Question

if q is the mid - point of pr, find the coordinates of r if p(11,-2) and q(4,3)
(7.5,0.5)
(-3,8)

find the value of x.
(7x - 1)°
(6x - 1)°
14
83
97

Explanation:

Step1: Use mid - point formula for x - coordinate

Let the coordinates of $R$ be $(x_R,y_R)$. The mid - point formula for two points $P(x_P,y_P)$ and $R(x_R,y_R)$ with mid - point $Q(x_Q,y_Q)$ is $x_Q=\frac{x_P + x_R}{2}$. Given $x_P = 11$, $x_Q=4$, then $4=\frac{11 + x_R}{2}$. Cross - multiply: $8 = 11+x_R$, so $x_R=8 - 11=-3$.

Step2: Use mid - point formula for y - coordinate

The mid - point formula for the y - coordinate is $y_Q=\frac{y_P + y_R}{2}$. Given $y_P=-2$, $y_Q = 3$, then $3=\frac{-2 + y_R}{2}$. Cross - multiply: $6=-2 + y_R$, so $y_R=6 + 2 = 8$. So the coordinates of $R$ are $(-3,8)$.

Step3: Solve for x in the angle problem

Since $(7x - 1)^{\circ}$ and $(6x - 1)^{\circ}$ are vertical angles, they are equal. Set up the equation $7x-1=6x - 1$. Subtract $6x$ from both sides: $7x-6x-1=6x-6x - 1$, which simplifies to $x-1=-1$. Add 1 to both sides: $x=-1 + 1=0$. But this is wrong. They are supplementary angles, so $(7x - 1)+(6x - 1)=180$. Combine like terms: $13x-2 = 180$. Add 2 to both sides: $13x=180 + 2=182$. Divide both sides by 13: $x = 14$.

Answer:

1. Coordinates of $R$: $(-3,8)$
2. Value of $x$: $14$