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, \(A=(160,350)\)

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

The length of \(BC\) is the distance from \(B(0,y)\) to \(C(x,0)\). But since \(AB\perp BC\) and \(MN\perp BC\), the slope of \(MN\) and \(AB\) are related. The slope of the line passing through \(M(120,430)\) and \(N(180,310)\) is \(m_{MN}=\frac{y_2-y_1}{x_2-x_1}=\frac{310 - 430}{180-120}=\frac{-120}{60}=-2\). Since \(MN\parallel AB\) (both perpendicular to \(BC\)), the slope of \(AB\) is also \(-2\). The equation of the line passing through \(A(160,350)\) with slope \(-2\) is \(y - 350=-2(x - 160)\). When \(x = 0\) (for point \(B\)), \(y-350=-2(0 - 160)\), \(y=350 + 320=670\). When \(y = 0\) (for point \(C\)), \(0-350=-2(x - 160)\), \(-350=-2x + 320\), \(2x=320 + 350=670\), \(x = 335\).

Step3: Calculate the length of \(BC\)

Using the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). For \(B(0,670)\) and \(C(335,0)\), \(d=\sqrt{(335 - 0)^2+(0 - 670)^2}=\sqrt{335^2+670^2}=\sqrt{112225+448900}=\sqrt{561125}\approx749.1\)

Answer:

\(749.1\)