# Town A is 240 km East and 70 km North of O. Town B is 480 km East and 250 km North of O. Town X is 339 km East and 238 km North of O. At A the airplane changes direction so it now flies towards...

Town A is 240 km East and 70 km North of O.

Town B is 480 km East and 250 km North of O.

Town X is 339 km East and 238 km North of O.

At A the airplane changes direction so it now flies towards B. Point Y is located between A and B, where the airplane comes closest to X.

Calculate the distance XY.

(The answer at the back of my book is 75 km but I don't know how to find that. Please help! Thanks)

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The problem is of coordinate geometry/algebra. Imagine the Origin as "O" and you get coordinates of point A, B, and X as under:

A is (240,70), B is (480,250) and X is (339,238).

The general equation of straight line is y = mx+c

For line passing through A and B:

m = (y2-y1)/(x2,x1) = (250-70)/(480-240) = 180/240 = 3/4

using coordinates of A and value of m in equation of line, we get:

70 = (3/4)*240 + c

70 = 180 + c

c = 70-180 = -110

The equation of line passing through A and B is y = (3/4)*x-110 or

3x-4y = 440 ........ (1)

The shotest distance between this line is the perpendicular distance of this line from point X.

The slope of line perpendicular to line line 3x-4y= 440 is -1/m or -4/3

Line passing through point (339,238) and slope -4/3 is:

(y-238) = (-4/3)*(x-339)

3*(y-238) = -4*(x-339)

4x+3y = 2070 ........ (2)

to solve (1) and (2) for the point of intersection, we multiply equations (1) by 3 and (2) by 4 to get:

9x-12y = 1320 ........ (3)

16x+12y = 8280 ....... (4)

adding (3) and (4)

25x = 9600

x = 9600/25 = 384, substituting this value of x in (2), we get:

y = (2070-4*384)/3 = 534/3 = 178

so the Y point nearest to point X is (384, 178)

**The point Y is 384 east and 178 North of O**

The shortest distance between xy (339,238) & (384,178) is:

sqrt[(384-339)^2+(178-238)^2) = sqrt(5625) = 75 Km

**The shortest distance XY is 75 Km**