Question

big idea questions
creating and carrying out a plan
one of the issues we all have with approaching complex application questions is we
sometimes forget to map out what we are intending to do before beginning the problem.
for the following question, plan out each of the steps you would carry out on the path to
the solution in words. once you have completed that, execute
those steps to determine the solution to the problem.

• in the graph provided
• ( overline{mn}perpoverline{ba} )
• ( overline{ab}perpoverline{bc} )
• point a divides segment ( overline{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 ( overline{bc} ) to the nearest hundredth of a unit?
numbered steps in words:
carrying out those steps:

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 \(m = 2\), \(n = 1\), \(M(120,430)\) and \(N(160,310)\).
\(x=\frac{2\times160+1\times120}{2 + 1}=\frac{320 + 120}{3}=\frac{440}{3}\approx146.67\)
\(y=\frac{2\times310+1\times430}{2 + 1}=\frac{620+430}{3}=\frac{1050}{3}=350\)
So, \(A(\frac{440}{3},350)\)

Step2: Find the slope of \(\overline{MN}\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For \(M(120,430)\) and \(N(160,310)\), \(m_{MN}=\frac{310 - 430}{160 - 120}=\frac{- 120}{40}=-3\)

Step3: Find the slope of \(\overline{BA}\)

Since \(\overline{MN}\perp\overline{BA}\), if the slope of \(\overline{MN}\) is \(m_1\) and the slope of \(\overline{BA}\) is \(m_2\), then \(m_1\times m_2=-1\). Given \(m_1=-3\), so \(m_2=\frac{1}{3}\)

Step4: Find the equation of line \(\overline{BA}\)

The line \(\overline{BA}\) passes through \(A(\frac{440}{3},350)\) and has slope \(m = \frac{1}{3}\). Using the point - slope form \(y - y_1=m(x - x_1)\), \(y-350=\frac{1}{3}(x-\frac{440}{3})\)

Since \(B\) is on the \(y\) - axis (\(x = 0\)), substitute \(x = 0\) into the equation:
\(y-350=\frac{1}{3}(0-\frac{440}{3})\)
\(y=350-\frac{440}{9}=\frac{3150 - 440}{9}=\frac{2710}{9}\approx301.11\)

Step5: Find the equation of line \(\overline{BC}\)

Since \(\overline{AB}\perp\overline{BC}\), the slope of \(\overline{AB}\) is \(\frac{1}{3}\), so the slope of \(\overline{BC}\) is \(-3\) (because if \(m_{AB}=m_1\) and \(m_{BC}=m_2\), \(m_1\times m_2=-1\)). The line \(\overline{BC}\) passes through \(B(0,\frac{2710}{9})\) and has slope \(m=-3\). Using the slope - intercept form \(y = mx + b\) (\(b=\frac{2710}{9}\)), the equation is \(y=-3x+\frac{2710}{9}\)

Since \(C\) is on the \(x\) - axis (\(y = 0\)), substitute \(y = 0\) into the equation:
\(0=-3x+\frac{2710}{9}\)
\(3x=\frac{2710}{9}\)
\(x=\frac{2710}{27}\approx100.37\)

Step6: Calculate the length of \(\overline{BC}\)

Using the distance formula between \(B(0,\frac{2710}{9})\) and \(C(\frac{2710}{27},0)\). The distance \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(d=\sqrt{(\frac{2710}{27}-0)^2+(0 - \frac{2710}{9})^2}=\sqrt{\frac{2710^{2}}{27^{2}}+\frac{2710^{2}}{81}}\)
\(d=\sqrt{\frac{2710^{2}(3 + 1)}{81}}=\frac{2710}{9}\sqrt{\frac{4}{3}}\approx100.37\)

Answer:

\(100.37\)