Answer :
Certainly! Let's look into how to determine the ratio into which point [tex]\( D(2, 4) \)[/tex] divides the line segment [tex]\(\overline{AB}\)[/tex] with given points [tex]\( A(-7, -14) \)[/tex] and [tex]\( B(5, 10) \)[/tex].
To find this ratio, we can use the section formula for internal division. The section formula states that a point [tex]\(D(x, y)\)[/tex] dividing the line segment [tex]\(\overline{AB}\)[/tex] into a ratio [tex]\(k : 1\)[/tex] can be found using the following formula:
[tex]\[ D(x, y) = \left( \frac{k \cdot x_2 + x_1}{k+1}, \frac{k \cdot y_2 + y_1}{k+1} \right) \][/tex]
Given:
- [tex]\(A(x_1, y_1) = (-7, -14)\)[/tex]
- [tex]\(B(x_2, y_2) = (5, 10)\)[/tex]
- [tex]\(D(x, y) = (2, 4)\)[/tex]
We need to solve for [tex]\(k\)[/tex].
First, let's use the x-coordinates to set up the equation:
[tex]\[ x = \frac{k \cdot x_2 + x_1}{k + 1} \][/tex]
[tex]\[ 2 = \frac{k \cdot 5 + (-7)}{k + 1} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ 2(k + 1) = 5k - 7 \][/tex]
[tex]\[ 2k + 2 = 5k - 7 \][/tex]
[tex]\[ 9 = 3k \][/tex]
[tex]\[ k = \frac{9}{3} \][/tex]
[tex]\[ k = 3 \][/tex]
Now let's use the y-coordinates to confirm:
[tex]\[ y = \frac{k \cdot y_2 + y_1}{k + 1} \][/tex]
[tex]\[ 4 = \frac{k \cdot 10 + (-14)}{k + 1} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ 4(k + 1) = 10k - 14 \][/tex]
[tex]\[ 4k + 4 = 10k - 14 \][/tex]
[tex]\[ 18 = 6k \][/tex]
[tex]\[ k = \frac{18}{6} \][/tex]
[tex]\[ k = 3 \][/tex]
Both coordinates x and y give us the same value for [tex]\(k\)[/tex], which is 3.
Therefore, the point [tex]\(D(2, 4)\)[/tex] divides the line segment [tex]\(\overline{AB}\)[/tex] in the ratio [tex]\(\mathbf{3:1}\)[/tex].
To find this ratio, we can use the section formula for internal division. The section formula states that a point [tex]\(D(x, y)\)[/tex] dividing the line segment [tex]\(\overline{AB}\)[/tex] into a ratio [tex]\(k : 1\)[/tex] can be found using the following formula:
[tex]\[ D(x, y) = \left( \frac{k \cdot x_2 + x_1}{k+1}, \frac{k \cdot y_2 + y_1}{k+1} \right) \][/tex]
Given:
- [tex]\(A(x_1, y_1) = (-7, -14)\)[/tex]
- [tex]\(B(x_2, y_2) = (5, 10)\)[/tex]
- [tex]\(D(x, y) = (2, 4)\)[/tex]
We need to solve for [tex]\(k\)[/tex].
First, let's use the x-coordinates to set up the equation:
[tex]\[ x = \frac{k \cdot x_2 + x_1}{k + 1} \][/tex]
[tex]\[ 2 = \frac{k \cdot 5 + (-7)}{k + 1} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ 2(k + 1) = 5k - 7 \][/tex]
[tex]\[ 2k + 2 = 5k - 7 \][/tex]
[tex]\[ 9 = 3k \][/tex]
[tex]\[ k = \frac{9}{3} \][/tex]
[tex]\[ k = 3 \][/tex]
Now let's use the y-coordinates to confirm:
[tex]\[ y = \frac{k \cdot y_2 + y_1}{k + 1} \][/tex]
[tex]\[ 4 = \frac{k \cdot 10 + (-14)}{k + 1} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ 4(k + 1) = 10k - 14 \][/tex]
[tex]\[ 4k + 4 = 10k - 14 \][/tex]
[tex]\[ 18 = 6k \][/tex]
[tex]\[ k = \frac{18}{6} \][/tex]
[tex]\[ k = 3 \][/tex]
Both coordinates x and y give us the same value for [tex]\(k\)[/tex], which is 3.
Therefore, the point [tex]\(D(2, 4)\)[/tex] divides the line segment [tex]\(\overline{AB}\)[/tex] in the ratio [tex]\(\mathbf{3:1}\)[/tex].
