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lf every point on the line \( \Large \left(a_{1}-a_{2}\right)x+ \left(b_{1}-b_{2}\right)y=c \) is equidistant from the points \( \Large \left(a_{1},\ b_{1}\right) \) then 2c is equal to:
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A) \( \Large a^{2}_{1}+b^{2}_{1}+a^{2}_{2}-b^{2}_{2} \) |
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B) \( \Large a^{2}_{1}+b^{2}_{1}+a^{2}_{2}+b^{2}_{2} \) |
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C) \( \Large a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2} \) |
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D) none of these |
Correct Answer:
| C) \( \Large a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2} \) |
Description for Correct answer:
Let \( \Large \left(h,\ k\right) \) be any point on the given line, then
\( \Large \left(h-a_{1}\right)^{2}+ \left(k-b_{1}\right)^{2}= \left(h-a_{2}\right)^{2}+ \left(k-b_{2}\right)^{2} \)
=> \( \Large 2 \left(a_{1}-a_{2}\right)h+2 \left(b_{1}-b_{2}\right)k = a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2} \)
=> \( \Large \left(a_{1}-a_{2}\right)h+ \left(b_{1}-b_{2}\right)k=\frac{1}{2} \left(a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2}\right) \)
Since (h, k) lies on the given line
\( \Large \left(a_{1}-a_{2}\right)h+ \left(b_{1}-b_{2}\right)k=c \)
On comparing equations (i) and (ii), we get
\( \Large c = \left(\frac{1}{2}\right) \left(a^{2}_{1}+b^{2}_{1}\right)-a^{2}_{2}-b^{2}_{2} \)
Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products
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