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in the graph provided ·mn ⊥ bc ·ab ⊥ bc ·point a divides segment mn in …

Question

in the graph provided
·mn ⊥ bc
·ab ⊥ bc
·point a divides segment mn in a 2:1 ratio from point m
·point b is on the y - axis
·point c is on the x - axis
what is the length of bc to the nearest hundredth of a unit?

Explanation:

Step1: Find the coordinates of point \(A\)

Use the section formula. If a point \(A(x,y)\) divides the line - segment joining \(M(x_1,y_1)\) and \(N(x_2,y_2)\) in the ratio \(m:n\) from \(M\), then \(x=\frac{mx_2 + nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\). Here, \(m = 2\), \(n = 1\), \(x_1=120\), \(y_1 = 430\), \(x_2=180\), \(y_2=310\).

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So, the coordinates of \(A\) are \((160,350)\).

Step2: Since \(B\) is on the \(y\) - axis (\(x = 0\)) and \(C\) is on the \(x\) - axis (\(y = 0\)), and \(AB\perp BC\) and \(MN\perp BC\). The slope of \(MN\) is \(m_{MN}=\frac{y_2 - y_1}{x_2 - x_1}=\frac{310-430}{180 - 120}=\frac{-120}{60}=-2\). Since \(MN\perp BC\), if the slope of \(MN\) is \(m_{MN}\) and the slope of \(BC\) is \(m_{BC}\), then \(m_{MN}\times m_{BC}=-1\). So, \(m_{BC}=\frac{1}{2}\).

Let \(B=(0,b)\) and \(C=(c,0)\). The slope of \(BC\) is \(m_{BC}=\frac{0 - b}{c-0}=\frac{-b}{c}=\frac{1}{2}\), so \(c=-2b\).
The vector \(\overrightarrow{AB}=(0 - 160,b - 350)=(-160,b - 350)\) and \(\overrightarrow{BC}=(c-0,0 - b)=(c,-b)\). Since \(AB\perp BC\), \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\). Substitute \(c=-2b\) into \(\overrightarrow{AB}\cdot\overrightarrow{BC}\):

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\(b = 0\) is not valid (as it will make the points degenerate). So, \(b = 670\), then \(c=-2\times670=-1340\).

Step3: Calculate the length of \(BC\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). For \(B=(0,670)\) and \(C=(-1340,0)\)

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Answer:

\(1508.12\)