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 coordinates of point A

Using the section formula, if a point \(A(x,y)\) divides the line - segment joining \(M(x_1,y_1)=(120,430)\) and \(N(x_2,y_2)=(180,310)\) in the ratio \(m:n = 2:1\) from point \(M\).
The formula for the \(x\) - coordinate is \(x=\frac{mx_2+nx_1}{m + n}\) and for the \(y\) - coordinate is \(y=\frac{my_2+ny_1}{m + n}\).
For \(x\) - coordinate: \(x=\frac{2\times180+1\times120}{2 + 1}=\frac{360+120}{3}=\frac{480}{3}=160\).
For \(y\) - coordinate: \(y=\frac{2\times310+1\times430}{2 + 1}=\frac{620 + 430}{3}=\frac{1050}{3}=350\). So, \(A=(160,350)\).

Step2: Since \(AB\perp BC\), \(B\) is on the \(y\) - axis (\(x = 0\)) and \(C\) is on the \(x\) - axis (\(y = 0\))

The slope of \(AB\) is \(m_{AB}=\frac{y - y_B}{x-0}\) (where \(A=(x,y)=(160,350)\) and \(B=(0,y_B)\)), and the slope of \(BC\) is \(m_{BC}=\frac{y_C - y_B}{x_C-0}\) (where \(C=(x_C,0)\) and \(B=(0,y_B)\)). Since \(AB\perp BC\), \(m_{AB}\times m_{BC}=-1\).
The equation of the line passing through \(A=(160,350)\) and \(B=(0,y_B)\) and \(C=(x_C,0)\) can also be found using the fact that the line \(ABC\) (right - angled at \(B\)) forms a right - triangle.
Another way: The line passing through \(A=(160,350)\) and perpendicular to \(MN\) (since \(MN\perp BC\) and \(AB\perp BC\), \(AB\) is parallel to \(MN\)).
The slope of \(MN\) is \(m_{MN}=\frac{y_N - y_M}{x_N - x_M}=\frac{310-430}{180 - 120}=\frac{-120}{60}=-2\).
The equation of the line passing through \(A=(160,350)\) with slope \(m = \frac{1}{2}\) (because if two lines with slopes \(m_1\) and \(m_2\) are perpendicular, \(m_1\times m_2=-1\), and since \(m_{MN}=-2\), \(m_{BC}=\frac{1}{2}\)) is \(y - 350=\frac{1}{2}(x - 160)\).
When \(x = 0\) (for point \(B\)), \(y-350=\frac{1}{2}(0 - 160)\), \(y=350-80 = 270\), so \(B=(0,270)\).
When \(y = 0\) (for point \(C\)), \(0 - 350=\frac{1}{2}(x - 160)\), \(-700=x - 160\), \(x=- 540\) (distance is non - negative).
Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(B=(0,270)\) and \(C=(540,0)\) (taking absolute value of \(x\) for distance), \(BC=\sqrt{(540 - 0)^2+(0 - 270)^2}=\sqrt{540^{2}+270^{2}}=\sqrt{270^{2}(4 + 1)}=270\sqrt{5}\approx270\times2.236 = 603.72\).

Answer:

\(603.7\)