1. **Problem Statement:** Given rhombus QRST with equations of sides QR: $2x + y = 7$, RS: $x = 1$, and TS: $2x + y = -1$, find: (a) Coordinates of Q and S. (b) Coordinates of M, the intersection of diagonals. (c) Coordinates of R and T. 2. **Key Properties and Formulas:** - A rhombus has all sides equal in length. - Opposite sides are parallel. - Diagonals bisect each other. - The intersection point M of diagonals is the midpoint of both diagonals. 3. **Step (a): Find Q and S** - S lies on RS: $x=1$ and on TS: $2x + y = -1$. - Substitute $x=1$ into $2x + y = -1$: $$2(1) + y = -1 \Rightarrow 2 + y = -1 \Rightarrow y = -3$$ - So, $S = (1, -3)$. - Q lies on QR: $2x + y = 7$ and on RS: $x=1$ (since QR and RS meet at R, but Q is on QR and adjacent to R, we need to find Q by using the fact that QR and TS are parallel lines with equations $2x + y = 7$ and $2x + y = -1$ respectively). - Since QR and TS are parallel, and QR passes through Q and R, and TS passes through T and S. - To find Q, note that Q lies on QR: $2x + y = 7$. - Also, since QRST is a rhombus, the distance between QR and TS lines is equal to the side length. - But easier is to find R first (see step (c)) or find Q by intersection of QR and the line perpendicular to QR passing through S. 4. **Step (b): Find M, the intersection of diagonals** - Diagonals bisect each other, so M is midpoint of QS and RT. - We need coordinates of Q, S, R, T to find M. 5. **Step (c): Find R and T** - R lies on QR: $2x + y = 7$ and on RS: $x=1$. - Substitute $x=1$ into $2x + y = 7$: $$2(1) + y = 7 \Rightarrow 2 + y = 7 \Rightarrow y = 5$$ - So, $R = (1, 5)$. - T lies on TS: $2x + y = -1$ and on line perpendicular to RS through Q. - RS is vertical line $x=1$, so perpendicular lines are horizontal lines $y = k$. - Since Q lies on QR: $2x + y = 7$, and we want T such that QT is parallel to RS (vertical), so T has $x=1$. - But T lies on TS: $2x + y = -1$. - Substitute $x=1$ into $2x + y = -1$: $$2(1) + y = -1 \Rightarrow 2 + y = -1 \Rightarrow y = -3$$ - So, $T = (1, -3)$, but this is same as S, so T must be different. - Since RS is $x=1$, and QR and TS are parallel lines $2x + y = 7$ and $2x + y = -1$, the points Q and T lie on $2x + y = 7$ and $2x + y = -1$ respectively. - We have S at $(1, -3)$ and R at $(1, 5)$. - The midpoint M of diagonal RS is: $$M = \left(\frac{1+1}{2}, \frac{5 + (-3)}{2}\right) = (1, 1)$$ - Since diagonals bisect each other, M is also midpoint of QT. - Let $Q = (x_q, y_q)$ and $T = (x_t, y_t)$. - Then: $$M = \left(\frac{x_q + x_t}{2}, \frac{y_q + y_t}{2}\right) = (1, 1)$$ - Q lies on $2x + y = 7$, T lies on $2x + y = -1$. - From midpoint formula: $$x_t = 2(1) - x_q = 2 - x_q$$ $$y_t = 2(1) - y_q = 2 - y_q$$ - Substitute $x_t$ and $y_t$ into $2x + y = -1$ for T: $$2(2 - x_q) + (2 - y_q) = -1$$ $$4 - 2x_q + 2 - y_q = -1$$ $$6 - 2x_q - y_q = -1$$ $$-2x_q - y_q = -7$$ $$2x_q + y_q = 7$$ - This matches the equation for Q: $2x_q + y_q = 7$. - So Q can be any point on $2x + y = 7$. - To find specific Q and T, use the fact that Q and R are adjacent vertices, so distance QR equals distance RS. 6. **Calculate side length RS:** $$RS = |5 - (-3)| = 8$$ 7. **Find Q on $2x + y = 7$ such that distance QR = 8, with R at (1,5):** - Let $Q = (x, y)$ with $2x + y = 7$. - Distance $QR = \sqrt{(x - 1)^2 + (y - 5)^2} = 8$. - Substitute $y = 7 - 2x$: $$\sqrt{(x - 1)^2 + (7 - 2x - 5)^2} = 8$$ $$\sqrt{(x - 1)^2 + (2 - 2x)^2} = 8$$ $$(x - 1)^2 + (2 - 2x)^2 = 64$$ $$(x - 1)^2 + 4(1 - x)^2 = 64$$ $$(x - 1)^2 + 4(x - 1)^2 = 64$$ $$5(x - 1)^2 = 64$$ $$(x - 1)^2 = \frac{64}{5}$$ $$x - 1 = \pm \frac{8}{\sqrt{5}}$$ - So, $$x = 1 \pm \frac{8}{\sqrt{5}}$$ - Corresponding $y$ values: $$y = 7 - 2x = 7 - 2\left(1 \pm \frac{8}{\sqrt{5}}\right) = 7 - 2 \mp \frac{16}{\sqrt{5}} = 5 \mp \frac{16}{\sqrt{5}}$$ - Thus two possible Q points: $$Q_1 = \left(1 + \frac{8}{\sqrt{5}}, 5 - \frac{16}{\sqrt{5}}\right), \quad Q_2 = \left(1 - \frac{8}{\sqrt{5}}, 5 + \frac{16}{\sqrt{5}}\right)$$ 8. **Find corresponding T points using midpoint M = (1,1):** - For $Q_1$: $$x_t = 2 - x_q = 2 - \left(1 + \frac{8}{\sqrt{5}}\right) = 1 - \frac{8}{\sqrt{5}}$$ $$y_t = 2 - y_q = 2 - \left(5 - \frac{16}{\sqrt{5}}\right) = -3 + \frac{16}{\sqrt{5}}$$ - For $Q_2$: $$x_t = 2 - x_q = 2 - \left(1 - \frac{8}{\sqrt{5}}\right) = 1 + \frac{8}{\sqrt{5}}$$ $$y_t = 2 - y_q = 2 - \left(5 + \frac{16}{\sqrt{5}}\right) = -3 - \frac{16}{\sqrt{5}}$$ 9. **Final answers:** - (a) $Q = \left(1 + \frac{8}{\sqrt{5}}, 5 - \frac{16}{\sqrt{5}}\right)$ or $\left(1 - \frac{8}{\sqrt{5}}, 5 + \frac{16}{\sqrt{5}}\right)$, $S = (1, -3)$. - (b) $M = (1, 1)$. - (c) $R = (1, 5)$, $T = \left(1 - \frac{8}{\sqrt{5}}, -3 + \frac{16}{\sqrt{5}}\right)$ or $\left(1 + \frac{8}{\sqrt{5}}, -3 - \frac{16}{\sqrt{5}}\right)$ corresponding to chosen Q. These points satisfy the rhombus properties and given line equations.