Question #83981

In a square ABCD, A(1,3) and C(4,2). AC is a diagonal. Express the coordinates of a point on the diagonal BD using a real parameter. Hence find the coordinates of the other two vertices.
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Expert's answer

2019-01-07T07:44:11-0500

Answer on Question #83981 – Math – Analytic Geometry

Question

In a square ABCDABCD, A(1,3)A(1,3) and C(4,2)C(4,2) are two vertices. AC is a diagonal. Express the coordinates of a point on the diagonal BD using a real parameter. Hence find the coordinates of the other two vertices.

Solution

The point M(Ax+Cx2,Ay+Cy2)=M(2.5,2.5)M\left(\frac{A_x + C_x}{2}, \frac{A_y + C_y}{2}\right) = M(2.5, 2.5) is the center of the square.

The parametric equations of ACAC are x=Mx+(CxAx)tx = M_x + (C_x - A_x)t, y=My+(CyAy)ty = M_y + (C_y - A_y)t.

The parametric equations of BDBD are x=Mx(CyAy)sx = M_x - (C_y - A_y)s, y=My+(CxAx)sy = M_y + (C_x - A_x)s.

There are relationships between vertexes:


rBD=rACrOA=rOM+tArACrOC=rOM+tCrAC=rOMtArAC,rOB=rOM+sBrBD=rOM+tArBD,rOD=rOM+sDrBD=rOMsBrBD\begin{array}{l} \overrightarrow{r_{BD}} = \overrightarrow{r_{AC}} \\ \overrightarrow{r_{OA}} = \overrightarrow{r_{OM}} + t_A \overrightarrow{r_{AC}} \\ \overrightarrow{r_{OC}} = \overrightarrow{r_{OM}} + t_C \overrightarrow{r_{AC}} = \overrightarrow{r_{OM}} - t_A \overrightarrow{r_{AC}}, \\ \overrightarrow{r_{OB}} = \overrightarrow{r_{OM}} + s_B \overrightarrow{r_{BD}} = \overrightarrow{r_{OM}} + t_A \overrightarrow{r_{BD}}, \\ \overrightarrow{r_{OD}} = \overrightarrow{r_{OM}} + s_D \overrightarrow{r_{BD}} = \overrightarrow{r_{OM}} - s_B \overrightarrow{r_{BD}} \\ \end{array}


The coordinates of the vertex B are


x=Mx(CyAy)sB=Mx(CyAy)tA=Mx(CyAy)(AxMx)(CxAx)=2.51(1.5)3=2,x = M_x - (C_y - A_y) s_B = M_x - (C_y - A_y) t_A = M_x - \frac{(C_y - A_y)(A_x - M_x)}{(C_x - A_x)} = 2.5 - \frac{-1*(-1.5)}{3} = 2,y=My+(CxAx)sB=Mx+(CxAx)tA=My+(CxAx)(AxMx)(CxAx)=2.5+3(1.5)3=1y = M_y + (C_x - A_x) s_B = M_x + (C_x - A_x) t_A = M_y + \frac{(C_x - A_x)(A_x - M_x)}{(C_x - A_x)} = 2.5 + \frac{3*(-1.5)}{3} = 1


The coordinates of the vertex D are


x=Mx(CyAy)sD=Mx+(CyAy)sB=2.5+1(1.5)3=3,x = M_x - (C_y - A_y) s_D = M_x + (C_y - A_y) s_B = 2.5 + \frac{-1*(-1.5)}{3} = 3,y=My+(CxAx)sD=My(CxAx)sB=2.53(1.5)3=4y = M_y + (C_x - A_x) s_D = M_y - (C_x - A_x) s_B = 2.5 - \frac{3*(-1.5)}{3} = 4

Answer:

The other two vertices are B(2,1)B(2,1) and D(3,4)D(3,4).

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