Determine the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$.

Question

Accepted Answer

Let the given points be $A(-3, 10)$ and $B(6, -8)$, and let the dividing point be $P(-1, 6)$. Assume that point $P$ divides the line segment $AB$ in the ratio $k : 1$. 
Using the section formula for the x-coordinate: $$x = \frac{k(x2) + 1(x1)}{k + 1}$$ 
Substitute the given values into the formula: $$-1 = \frac{k(6) + 1(-3)}{k + 1}$| $$-1 = \frac{6k - 3}{k + 1}$| 
Now, solve for $k$: $$-(k + 1) = 6k - 3$$ $$-k - 1 = 6k - 3$$ $$3 - 1 = 6k + k$$ $$2 = 7k$$ $$k = \frac{2}{7}$| 
To verify, let us check with the y-coordinate using $k = 2/7$: $$y = \frac{\frac{2}{7}(-8) + 1(10)}{\frac{2}{7} + 1} = \frac{-\frac{16}{7} + 10}{\frac{9}{7}} = \frac{\frac{-16 + 70}{7}}{\frac{9}{7}} = \frac{54}{9} = 6$$ 
Since the y-coordinate matches the given value ($6$), our value of $k$ is correct. Thus, the required ratio is $k : 1 = \frac{2}{7} : 1 = 2 : 7$ internally.

MP Board · Class 10 · Mathematics · Coordinate GeometryDetermine the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$.