Question

One endpoint of a line segment is (8, −1). The point (5, −2) is one-third of the way from that endpoint to the other endpoint. Find the other endpoint

Answer 1

Answer:

The other endpoint is (-4, -5)

Step-by-step explanation:

We know the formula for the coordinates of the point dividing the line segment in the ratio of a : b is

x = x' + $\frac{a}{a+b}(x"-x')$

y = y' + $\frac{a}{a+b}(y"-y')$

Now we plug in the values of endpoints (8, -1) and (x, y) with a point (5, -2) dividing the segment in 1 : 3 ratio.

5 = 8 + $\frac{1}{1+3}(x-8)$

5 - 8 = $\frac{x-8}{4}$

(x - 8) = -3×4 = -12

x = -12 + 8 = -4

Similarly, -2 = -1 + $\frac{1}{1+3}(y+1)$

-2 + 1 = $\frac{1}{4}(y+1)$

y + 1 = -1(4)

y + 1 = -4

y = -1 - 4 = -5

Therefore, another endpoint of the segment is (-4, -5).

Answer 2

Answer:

Other end point is (-4,-5)

Step-by-step explanation:

One end point is (8,-1) is (x1,y1)

Other end point is (x2,y2)

The point (5, −2) is one-third of the way from that endpoint to the other endpoint.

the ratio the line is divided is 1:3 (m:n)

Apply section formula to find the other end point

$(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n})$

Plug in the values in the formula

$(\frac{1x_2+3(8)}{1+3} , \frac{1y_2+3(-1)}{1+3})$

$(\frac{1x_2+24}{4} , \frac{1y_2-3}{4})=(5,-2)$

$\frac{1x_2+24}{4}=5$

$x_2+24= 20$

$x_2=-4$

$\frac{1y_2-3}{4}=-2$

$y_2-3=-8$

$x_2=-5$

Other end point is (-4,-5)