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Chapter 14: Problem 1314

If \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\) and \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) both are in GP with the same common ratio, then the points \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)\) (a) lie on a straight line (b) lie on a ellipse (c) lie on a circle (d) are vertices of a triangle

Short Answer

Expert verified
The points \((x_{1}, y_{1})\), \((x_{2}, y_{2})\), and \((x_{3}, y_{3})\) are vertices of a triangle.

Step by step solution

01

Express coordinates with geometric progression property

Let the common ratio of two geometric progressions be \(r\). The x coordinates, \(x_{1},\) \(x_{2}\), and \(x_{3}\), will then be a, ar, ar^2 where a is the first term in the GP. Similarly, the y coordinates, \(y_{1},\) \(y_{2}\), and \(y_{3}\), will be b, br, br^2 where b is the first term of the other GP. The given points will then be P1 \(( a, b)\), P2 \(( ar, br)\), and P3 \((ar^2, br^2)\).
02

Check if the points lie on a straight line

Three points are collinear (lie on a straight line) if the area of the triangle formed by them is 0. We will use the determinant method to find the area of the triangle and check if it's equal to 0. Let the points be A \((x_{A}, y_{A})\), B \((x_{B}, y_{B})\), and C \((x_{C}, y_{C})\). The area of the triangle is given by: \[Area = \dfrac{1}{2}|x_{A}(y_{B}-y_{C}) + x_{B}(y_{C}-y_{A}) + x_{C}(y_{A}-y_{B})|\] Plugging in our given points, we get: \[Area = \dfrac{1}{2}|a(br^2 - br) + ar(br^2 - b) + ar^2(b - br)|\]
03

Simplify the expression for the area

Now we'll simplify the expression for the area: Area \(= \dfrac{1}{2}|a\cdot b\cdot r^2 - a\cdot b\cdot r + a\cdot b\cdot r^2 - a\cdot b + a\cdot b\cdot r^2 - a\cdot b\cdot r\cdot r|\) Area \(= \dfrac{1}{2}|a\cdot b\cdot r^2 - a\cdot b\cdot r + a\cdot b\cdot r^2 - a\cdot b + a\cdot b\cdot r^2 - a\cdot b\cdot r^3|\) Now, we'll factor out the \(a\cdot b\) term: Area \(= \dfrac{1}{2} |ab \cdot (r^2 - r + r^2 - 1 + r^2 - r^3)|\) Area \(= \dfrac{1}{2} |ab \cdot (r^2 + r^2 + r^2 - r - r^3 - r - r^3 + 1)|\) Area \(= \dfrac{1}{2} |ab \cdot (3r^2 - 2r - 2r^3 + 1)|\)
04

Determine the shape based on the area

The area of the triangle appears to be non-zero as the expression inside the absolute value brackets depends on the common ratio r. This means that these three points do not lie on a straight line and form a triangle. Hence, we can choose option (d) as the correct answer. So, the points \((x_{1}, y_{1})\), \((x_{2}, y_{2})\), and \((x_{3}, y_{3})\) are vertices of a triangle.

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Most popular questions from this chapter

Find the equation of line passing through the point \((\sqrt{3},-1)\) and at a distance \(\sqrt{2}\) units from the origin. (a) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}-(\sqrt{3}+1) \mathrm{y}=4\) (b) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (c) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (d) \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\)

If \((1 / a),(1 / b),(1 / c)\) are in arithmetic sequence, then the line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})+(1 / \mathrm{c})=0\) passes through the fixed point \(\ldots \ldots\) (a) \((-1,-2)\) (b) \((-1,2)\) (c) \([1,-(1 / 2)]\) (d) \((1,-2)\)

Find the slope of the line passing through the point \((1,2)\) and the point of intersection of this line with the line \(\mathrm{x}+\mathrm{y}+3=0\) is at a distance \(3 \sqrt{2}\) units from \((1,2)\). (a) \((1 / \sqrt{3})\) (b) 1 (c) \(\sqrt{3}\) (d) \([(\sqrt{3}-1) / 2]\)

The y-intercept of the line \(\mathrm{y}+\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) is \(\ldots \ldots\) (a) \(-\left(\mathrm{y}_{1}+\mathrm{mx}_{1}\right)\) (b) \(\mathrm{y}_{1}-\mathrm{mx}_{1}\) (c) \(\left[\left(\mathrm{y}_{1}+\mathrm{mx}_{1}\right) / \mathrm{m}\right]\) (d) None of these

In a triangle \(\mathrm{ABC}\), coordinates of \(\mathrm{A}\) are \((1,2)\) and the equations of the medians through \(\mathrm{B}\) and \(\mathrm{C}\) are \(\mathrm{x}+\mathrm{y}=5\) and \(\mathrm{x}=4\) respectively. Then coordinates of \(\mathrm{B}\) and \(\mathrm{C}\) will be (a) \((-2,7),(4,3)\) (b) \((7,-2),(4,3)\) (c) \((2,7),(-4,3)\) (d) \((2,-7),(3,-4)\)
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