

P = 1/(5 + 1) * (5 * -2 + 1 * 1, 5 * -3 + 1 * 6)

This gives

P = 1/6 * (-9, -9)

So, we have

P = (-3/2, -3/2)

Hence, the coordinate is P = (-3/2, -3/2)

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To find the coordinates of point P along the directed line segment AB, we can use the concept of section formula and plug in the given values to find the coordinates (-3/2, -3/2).

To find the coordinates of point P along the directed line segment AB, we can use the concept of section formula. The section formula states that if a point P divides a line segment AB in the ratio m:n, then the coordinates of P can be found using the formula:

P(x,y) = [(mx2 + nx1)/(m+n), (my2 + ny1)/(m+n)]

Using this formula, we can plug in the values: m=5, n=1, x1=1, y1=6, x2=-2, y2=-3 to find the coordinates of point P.

P(x,y) = [(5*(-2) + 1*1)/(5+1), (5*(-3) + 1*6)/(5+1)] = [(-10+1)/6, (-15+6)/6] = [-9/6, -9/6] = (-3/2, -3/2)

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