A) 5
B) 6
C) 8
D) 7
Correct Answer: D
Solution :
| Given, vertices of a parallelogram are A (6, 1), 5 (8, 2), C (9, 4) and D (p, 3). Here, we have to find the value of p. We know that, diagonals of a parallelogram bisect each other. |
| \[\therefore\] Coordinates of mid-point of diagonal AC = Coordinates of mid-point of diagonal BD |
| \[\Rightarrow \,\,\,\left( \frac{6+9}{2},\,\frac{1+4}{2} \right)=\left( \frac{8+p}{2},\,\frac{2+3}{2} \right)\] |
| \[\left[ \because \,\,mid-po\operatorname{int}\,=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \right]\] |
| \[\Rightarrow \,\,\,\,\left( \frac{15}{2},\,\frac{5}{2} \right)=\left( \frac{8+p}{2},\,\frac{5}{2} \right)\] |
| On equating .x-coordinate from both sides, |
| we get |
| \[\frac{15}{2}=\frac{8+p}{2}\Rightarrow \,\,\,15=8+p\] |
| \[\Rightarrow \,\,\,p=15-8\,\Rightarrow \,p=7\] |
| Hence, the required value of p is 7. |
| Given, vertices of a parallelogram are A (6, 1), 5 (8, 2), C (9, 4) and D (p, 3). Here, we have to find the value of p. We know that, diagonals of a parallelogram bisect each other. |
| \[\therefore\] Coordinates of mid-point of diagonal AC = Coordinates of mid-point of diagonal BD |
| \[\Rightarrow \,\,\,\left( \frac{6+9}{2},\,\frac{1+4}{2} \right)=\left( \frac{8+p}{2},\,\frac{2+3}{2} \right)\] |
| \[\left[ \because \,\,mid-po\operatorname{int}\,=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \right]\] |
| \[\Rightarrow \,\,\,\,\left( \frac{15}{2},\,\frac{5}{2} \right)=\left( \frac{8+p}{2},\,\frac{5}{2} \right)\] |
| On equating .x-coordinate from both sides, |
| we get |
| \[\frac{15}{2}=\frac{8+p}{2}\Rightarrow \,\,\,15=8+p\] |
| \[\Rightarrow \,\,\,p=15-8\,\Rightarrow \,p=7\] |
| Hence, the required value of p is 7. |
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