In the diagram, the length of segment TQ is 40 units.

Line m is a perpendicular bisector of line segment S Q. It intersects line segment S Q at point R. Line m also contains points T and V. Line segment T S is 2 x + 8. Line segment S V is 3 x minus 4.
What is the length of segment QV?

32 units
36 units
40 units
44 units

1 answer

To solve for the length of segment QV, we start by analyzing the information given in the problem. We know that segment TQ measures 40 units, and we also have expressions for segments TS and SV depending on the value of \( x \):

  • \( TS = 2x + 8 \)
  • \( SV = 3x - 4 \)

Given that line m is the perpendicular bisector of segment SQ and intersects it at point R, we can infer that \( SR = RQ \).

Since TQ is 40 units long, we have that:

\[ TS + SR = TQ \]

From the properties of a bisector, we also know that:

\[ SR = RQ \]

Let's denote \( SR \) (and thus \( RQ \)) as \( y \). Therefore, we can express the length of segment SQ as:

\[ SQ = SR + RQ = 2y \]

Next, we can find the length of segment TS in terms of \( y \):

\[ TS = TQ - SR = 40 - y \]

Now, we set the two equations for \( TS \) equal to each other:

\[ 2x + 8 = 40 - y \quad \text{(1)} \]

Next, we can express the length of segment SV in terms of \( SR \):

\[ SV = y + SV = y + (3x - 4) = 40 \]

Using the equations we have, to find the values of \( x \) and then the corresponding lengths of each segment.

Now let’s isolate \( y \):

From Equation (1):

\[ y = 40 - (2x + 8) \] \[ y = 32 - 2x \quad \text{(2)} \]

Next, let's substitute value of y into \( SV \):

\[ SV = 3x - 4 \]

We know \( TQ = 40 \), and therefore we set the two equations for \( SV \):

\[ y + SV = 40 \]

Substituting:

\[ (32 - 2x) + (3x - 4) = 40 \]

Solving this gives us:

\[ 32 - 2x + 3x - 4 = 40 \] \[ 32 - 4 + x = 40 \] \[ x = 40 - 28 \] \[ x = 12 \]

Now substituting \( x = 12 \) back into the segment lengths:

Calculate \( TS \): \[ TS = 2(12) + 8 = 24 + 8 = 32 \]

Calculate \( SV \): \[ SV = 3(12) - 4 = 36 - 4 = 32 \]

Finally, since \( SQ = SR + RQ = 2y = TS + SV \) must hold due to properties of the bisector:

Since both lengths are equal, we can calculate \( RQ = 20 \) thus we get \( QV = SV + RQ \) thus:

\[ QV = 32 + 20 = 32. \]

Thus, the options given are ( 32 \text{ units,} 36 \text{ units,} 40 \text{ units,} 44 \text{ units.}] Thus the answer is:

32 units.

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