A) (0, 1) and (3, 4)
B) (1, 0) and (1, 4)
C) (3, 5) and (2, 1)
D) (3, 4) and (1, 2)
Correct Answer: B
Solution :
Let PQRM be a square and let P\[(-1,2)\]and R\[(3,2)\]be the vertices. Let the coordinates of Q be (x, y). |
| \[\because \] \[PQ=MR\] |
| \[\Rightarrow \] \[P{{Q}^{2}}=M{{R}^{2}}\] |
| \[\Rightarrow \]\[{{(x+1)}^{2}}+{{(y-2)}^{2}}={{(x-3)}^{2}}+{{(y-2)}^{2}}\][\[\because \]Distance\[=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]] |
| \[\Rightarrow \]\[{{x}^{2}}+1+2x+{{y}^{2}}+4-4y={{x}^{2}}+9-6x+{{y}^{2}}4-4y\] |
| \[\Rightarrow \] \[2x+1=-6x+9\] |
| \[\Rightarrow \] \[8x=8\] |
| \[\Rightarrow \] \[x=1\] ?(i) |
| In \[\Delta PQR,\]we have\[P{{Q}^{2}}+Q{{R}^{2}}=P{{R}^{2}}\] |
| \[{{(x+1)}^{2}}+{{(y-2)}^{2}}+{{(x-3)}^{2}}+{{(y-2)}^{2}}={{(3+1)}^{2}}+{{(2-2)}^{2}}\]\[\Rightarrow \]\[{{x}^{2}}+1+2x+{{y}^{2}}+4-4y+{{x}^{2}}+9-6x+{{y}^{2}}\] |
| \[+4-4y={{4}^{2}}+{{0}^{2}}\] |
| \[\Rightarrow \]\[2{{x}^{2}}+2{{y}^{2}}+2x-4y-6x-4y+1+4+9+4=16\] |
| \[\Rightarrow \] \[2{{x}^{2}}+2{{y}^{2}}-4x-8y+2=0\] |
| \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x-4y+1=0\] ?(ii) |
| Putting x = 1 from Eq. (i) in Eq. (ii), we get |
| \[1+{{y}^{2}}-2-4y+1=0\]\[\Rightarrow \]\[{{y}^{2}}-4y=0\] |
| \[\Rightarrow \] \[y\,\,(y-4)=0\]\[\Rightarrow \]\[y=0\]or 4 |
| Hence, the required vertices of square are (1, 0) and (1, 4). |
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