Division of Segments and Angles Level 5 In this video we will go over more challenging
examples involving midpoints and segment bisectors.
Let’s take a look at the first example.
Point E is the midpoint of segment DF.
Find the value of x if the measurement of segment DE equals 2x minus 3 and the measurement
of segment EF equals 5x minus 24.
In this problem we are provided with a graph of two line segments intersecting at a common
point.
We are given algebraic expressions for the lengths of two line segments, we are asked
to find the value of x.
Let’s start by marking the diagram with the given information.
We know that point E is the midpoint of segment DF this means that segment DE and segment
EF are congruent so we mark these two segments with a single tick mark.
We also know that the length of segment DE equals 2x minus 3 and the length of segment
EF equals 5x minus 24.
Since we know that both segment DE and segment EF are congruent, we can go ahead and set
these two algebraic expressions equal to one another.
Now we have an equation for which we can solve for x.
So we subtract 2x from both sides of the equation.
Then we add 24 to both sides of the equation.
Lastly we divide both sides of the equation by 3, simplifying the expression we obtain
x equals 7 as the final answer.
All right let’s take a look at the next example.
Given that line TP bisects segment VS and segment MR.
Segment VM is congruent to segment SR.
The length of segment MP is equal to 9.
The length of segment VT is equal to 6.
The perimeter of figure MRSV is equal to 62.
Find the length of segment VM.
In this problem we are provided with a diagram of a figure along with various geometric relations.
We are asked to find the length of a particular segment.
We will start solving this problem by marking the diagram with all the given information;
this will help plan our approach in solving this problem.
We know that line TP bisects segment VS and segment MR so we can go ahead and draw single
tick marks on segment VT and segment TS and double tick marks on segment MP and segment
PR since these two pairs of segments are congruent to one another.
In addition we are also given that segment VM is congruent to segment SR so we will use
triple tick marks to illustrate this on the diagram.
Next we go ahead and label the measurement of segment MP and label the measurement of
segment VT in this case they are equal to 9 and 6 respectively.
Lastly we can write a geometric expression for the perimeter of the figure in this case
segment VM plus segment MR plus segment RS plus segment SV is equal to 62.
Now let’s figure out how we can determine the length of segment VM.
We will start by assigning a variable to our unknown length so let’s label the length
of segment VM with the variable x.
Since segment SR is congruent to segment VM we can also label this length with the variable
x.
We also know that the length of segment PR is equal to 9 since it is congruent to segment
MP.
Similarly we also know that the length of segment TS is equal to 6 since it is congruent
to segment VT.
Now we can take all of these measurements and substitute them into our perimeter expression
so we replace the length of segment VM and segment SR with x and replace the length of
segment VS and MR with 12 and 18 respectively.
Now it is just a matter of solving for x so we collect like terms, subtract the constant,
and divide both sides of the equation by 2 doing that and simplifying we obtain x equals
16.
Since this value of x also represents the length of segment VM we are essentially done
with the problem so this is our final answer.
All right let’s end the video with the final example.
Given the diagram find the value of x.
Determine if point Q is the midpoint of segment PR.
In this problem we are provided with a diagram of a line segment broken into two smaller
line segments.
We are also given the measurement of line segment PR and algebraic expressions representing
the measurement of the smaller line segments.
We are asked to determine the value of x and determine if point Q is the midpoint of segment
PR.
Let’s start by determining the value of x.
Notice that we have expressions for the measurement of all three line segments we also know from
the diagram that point P, Q and R are collinear this also allows us to set up the following geometric
relation, the length of segment PQ plus the length of segment QR is equal to the length
of segment PR.
We can now substitute the algebraic and numerical expressions into this geometric relation,
doing that we obtain the following equation.
From here it just a matter of solving for x.
Notice that we are face to face with a quadratic equation so let’s move every single term
to one side of the equation so that we can set the equation equal to 0.
Making sure we add like terms along the way.
We also want to write the quadratic equation in standard form.
Next we need to see if this quadratic expression is factorable so we use the factors of the
constant term in this case 4 and negative 2 and use them to set up two binomials in
this case the quantity x plus 4 times the quantity x minus 2.
Having factored the quadratic expression we go ahead and use the zero product property
and set each factor equal to 0.
We then solve for x for each factor, doing that we obtain x equals negative 4 and x equals
positive 2.
Now we need figure out if we can use both of these solutions, we can figure this out
by substituting each value of x into the algebraic expression representing the two smaller line
segments if we obtain positive measurements then we will know that it is a solution to
the problem.
If we obtain negative measurements we need to eliminate the value of x since measurements
must be positive and not negative.
Let’s first substitute negative 4 into each algebraic expression and simplify.
Notice that the measurement of segment QR ends up being negative so we have to discard
negative 4 as a potential solution.
Now let’s substitute positive 2 into each algebraic expression and simplify.
In this case we obtain positive measurements so let’s use this value of x to finish the
problem.
We need to determine if point Q is the midpoint of segment PR.
This will be the case if the measurement of segment PQ is equal to the measurement of
segment QR.
Looking the measurements for each segment when x equals positive 2 we conclude that
point Q is not the midpoint of segment PR since both measurements are not equal and
this is our final answer.
All right in our next video we will go over examples that involve adjacent angles, angle
bisectors, and angle trisectors.