Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s). in the given diagram, △abc is a right triangle with (overline{dg}perpoverline{cb}). segment ab is divided into four equal parts. (a(-3, -1)) (c(-3, -3)) (b(6, -3)) the coordinates of point f are ( , ), and the coordinates of point g are ( , ).

Explanation:

Step1: Find the x - coordinate of point F

Since segment AB is divided into four equal parts and \(A(-3,-1)\), \(B(6,-3)\), the x - coordinate of F can be found using the section formula for equally - divided line segments. The x - coordinate of a point that divides the line segment joining \((x_1,y_1)\) and \((x_2,y_2)\) into \(n\) equal parts at the \(k^{th}\) division is \(x=x_1+\frac{k}{n}(x_2 - x_1)\). Here, \(n = 4\), \(k = 3\), \(x_1=-3\), \(x_2 = 6\). So \(x=-3+\frac{3}{4}(6+3)=-3+\frac{27}{4}=\frac{-12 + 27}{4}=\frac{15}{4}=3.75\).

Step2: Find the y - coordinate of point F

The y - coordinate of points on line AB can be found using the fact that the line AB is a non - vertical line. Since the y - coordinate of points on the line segment AB changes linearly, and the y - coordinate of \(A(-3,-1)\) and \(B(6,-3)\), the y - coordinate of F is \(y=-1+\frac{3}{4}(-3 + 1)=-1+\frac{3}{4}\times(-2)=-1-\frac{3}{2}=-\frac{2 + 3}{2}=-\frac{5}{2}=-2.5\).

Step3: Find the x - coordinate of point G

Since \(DG\perp CB\) and \(C(-3,-3)\), \(B(6,-3)\), and D is a point on AB such that \(AD:DB = 1:3\). The x - coordinate of D (and G since \(DG\perp CB\)) can be found using the section formula for the x - coordinate. The x - coordinate of D (and G) is \(x=-3+\frac{1}{4}(6 + 3)=-3+\frac{9}{4}=\frac{-12+9}{4}=-\frac{3}{4}=-0.75\).

Step4: Find the y - coordinate of point G

Since \(DG\perp CB\) and \(CB\) is a horizontal line with \(y=-3\), the y - coordinate of G is \(y=-3\).

Answer:

The coordinates of point F are \((3.75,-2.5)\), and the coordinates of point G are \((-0.75,-3)\)