Geometry

[IloveManUtd]

Hi, I've got a geometry question that goes like this:

The coordinates of A, B and C are (12,6), (16,-8) and (-4,-14) respectively. P is the point on AC such that CP= 3PA. The line through P and perpendicular to AC cuts the y-axis at the point d. Find:
a) the coordinates of P
b) the equation of the line PD
c) the coordinates of D
d) the area of the quadrilateral ABCD

Thx!

 

[mmm4444bot]

What have you done so far?

I began by calculating the distance AC, followed by dividing it into fourths.

 

[soroban]

Hello, IloveManUtd!

Three points: .\(\displaystyle A(12,6),\; B(16,-8),\;C(-4,-14).\)
\(\displaystyle P\) is the point on \(\displaystyle AC\) such that: \(\displaystyle CP = 3\!\cdot\!PA.\)
The line through \(\displaystyle P\) and perpendicular to \(\displaystyle AC\) cuts the \(\displaystyle y\)-axis at the point \(\displaystyle D.\)

Find:

(a) the coordinates of \(\displaystyle P\)

Going from \(\displaystyle C\) to \(\displaystyle A\), we move 16 units right and 20 units up.

                   (12,6)
                      o A
                    * :
                  *   :
                *     :
              *       : 20
            *         :
          *           :
        *             :
    C o - - - - - - - +
  (-4,-14)    16

\(\displaystyle \text{Since }CP = \tfrac{3}{4}CA\text{, going from }C\text{ to }P,\)

. . \(\displaystyle \text{we move }\tfrac{3}{4}(16) = 12\text{ units right, and }\tfrac{3}{4}(20) = 15\text{ units up.}\)

                      A
                      o
                  P *
                  o 
                * : 
              *   : 
            *     : 15
          *       :
        *         :
    C o - - - - - +
  (-4,-14)  12

Therefore: .\(\displaystyle P(8,\,1)\)

(b) the equation of the line \(\displaystyle PD\)

\(\displaystyle \text{The slope of }AC\text{ is: }\:\frac{6-(\text{-}14)}{12-(\text{-}4)} \:=\:\frac{20}{16} \:=\:\frac{5}{4}\)

\(\displaystyle \text{Line }PD\text{ contains point }(8,\,1)\text{ and has slope }\,m \,=\,-\tfrac{4}{5}\)

\(\displaystyle \text{Write its equation.}\)

(c) the coordinates of \(\displaystyle D\)

\(\displaystyle \text{This is the }y\text{-intercept of the line in part (b).}\)

(d) the area of the quadrilateral \(\displaystyle ABCD\)

\(\displaystyle \text{The area of a triangle with vertices: }\,(x_1,y_1),\;(x_2,y_2),\;(x_3,y_3)\)

. . \(\displaystyle \text{is given by: }\;A\;=\;\tfrac{1}{2}\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3\end{vmatrix} \qquad\left(\begin{array}{c}\text{Take the absolute value} \\ \text{of the determinent.}\end{array}\right)\)


\(\displaystyle \text{Apply this formula to }\Delta ABC\text{ and }\Delta ACD.\)

 

[BigGlenntheHeavy]

\(\displaystyle r \ = \ \frac{CP}{PA} \ = \ \frac{3}{1} \ = \ 3\)

\(\displaystyle Hence, \ x \ = \ \frac{-4+3(12)}{1+3} \ = \ 8 \ and \ y \ = \ \frac{-14+3(6)}{1+3} \ = \ 1\)

\(\displaystyle Ergo, \ P \ = \ (8,1) \ and \ PD \ = \ [4x+5y \ = \ 37], \ \implies \ D \ = \ (0,37/5)\)

\(\displaystyle Area[ABCD] \ = \ \frac{1}{2}[(0)(6)+(12)(-8)+(16)(-14)+(-4)(37/5)-(0)(-14)-(-4)(-8)-(16)(6)\)

\(\displaystyle -(12)(37/5)] \ = \ 283.2 \ sq. \ units\)

 

[mmm4444bot]

I got Soroban's and Glenn's results.

I used a different approach, to find the coordinates of P.

Through the distance formula and the relationship CP = 3*AP:

x^2 - 28x + y^2 - 17y + 176 = 0

The line through CA is y = 5/4 x - 9

Substituting the latter into the former gives a quadratic equation.

41x^2 - 1148x + 6560 = 0.

The solutions are x = 8 or x = 20, so there are two points on that line such that CP = 3AP. We want x = 8 because P is on the segment CA.

I finished with the others' methods.

For those not familiar with determinants, here is the formula for the area of a triangle with vertex coordinates (a, b), (c, d), and (e, f).

\(\displaystyle A = \left | \frac{(a - c) \cdot (d - f) \; - \; (c - e) \cdot (b - d)}{2} \right |\)