Question

question 7 of 10
in line segment ab with endpoints a(5,9) and b(-2,16), the point p is located $\frac{6}{7}$ of the way along ab from point b.
determine the coordinates of point p.

Explanation:

Step1: Recall the section - formula

If a point \(P(x,y)\) divides the line - segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\) from \(B\), then \(x=\frac{mx_1+nx_2}{m + n}\) and \(y=\frac{my_1+ny_2}{m + n}\). Here, \(A(5,0)\), \(B(-2,16)\), and the ratio of \(BP\) to \(BA\) is \(\frac{6}{7}\), so \(m = 6\) and \(n=7 - 6=1\), \(x_1 = 5\), \(y_1 = 0\), \(x_2=-2\), \(y_2 = 16\).

Step2: Calculate the \(x\) - coordinate of \(P\)

\[x=\frac{6\times5+1\times(-2)}{6 + 1}=\frac{30-2}{7}=\frac{28}{7}=4\]

Step3: Calculate the \(y\) - coordinate of \(P\)

\[y=\frac{6\times0+1\times16}{6 + 1}=\frac{0 + 16}{7}=\frac{16}{7}\]

Answer:

\((4,\frac{16}{7})\)