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# Several Problems

+1
813
6
+1432

1:  A line with slope 6 bisects the area of a unit square with vertices (0,0), (1,0), (1,1), and (0,1). What is the y-intercept of this line?

2:  The line $$y = (3x + 7)/4$$ intersects the circle $$x^2 + y^2 = 25$$ at A and B. Find the length of chord $$\overline{AB}$$.

3:  Find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y.

4:  The lines $$y = \frac{5}{12} x$$ and $$y = \frac{4}{3} x$$ are drawn in the coordinate plane. Find the slope of the line that bisects the angle between these lines.

5:  Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 2(AD) = DB, point E is on segment BC such that 2(BE) = EC and point F is on segment CA such that 2(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

Thank you for all of your help!

Jun 13, 2018

#1
+21978
+1

3:  Find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y.

$$\begin{array}{|rcll|} \hline x^2+y^2 &=& 2x+2y \\ x^2-2x + y^2-2y &=& 0 \\ (x-1)^2 -1 + (y-1)^2 -1 &=& 0 \\ (x-1)^2 + (y-1)^2 &=& 2 \\\\ \text{circle}_{\text{center}} &=& (1,1) \\ \text{circle}_{\text{radius}} &=& \sqrt{2} \\ x_{\text{max}} &=& x_{\text{center}} + \text{ radius} \\ x_{\text{max}} &=& 1 + \sqrt{2} \\ x_{\text{max}} &=& 2.41421356237 \\ \hline \end{array}$$

Jun 14, 2018
#6
+1432
+1

Thanks so much for your answers Heureka!  I think I'm having an issue with the website though.  I see on the side of my screen that there were 5 answers posted to the thread in the little (+5) notification bubble, and I had 5 questions posted here which makes sense, but on the actual post itself I only see your answers to 3, 4, and 1, in that order.  I think this might be a problem with the website but does anyone know how I can view the other answers?  I have tried reloading the site.

Thanks

AnonymousConfusedGuy  Jun 14, 2018
#2
+21978
+1

4:  The lines

$$y = \frac{5}{12} x$$

and

$$y = \frac{4}{3} x$$

are drawn in the coordinate plane.

Find the slope of the line that bisects the angle between these lines.

$$\text{Let \tan(\varphi_1) = \frac{5}{12} } \\ \text{Let \tan(\varphi_2) = \frac{4}{3} }$$

$$\begin{array}{|rcll|} \hline \text{slope}_{bisect} &=&\tan \left( \dfrac{\arctan(\varphi_1)+\arctan(\varphi_2) } {2} \right) \\\\ &=&\tan \left( \dfrac{22.6198649480^{\circ}+53.1301023542^{\circ} } {2} \right) \\\\ &=&\tan \left( \dfrac{75.7499673022^{\circ} } {2} \right) \\\\ &=&\tan (37.8749836511^{\circ} ) \\\\ &=& 0.77777777778 \\\\ &=& \dfrac{7}{9} \\ \hline \end{array}$$

Jun 14, 2018
#3
+21978
+1

1:  A line with slope 6 bisects the area of a unit square with vertices (0,0), (1,0), (1,1), and (0,1).
What is the y-intercept of this line?

$$\begin{array}{|lrcll|} \hline & y &=& mx + b \quad & | \quad m = 6 \\ & y &=& 6x + b \\\\ P_1 = (a,0): & 0 &=& 6a+b \\ & -6a &=& b \\\\ P_2 = (1-a,1): & 1 &=& 6(1-a)+b \\ & 1 &=& 6-6a + b \quad & | \quad -6a = b \\ & 1 &=& 6 +b + b \\ & 1 &=& 6 +2b \\ & -5 &=& 2b \\ & 2b &=& -5 \\\\ & b &=& -\dfrac{5}{2} \\ & b &=& -2.5 \\ \hline \end{array}$$

The y-intercept of this line is -2.5

Jun 14, 2018