Question

chapter ii.4
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the line segment (l) has endpoints ((-11, -9)) and ((-4, 10)). determine the point (p) in (l) so that the distance from ((-11, -9)) to (p) is (\frac{4}{13}) times the distance from ((-11, -9)) to ((-4, 10)).

Explanation:

Step1: Use section - formula

The section - formula for a point $P(x,y)$ that divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$ is given by $x=\frac{mx_2+nx_1}{m + n}$ and $y=\frac{my_2+ny_1}{m + n}$. Here, $x_1=-11,y_1=-9,x_2=-4,y_2 = 10$, and $m=\frac{4}{13}$, $n = 1-\frac{4}{13}=\frac{9}{13}$.

Step2: Calculate the x - coordinate of point $p$

$x=\frac{\frac{4}{13}\times(-4)+\frac{9}{13}\times(-11)}{\frac{4}{13}+\frac{9}{13}}=\frac{\frac{-16-99}{13}}{\frac{13}{13}}=\frac{-115}{13}=- \frac{115}{13}\approx - 8.85$.

Step3: Calculate the y - coordinate of point $p$

$y=\frac{\frac{4}{13}\times10+\frac{9}{13}\times(-9)}{\frac{4}{13}+\frac{9}{13}}=\frac{\frac{40 - 81}{13}}{\frac{13}{13}}=\frac{-41}{13}\approx - 3.15$.

Answer:

$(-\frac{115}{13},-\frac{41}{13})$