Problem Solving

GMATH practice exercise (Quant Class 20)

The circle with center (1,0) and radius 5 intercepts at points A and B the straight line defined by C = (0,0) and D= (4,8). If M = (w,z) is the midpoint of the line segment AB, what is the value of w+z ?

(A) 3/7
(B) 1/2
(C) 3/5
(D) 3/4
(E) 1

Answer: [spoiler]_____(C)___[/spoiler]

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 20)

The circle with center (1,0) and radius 5 intercepts at points A and B the straight line defined by C = (0,0) and D= (4,8). If M = (w,z) is the midpoint of the line segment AB, what is the value of w+z ?

(A) 3/7
(B) 1/2
(C) 3/5
(D) 3/4
(E) 1

$$? = w + z = {x_M} + {y_M}\,\,\,\,\,\left[ {M = \left( {{x_M};{y_M}} \right)} \right]$$

$$A,B\,\,\,\,\left\{ \matrix{
\,{\left( {x - 1} \right)^2} + {y^2} = {5^2}\,\,\,\,\,\left[ {{\rm{circle}}} \right] \hfill \cr
\,y = 2x\,\,\,\,\,\left[ {{\rm{line}}} \right] \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\left( {x - 1} \right)^2} + {\left( {2x} \right)^2} = {5^2}\,\,\,\, \Rightarrow \,\,\,\,5{x^2} - 2x - 24 = 0\,\,\,\,\left( * \right)$$

$$M = \left( {{{{x_A} + {x_B}} \over 2}\,\,;\,\,{{{y_A} + {y_B}} \over 2}} \right)\,\,\,\mathop = \limits^{M\, \in \,\,{\rm{line}}} \,\,\,\left( {{{{x_A} + {x_B}} \over 2}\,\,;\,\,{{2\,{x_A} + 2\,{x_B}} \over 2}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = {3 \over 2}\left( {{x_A} + {x_B}} \right)$$

$$\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{x_A} + {x_B} = {\rm{sum}}\,\,{\rm{roots}} = - {{\left( { - 2} \right)} \over 5} = {2 \over 5}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = {3 \over 2}\left( {{2 \over 5}} \right) = {3 \over 5}$$

The correct answer is therefore (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

$$m=\left(w,2\right)=\left(x_m;y_m\right)$$
$$=w+2=x_m+y_m$$
At point AB for the circle
$$\left(x-a\right)^2+\left(y-b\right)^2=r^2$$
$$\left(x-1\right)^2+\left(y\right)^2=5^2$$
$$\left(x-1\right)^2+\left(y\right)^2=5^2............eq\ i$$
For the line y = 2x ............ equ (2)

The line pass through (0,0) to (4,8)
substituting y=2x in equ (i)
$$\left(x-1\right)^2+\left(2x\right)^2=25$$
$$\left(5x^2-2x-24\ \right)=0$$
$$where\ the\ sum\ of\ roots\ \ \frac{b}{a}=-\frac{\left(-2\right)}{5}$$
$$M=\left(\frac{x_A+x_B}{2};\frac{y_A+y_B}{2}\right)$$
$$M=\left(\frac{x_A+x_B}{2}+\frac{2x_A+2x_B}{2}\right)$$
$$=\frac{3}{2}\left(x_A+x_B\right)$$
$$where\left(x_A+x_B\right)=sum\ of\ roots=+\frac{\left(+2\right)}{5}=\frac{2}{5}$$
$$=\frac{3}{2}\left(\frac{2}{5}\right)=\frac{3}{5}$$

$$answer\ is\ Option\ C$$