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unit 1 exam review what is the location of the point that is 1/4 of the…

Question

unit 1 exam review
what is the location of the point that is 1/4 of the distance between the points u(-1,2) and b(5,10) starting from point u.
** enter you answer in fractional form. no spaces
** dont for get the parentheses for the coordinate.
point 1/4 the distance from u to b =
what if we start from point b instead?
point 1/4 the distance from b to u =

Explanation:

Step1: Find the difference in x - coordinates

The x - coordinate of U is $x_U=-1$ and of B is $x_B = 5$. The difference $\Delta x=x_B - x_U=5-(-1)=6$.

Step2: Find the difference in y - coordinates

The y - coordinate of U is $y_U = 2$ and of B is $y_B=10$. The difference $\Delta y=y_B - y_U=10 - 2=8$.

Step3: Calculate the x - coordinate of the point 1/4 from U to B

The x - coordinate of the point P that is 1/4 of the distance from U to B is $x=x_U+\frac{1}{4}\Delta x=-1+\frac{1}{4}\times6=-1+\frac{6}{4}=\frac{-4 + 6}{4}=\frac{2}{4}=\frac{1}{2}$.

Step4: Calculate the y - coordinate of the point 1/4 from U to B

The y - coordinate of the point P is $y=y_U+\frac{1}{4}\Delta y=2+\frac{1}{4}\times8=2 + 2=4$. So the point 1/4 of the distance from U to B is $(\frac{1}{2},4)$.

Step5: Calculate the point 1/4 of the distance from B to U

For the point 1/4 of the distance from B to U, we use the formulas $x=x_B-\frac{1}{4}\Delta x$ and $y=y_B-\frac{1}{4}\Delta y$.
$x = 5-\frac{1}{4}\times6=5-\frac{6}{4}=\frac{20 - 6}{4}=\frac{14}{4}=\frac{7}{2}$
$y=10-\frac{1}{4}\times8=10 - 2=8$. So the point 1/4 of the distance from B to U is $(\frac{7}{2},8)$.

Answer:

$(\frac{1}{2},4)$
$(\frac{7}{2},8)$