Question

Trisect the segment connecting P(-5,2) and Q(3,-4).

Answer 1

Answer on Question #56705 – Math – Analytic Geometry

Question

Trisect the segment connecting $P(-5,2)$ and $Q(3,-4)$.

Solution

Segment PQ, that connects points $P(-5,2)$ and $Q(3,4)$, is divided into three equal parts by points $A(x_{a},y_{a})$, $B(x_{b},y_{b})$.

Besides, $x_{p} < x_{a} < x_{q}$, $x_{p} < x_{b} < x_{q}$, $y_{q} < y_{a} < y_{p}$, $y_{q} < y_{b} < y_{p}$, that is, $-5 < x_{a} < 3$, $-5 < x_{b} < 3$, $-4 < y_{a} < 2$, $-4 < y_{b} < 2$.

Points P, A, B, Q lie on the one line, so

$$\frac {x _ {a} - x _ {p}}{y _ {a} - y _ {p}} = \frac {x _ {b} - x _ {p}}{y _ {b} - y _ {p}} = \frac {x _ {q} - x _ {p}}{y _ {q} - y _ {p}} = \frac {8}{- 6} = - \frac {4}{3}$$

Length of original segment $L_{PQ} = \sqrt{(x_q - x_p)^2 + (y_q - y_p)^2} = \sqrt{8^2 + 6^2} = 10$,

Length of $L_{PA} = \frac{1}{3} L_{PQ} = \frac{10}{3} = \sqrt{(x_a - x_p)^2 + (y_a - y_p)^2} = \sqrt{\left(\frac{4^2}{3^2} + 1\right)(y_a - y_p)^2} = \pm \frac{5}{3}(y_a - y_p)$, we obtain $y_a - y_p = 2$ or $y_a - y_p = -2$, but the case of $y_a - y_p = 2$ is impossible, because $y_a - y_p < 0$. Then $y_a - y_p = -2$, hence $y_a = y_p - 2 = 2 - 2 = 0$.

It follows from $\frac{x_a - x_p}{y_a - y_p} = -\frac{4}{3}$ and $y_a - y_p = -2$ that

$$x _ {a} - x _ {p} = - \frac {4}{3} (- 2) = \frac {8}{3}, \text{ hence}$$ $$x _ {a} = \frac {8}{3} + x _ {p} = \frac {8}{3} - 5 = - \frac {7}{3}.$$

Similarly for

$$L _ {P B} = \frac {2}{3} L _ {P Q} = \frac {2 0}{3} = \sqrt {(x _ {b} - x _ {p}) ^ {2} + (y _ {b} - y _ {p}) ^ {2}} = \sqrt {\left(\frac {4 ^ {2}}{3 ^ {2}} + 1\right) (y _ {b} - y _ {p}) ^ {2}} = \pm \frac {5}{3} (y _ {b} - y _ {p}), \text{ we obtain that } y _ {b} - y _ {p} = 4 \text{ or } y _ {b} - y _ {p} = - 4, \text{ but the case of}$$ $$y _ {b} - y _ {p} = 4 \text{ is impossible, because } y _ {b} - y _ {p} < 0.$$

Then $y_{b} - y_{p} = -4$, hence $y_{b} = y_{p} - 4 = 2 - 4 = -2$.

It follows from $\frac{x_b - x_p}{y_b - y_p} = -\frac{4}{3}$ and $y_a - y_p = -2$ that

$$x _ {b} - x _ {p} = - \frac {4}{3} (- 4) = \frac {1 6}{3}, \text{ hence}$$ $$x _ {b} = \frac {1 6}{3} + x _ {p} = \frac {1 6}{3} - 5 = \frac {1}{3}. \mathrm {T h u s}, (x _ {a}, y _ {a}) = \left(\frac {- 7}{3}, 0\right), (x _ {b}, y _ {b}) = \left(\frac {1}{3}, - 2\right).$$

Answer: the points that divided PQ into three equal parts are $A\left( {-\frac{7}{3},0}\right)$ and $B\left( {\frac{1}{3}, - 2}\right)$ .

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