Question

in exercises 1 and 2, find the coordinates of point p along the directed line segment st so that sp to pt is the given ratio.

1. ( s(8,4), t(-4,-8) ); 1 to 3
2. ( s(-6,7), t(9,25) ); 2 to 3

in exercises 3 and 4, find the coordinates of point q along the directed line segment lm so that lq to qm is the given ratio.

1. ( l(-1,-2), m(3,6) ); 5 to 3
2. ( l(2,7), m(-1,1) ); 2 to 1

Explanation:

Step1: Use the section formula

If a point \(P(x,y)\) divides the line segment joining \(S(x_1,y_1)\) and \(T(x_2,y_2)\) in the ratio \(m:n\), then the coordinates of \(P\) are given by \(x=\frac{mx_2 + nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\)

For Exercise 1: \(S(6,4)\), \(T(-4,-8)\); ratio \(m:n = 1:3\)

Step2: Calculate \(x\) - coordinate

\(x=\frac{1\times(-4)+3\times6}{1 + 3}=\frac{-4 + 18}{4}=\frac{14}{4}=\frac{7}{2}\)

Step3: Calculate \(y\) - coordinate

\(y=\frac{1\times(-8)+3\times4}{1+3}=\frac{-8 + 12}{4}=\frac{4}{4} = 1\)

For Exercise 2: \(S(-6,7)\), \(T(9,25)\); ratio \(m:n=2:3\)

Step2: Calculate \(x\) - coordinate

\(x=\frac{2\times9+3\times(-6)}{2 + 3}=\frac{18-18}{5}=0\)

Step3: Calculate \(y\) - coordinate

\(y=\frac{2\times25+3\times7}{2+3}=\frac{50 + 21}{5}=\frac{71}{5}=14.2\)

For Exercise 3: \(L(-1,-2)\), \(M(3,6)\); ratio \(m:n = 5:3\)

Step2: Calculate \(x\) - coordinate

\(x=\frac{5\times3+3\times(-1)}{5+3}=\frac{15-3}{8}=\frac{12}{8}=\frac{3}{2}\)

Step3: Calculate \(y\) - coordinate

\(y=\frac{5\times6+3\times(-2)}{5+3}=\frac{30 - 6}{8}=\frac{24}{8}=3\)

For Exercise 4: \(L(2,7)\), \(M(-1,1)\); ratio \(m:n=2:1\)

Step2: Calculate \(x\) - coordinate

\(x=\frac{2\times(-1)+1\times2}{2+1}=\frac{-2 + 2}{3}=0\)

Step3: Calculate \(y\) - coordinate

\(y=\frac{2\times1+1\times7}{2+1}=\frac{2+7}{3}=3\)

Answer:

1. \(P(\frac{7}{2},1)\)
2. \(P(0,14.2)\)
3. \(Q(\frac{3}{2},3)\)
4. \(Q(0,3)\)