Dividing an interval externally in a given ratio (1 Viewer)

[ashbee28]

I need some help with this question, I have no idea how to solve it :s

"The point P(5,7) divides the interval joining the point A(-1,1) and B(3,5) externally in the ratio k:1. Find the value of k."

How do I find k? What formula do I use?

Thanks

 

[Timske]

Same formula plug in the values and solve for k.

<a href="http://www.codecogs.com/eqnedit.php?latex=P = \left ( \frac{mx_{2} - nx_{1}}{m-n} , \frac{my_{2} -ny_{1}}{m-n}\right )\\\\ (5 ,7) = \left ( \frac{k(3) -1(-1)}{k-1}, \frac{k(5)-1(1)}{k-1} \right )\\\\ (5.7) = \left ( \frac{3k @plus; 1}{k-1}, \frac{5k -1}{k-1} \right ) \\\\\therefore 5 = \frac{3k@plus;1}{k-1} ~and~ 7= \frac{5k-1}{k-1} \\\\ 5k-5=3k@plus;1\\\2k = 6 \\\ \therefore k = 3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?P = \left ( \frac{mx_{2} - nx_{1}}{m-n} , \frac{my_{2} -ny_{1}}{m-n}\right )\\\\ (5 ,7) = \left ( \frac{k(3) -1(-1)}{k-1}, \frac{k(5)-1(1)}{k-1} \right )\\\\ (5.7) = \left ( \frac{3k + 1}{k-1}, \frac{5k -1}{k-1} \right ) \\\\\therefore 5 = \frac{3k+1}{k-1} ~and~ 7= \frac{5k-1}{k-1} \\\\ 5k-5=3k+1\\\2k = 6 \\\ \therefore k = 3" title="P = \left ( \frac{mx_{2} - nx_{1}}{m-n} , \frac{my_{2} -ny_{1}}{m-n}\right )\\\\ (5 ,7) = \left ( \frac{k(3) -1(-1)}{k-1}, \frac{k(5)-1(1)}{k-1} \right )\\\\ (5.7) = \left ( \frac{3k + 1}{k-1}, \frac{5k -1}{k-1} \right ) \\\\\therefore 5 = \frac{3k+1}{k-1} ~and~ 7= \frac{5k-1}{k-1} \\\\ 5k-5=3k+1\\\2k = 6 \\\ \therefore k = 3" /></a>

 

[FluffyPuma]

Sorry that's wrong. It's mx+nx and m+n not minus.

 

[Kurosaki]

It's external

 

[lambertn01]

What about if they dont give you the 1? its two unknowns?

 

[cricketfan1997]

use cross method

 

[RealiseNothing]

What about if they dont give you the 1? its two unknowns?

If they give you the ratio as m:n there are two ways.

1) You can use the formula and find the x co-ordinate and y co-ordinate, which will be in terms of m and n. Then solve these two simultaneously.

2) We can consider the y co-ordinates (or x, but in this case y is easier). Height of BP is 7-5=2 and height of AP is 7-1=6. So the ratio of AP:BP is 6:2 = 3:1. Therefore m=3 and n=1.

Method 2 works as the line ABP is made up of similar triangles, and so the ratio of the hypotenuses is the same as the ratio of the heights and widths.