Question

6 if point d in the triangle bcd is located on the y - axis, what is the length of bd?
7 two boats head out on straight paths. a sailboat starts at point s, a motorboat starts at point m and they both pass through the point p.
approximately how far apart were they when they started their trips (the distance between points s and m)?
round the answer you calculated to the nearest km.

Explanation:

Problem 6 Solution:

Step1: Determine coordinates of D

Since D is on the y - axis, its x - coordinate is 0. Let \(D=(0,d)\). \(B=(2,13)\), \(C=(6,5)\). The slope of \(BC\): \(m_{BC}=\frac{5 - 13}{6 - 2}=\frac{-8}{4}=-2\). Since \(CD\perp BC\) (right angle at C), the slope of \(CD\) is the negative reciprocal of \(m_{BC}\), so \(m_{CD}=\frac{1}{2}\). The slope of \(CD\) using points \(C(6,5)\) and \(D(0,d)\) is \(m_{CD}=\frac{d - 5}{0 - 6}=\frac{d - 5}{-6}\). Set \(\frac{d - 5}{-6}=\frac{1}{2}\), cross - multiply: \(2(d - 5)=- 6\), \(2d-10=-6\), \(2d = 4\), \(d = 2\). So \(D=(0,2)\).

Step2: Calculate length of BD

Using distance formula between \(B(2,13)\) and \(D(0,2)\): \(BD=\sqrt{(2 - 0)^{2}+(13 - 2)^{2}}=\sqrt{4 + 121}=\sqrt{125}=5\sqrt{5}\approx11.18\) (or if we consider vertical distance? Wait, no, \(B\) has \(x = 2\), \(D\) has \(x = 0\), but wait, maybe a simpler way: since \(D\) is on \(y\) - axis \((x = 0)\) and \(B\) is at \((2,13)\), and we found \(D=(0,2)\), then the distance \(BD=\sqrt{(2 - 0)^2+(13 - 2)^2}=\sqrt{4 + 121}=\sqrt{125}=5\sqrt{5}\approx11.2\)

Problem 7 Solution:

Step1: Find slope of SM

The line \(y=\frac{1}{2}x + 8\) has slope \(m_1=\frac{1}{2}\). Since the motorboat's path is perpendicular to \(y=\frac{1}{2}x + 8\) (right angle at \(P\)), the slope of the motorboat's path (and thus the slope of \(PM\)) is the negative reciprocal of \(\frac{1}{2}\), so \(m_{PM}=-2\). The sailboat's path (SM) is along the line with slope \(\frac{1}{2}\) (since it passes through \(P(12,14)\) and \(S\) and is parallel to \(y=\frac{1}{2}x + 8\) in terms of slope? Wait, no, \(SM\) is along the line with slope \(\frac{1}{2}\). Let the coordinates of \(S\) be \((0,y)\) (since it's on the \(y\) - axis). Using the point - slope form for the line with slope \(\frac{1}{2}\) passing through \(P(12,14)\): \(y - 14=\frac{1}{2}(x - 12)\). When \(x = 0\) (since \(S\) is on \(y\) - axis), \(y-14=\frac{1}{2}(0 - 12)=-6\), \(y=14 - 6 = 8\). So \(S=(0,8)\).

Step2: Find coordinates of M

The line \(PM\) has slope \(-2\) and passes through \(P(12,14)\). Equation: \(y - 14=-2(x - 12)\), \(y-14=-2x + 24\), \(y=-2x + 38\). Since \(M\) is on \(x\) - axis, \(y = 0\). Set \(0=-2x + 38\), \(2x=38\), \(x = 19\). So \(M=(19,0)\).

Step3: Calculate distance between S(0,8) and M(19,0)

Using distance formula: \(SM=\sqrt{(19 - 0)^{2}+(0 - 8)^{2}}=\sqrt{361+64}=\sqrt{425}\approx20.61\approx21\) (rounded to nearest km). Or we can use distance between \(S\) and \(P\) and \(P\) and \(M\) and Pythagoras. Distance \(SP\): between \(S(0,8)\) and \(P(12,14)\): \(\sqrt{(12 - 0)^{2}+(14 - 8)^{2}}=\sqrt{144 + 36}=\sqrt{180}\approx13.42\). Distance \(PM\): between \(P(12,14)\) and \(M(19,0)\): \(\sqrt{(19 - 12)^{2}+(0 - 14)^{2}}=\sqrt{49+196}=\sqrt{245}\approx15.65\). Then by Pythagoras (since \(\angle SPM = 90^{\circ}\)), \(SM=\sqrt{SP^{2}+PM^{2}}=\sqrt{180 + 245}=\sqrt{425}\approx20.6\approx21\) km.

Problem 6 Answer:

The length of \(BD\) is \(\boldsymbol{5\sqrt{5}\approx11.2}\) (or if we made a mistake in \(D\)'s \(y\) - coordinate? Wait, let's re - check. The slope of \(BC\): \(B(2,13)\), \(C(6,5)\), \(\frac{5 - 13}{6 - 2}=\frac{-8}{4}=-2\). So the slope of \(CD\) should be \(\frac{1}{2}\) (negative reciprocal). The line \(CD\): passes through \(C(6,5)\), slope \(\frac{1}{2}\), equation: \(y - 5=\frac{1}{2}(x - 6)\), \(y=\frac{1}{2}x-3 + 5=\frac{1}{2}x + 2\). Since \(D\) is on \(y\) - axis (\(x = 0\)), when \(x = 0\), \(y=\frac{1}{2}(0)+2 = 2\). So \(D=(0,2)\). Then \(BD\): \(B(2,13)\), \(D(0,2)\), distance is \(\sqrt{(2 - 0)^2+(13 - 2)^2}=\sqrt{4 + 121}=\sqrt{125}=5\sqrt{5}\approx11.2\)

Problem 7 Answer:

The distance between \(S\) and \(M\) is approximately \(\boldsymbol{21}\) km.

For Problem 6:

Answer:

The length of \(BD\) is \(5\sqrt{5}\) (or approximately \(11.2\) units).

For Problem 7: