Calculus

[jenna.fernandes]

Given A(1,3) , B(5,1) and P|AB in ratio 3:1 , determine point P

 

[galactus]

It would help if you worded this problem so we can understand what you mean.

I assume you want to find the coordinates of a point, P, 3/4 the way up the line formed by (1,3) and (5,1).

The ratio 3:1 means it is 3/4 the way between point A(1,3) and B(5,1).

The equation of the line given by the two points A and B is \(\displaystyle y=\frac{-1}{2}x+\frac{7}{2}\) and per the distance formula the distance between them is

\(\displaystyle \sqrt{(1-5)^{2}+(3-1)^{2}}=2\sqrt{5}\)

The distance between the point P with coordinates (x,y) and point A is given by the distance formula, \(\displaystyle d^{2}=(x-1)^{2}+(y-3)^{2}\)

From the ratio this distance is \(\displaystyle \frac{3\sqrt{5}}{2}\)

But, subbing in the line equation for y gives \(\displaystyle (x-1)^{2}+(\frac{-1}{2}x+\frac{1}{2})^{2}=\frac{45}{4}\)

Now, solving for x we get x=4 and y=3/2

Vectors could be used to solve this as well. Which is much easier.

\(\displaystyle AP=\frac{3}{4}AB\)

\(\displaystyle (x-1, \;\ y-3)=\frac{3}{4}(4,-2)=(3, \;\ \frac{-3}{2})\)

So, \(\displaystyle x-1=3, \;\ y-3=\frac{-3}{2}\)

\(\displaystyle x=4, \;\ y=\frac{3}{2}\)

 

[BigGlenntheHeavy]

\(\displaystyle A \ = \ (x_1,y_1) \ = \ (1,3), \ B \ = \ (x_2,y_2) \ = \ (5,1), \ P \ = \ (x,y), \ ratio \ = \ 3:1\)

\(\displaystyle Ergo, \ x \ = \ \frac{x_1+rx_2}{1+r} \ = \ \frac{1+3(5)}{1+3} \ = \ 4 \ and \ y \ = \ \frac{y_1+ry_2}{1+r} \ = \ \frac{3+3(1)}{1+3} \ = \ \frac{3}{2}.\)

\(\displaystyle Therefore, \ P \ = \ (x,y) \ = \ (4,\frac{3}{2})\)

 

[soroban]

\(\displaystyle \text{Hello, Jenna!}\)

\(\displaystyle \text{You can }talk\text{ your way through this one . . .}\)

Given A(1,3) , B(5,1) and P|AB in ratio 3:1 , determine point P.

        |   A
      3 + - *
        |   :   *
        |   :       *   P
        |   :           o   B
      1 + - : - - - - - - - * 
        |   :               :
     ---+---+---------------+----
        |   1               5

\(\displaystyle P\text{ divides }AB\text{ in the ratio }3:1 \quad\Rightarrow\quad \frac{AP}{PB} = \frac{3}{1}\)
. . \(\displaystyle \text{Hence, }P \:=\:\tfrac{3}{4}(AB)\)
\(\displaystyle \text{That is, }P\text{ is }\tfrac{3}{4}\text{ of the distance from }A\text{ to }B.\)


\(\displaystyle \text{Then }P_x\text{ is }\tfrac{3}{4}\text{ of the horizontal distance from }A\text{ to }B.\)

\(\displaystyle AP_x \:=\:\tfrac{3}{4}(5-1) \:=\:3\)

. . \(\displaystyle \text{So: }\_x \:=\:1 + 3 \:=\:4\)


\(\displaystyle \text{And }P_y\text{ is }\tfrac{3}{4}\text{ of the vertical distance from }A\text{ to }B.\)

\(\displaystyle AP_y \:=\:\tfrac{3}{4}(1-3) \:=\:-\tfrac{3}{2}\)

. . \(\displaystyle \text{So: }\_y \:=\:3 + \left(-\tfrac{3}{2}\right) \:=\:\tfrac{3}{2}\)


\(\displaystyle \text{Therefore: }\\left(4,\tfrac{3}{2}\right)\)

 

[pka]

Given A(1,3) , B(5,1) and P|AB in ratio 3:1 , determine point P

Here is a vector solution.
\(\displaystyle \overrightarrow {AB} = B - A = \left\langle {4, - 2} \right\rangle \;\& \,\overline {AB} (t) = A + t\overrightarrow {AB} = \left\langle {1 + 4t,3 - 2t} \right\rangle ,\;0 \le t \le 1\).

The point you want is \(\displaystyle \overline {AB} \left( {\frac{3}{4}} \right) = \left( {4,\frac{3}{2}} \right)\).