Question

in triangle pmt, $gh = x + 3$, $pg = 4x + 1$, and $gj = 2x - 1$
what is the length of segment pg?
$x=$ type your answer...
pg= type your answer...
do the following points create a parallelogram: k(-3,2), l(4,2), n(-4,-2), m(3,-2)?
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Explanation:

Step1: Identify segment relationship

In a triangle, the intersection of medians (centroid) divides each median into a ratio of $2:1$, so $PG = 2 \times GJ$.

Step2: Substitute given expressions

Substitute $PG = 4x+1$ and $GJ = 2x-1$ into the ratio equation.
$$4x + 1 = 2(2x - 1)$$

Step3: Solve for $x$

Expand and simplify the equation.
$$4x + 1 = 4x - 2$$
Subtract $4x$ from both sides:
$$1 = -2$$
This is a contradiction, which means we use the other median relationship: $PG = 2 \times GH$ (since $GH$ is the shorter segment of the median containing $PG$)

Step4: Set up correct equation

$$4x + 1 = 2(x + 3)$$

Step5: Expand and solve for $x$

$$4x + 1 = 2x + 6$$
Subtract $2x$ from both sides:
$$2x + 1 = 6$$
Subtract 1 from both sides:
$$2x = 5$$
$$x = \frac{5}{2} = 2.5$$

Step6: Calculate length of $PG$

Substitute $x=2.5$ into $PG = 4x+1$.
$$PG = 4(2.5) + 1 = 10 + 1 = 11$$

A quadrilateral is a parallelogram if both pairs of opposite sides are parallel (equal slopes) or both pairs of opposite sides are congruent (equal lengths).

1. Calculate slopes:

Slope of $KL$: $\frac{2-2}{4-(-3)} = 0$
Slope of $NM$: $\frac{2-(-2)}{3-(-4)} = \frac{4}{7}$
Slope of $KN$: $\frac{-2-2}{-4-(-3)} = 4$
Slope of $LM$: $\frac{2-2}{3-4} = 0$
Opposite slopes are not equal, so check side lengths:
Length of $KL$: $\sqrt{(4-(-3))^2 + (2-2)^2} = 7$
Length of $NM$: $\sqrt{(3-(-4))^2 + (2-(-2))^2} = \sqrt{49+16} = \sqrt{65}$
Length of $KN$: $\sqrt{(-4-(-3))^2 + (-2-2)^2} = \sqrt{1+16} = \sqrt{17}$
Length of $LM$: $\sqrt{(3-4)^2 + (2-2)^2} = 1$
Opposite sides are not congruent.

Answer:

$x=2.5$
$PG=11$

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For the parallelogram question: