+1452 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!
+26393 3: Find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y.
\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 2x+2y \\ x^2-2x + y^2-2y &=& 0 \\ (x-1)^2 -1 + (y-1)^2 -1 &=& 0 \\ (x-1)^2 + (y-1)^2 &=& 2 \\\\ \text{circle}_{\text{center}} &=& (1,1) \\ \text{circle}_{\text{radius}} &=& \sqrt{2} \\ x_{\text{max}} &=& x_{\text{center}} + \text{ radius} \\ x_{\text{max}} &=& 1 + \sqrt{2} \\ x_{\text{max}} &=& 2.41421356237 \\ \hline \end{array}\)
+1452 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
+26393 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.
\(\text{Let $\tan(\varphi_1) = \frac{5}{12}$ } \\ \text{Let $\tan(\varphi_2) = \frac{4}{3}$ }\)
\(\begin{array}{|rcll|} \hline \text{slope}_{bisect} &=&\tan \left( \dfrac{\arctan(\varphi_1)+\arctan(\varphi_2) } {2} \right) \\\\ &=&\tan \left( \dfrac{22.6198649480^{\circ}+53.1301023542^{\circ} } {2} \right) \\\\ &=&\tan \left( \dfrac{75.7499673022^{\circ} } {2} \right) \\\\ &=&\tan (37.8749836511^{\circ} ) \\\\ &=& 0.77777777778 \\\\ &=& \dfrac{7}{9} \\ \hline \end{array}\)
+26393 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
