Chapter 13, Problem 12CR

To determine

To find the coordinate of point U

Answer to Problem 12CR

U(3,1)

Explanation of Solution

Given,

U is on $\overline{\text{QT}}$

  $\overline{UR}\left|\right|\overline{\text{ST}}$ (Line UR is parallel to line ST)

Given Points are

  $\text{T(7,4) S(10,1) R(6,-2)}$

If two lines are parallel then their slopes are equal.

Let coordinate of U be $\text{(}x,y\text{)}$

Slope of a line , $\text{m =}\frac{y_2-y_1}{x_2-x_1}$

Therefore , slope of $\overline{\text{ST}}=\frac{4-1}{7-10}=\frac{3}{-3}=-1$

Slope of $\overline{\text{UR}}=\frac{-2-y}{6-x}$

Since both slopes are equal,

  $\frac{-2-y}{6-x}=-1$

  $-2-y=-6+x$

  $\begin{array}{l}-2+6=x+y\\ x+y=4\cdots \cdots \cdots (a)\end{array}$

Now,

Given $\text{ U}$ is on line $\overline{\text{QT}}$

Slope of $\overline{\text{QT}}\text{ =}\frac{y_2-y_1}{x_2-x_1}=\frac{4-(-2)}{7-(-1)}=\frac{6}{8}=\frac{3}{4}$

Point on $\overline{\text{QT}}$ is $\text{(-1,-2)}$

Therefore equation of line $\text{ QT : }y-(-2)=\frac{3}{4}(x-(-1))$

[Equation of line : $y-y_o=m(x-x_o)$ ]

Equation of line $\text{QT}$ is

  $\begin{array}{l}\text{4(}y+2)=3(x+1)\\ \text{4y+8 = 3x+3}\\ \text{3}x-4y=5\cdots \cdots (b)\end{array}$

Solving equations (a) & (b), we get,

  $\begin{array}{l}x+y=4\cdots \cdots \cdots (a)\\ \text{ 3}x-4y=5\cdots \cdots (b)\\ (a)*3\Rightarrow \\ \text{ 3}x+3y=12\cdots \cdots (c)\\ (c)-(b)\Rightarrow \\ \text{ 7}y=12-5=7\\ y=1\end{array}$

Substituting $\text{y=1}$ in (a)

  $\begin{array}{l}x+1=4\\ x=3\end{array}$

Therefore, the coordinate $\text{U(}x,y)\text{ = U(3,1)}$