MHT CET · Maths · Straight Lines
The points \(A(-a,-b)\), B \((0,0), C(a, b)\) and \(D\left(a^{2}, a b\right)\) are
- A collinear
- B vertices of a parallelogram
- C vertices of a square
- D vertices of a rectangle
Answer & Solution
Correct Answer
(A) collinear
Step-by-step Solution
Detailed explanation
Distance between the points \(A(-a,-b)\) and \(B(0,0)=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}\)
Distance between the points \(B(0,0)\) and \(C(a, b), \sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\)
Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\)
\(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\)
\(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\)
Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)\)
\(=\sqrt{\left(a^{2}+a\right)+(a b+b)^{2}} =(a+1) \sqrt{a^{2}+b^{2}} \)
\(A B+B C+C D =\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}\)\(+(a-1) \sqrt{a^{2}+b^{2}} \)
\(=(a+1) \sqrt{a^{2}+b^{2}}=A D\)
Hence the point are collinear.
Distance between the points \(B(0,0)\) and \(C(a, b), \sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}\)
Distance between the points \(C(a, b)\) and \(D\left(a^{2}, a b\right)\)
\(=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}}\)
\(=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}\) \(=(a-1) \sqrt{a^{2}+b^{2}}\)
Similarly, distance between the points \(A(-a,-b)\) and \(D\left(a^{2}, a b\right)\)
\(=\sqrt{\left(a^{2}+a\right)+(a b+b)^{2}} =(a+1) \sqrt{a^{2}+b^{2}} \)
\(A B+B C+C D =\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}\)\(+(a-1) \sqrt{a^{2}+b^{2}} \)
\(=(a+1) \sqrt{a^{2}+b^{2}}=A D\)
Hence the point are collinear.
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