Question

question 8 of 10
what is the distance from point y to \overleftrightarrow{wx} in the figure below?

a. 3
b. 15
c. 15\sqrt{2}
d. 7.5
e. 15\sqrt{3}
f. cannot be determined

Explanation:

Step1: Recall the definition of the distance from a point to a line

The distance from a point to a line is the length of the perpendicular segment from the point to the line. In the given figure, \(YZ\) is perpendicular to \(\overleftrightarrow{WX}\).

Step2: Analyze the given lengths

We are given that \(YZ\) is the perpendicular segment from \(Y\) to \(\overleftrightarrow{WX}\). The length of \(YZ\) is not directly affected by the lengths \(WY = 39\), \(WZ=36\) (since we can use the right - triangle relationship for \(\triangle YZX\) where \(YX = 15\sqrt{2}\), \(ZX = 15\). By the Pythagorean theorem \(YZ^{2}+ZX^{2}=YX^{2}\). Let \(YZ = h\), then \(h^{2}+15^{2}=(15\sqrt{2})^{2}\).

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Another way: Since \(YZ\) is the perpendicular from \(Y\) to \(\overleftrightarrow{WX}\), and we know that in a right - triangle formed by the perpendicular, the segment on the line and the hypotenuse (connecting the external point to a point on the line), and we can also note that for the right - triangle with hypotenuse \(YX=15\sqrt{2}\) and one leg \(ZX = 15\), using the Pythagorean triple or direct calculation \(YZ = 15\) (because if \(a^{2}+b^{2}=c^{2}\), and \(c = 15\sqrt{2}\), \(b=15\), then \(a=\sqrt{(15\sqrt{2})^{2}-15^{2}}=\sqrt{450 - 225}=\sqrt{225}=15\))

Answer:

B. 15