Question

find the coordinates of the points of trisection of the line segment AB, whose end points are A(2,1) and B(5,-8).

Answer 1

using distance formula ..
$\sqrt{(x2 - x1) {}^{2} } (y2 - y1) {}^{2}$

Answer 2

Coordinates of points are C(4,-5) and D(3,-2), which trisected the line joining the point A and B.

Given:

• A line segment AB.
• A(2,1) and B(5,-8).

To find:

• Find the coordinates of the points of trisection of the line segment.

Solution:

Concept to be used:

A line segment joining by end points

$A(x_1,y_1 )\: and \: B(x_2,y_2)$, if divided by point P(x,y) in m:n, then coordinates of P are

$\bf x = \frac{mx_2 + nx_1}{m + n}$

$\bf y = \frac{my_2 + ny_1}{m + n}$

Step 1:

Find the coordinates of point C.

As shown in attached figure point C and D trisected the line segment AB.

As line is trisected by C and D, so

AD:DC:CB=1:1:1

So,

AC:CB=2:1

Apply section formula to calculate coordinates of point C.

$x = \frac{10 + 2}{2 + 1}$

$\bf x =4$

and

$y = \frac{ - 16 + 1}{3}$

$\bf y = - 5$

Thus,

Coordinates of C are (4,-5).

Step 2:

Find the coordinates of point D.

See the second attachment.

$x= \frac{5 + 4}{3}$

$\bf x = 3$

and

$y = \frac{ - 8 + 2}{3}$

$\bf y = - 2$

Thus,

Coordinates of D are (3,-2).

Final answer:

Coordinates of points are (4,-5) and (3,-2), which trisected the line joining the point A and B.

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