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Chapter 14: Problem 1270

If \(\mathrm{P}(1,2), \mathrm{Q}(4,6), \mathrm{R}(5,7)\) and \(\mathrm{S}(\mathrm{a}, \mathrm{b})\) are the vertices of a parallelogram PQRS then (a) \(a=2, b=4\) (b) \(a=3, b=4\) (c) \(\mathrm{a}=2, \mathrm{~b}=3\) (d) \(a=2, b=5\)

Short Answer

Expert verified
The coordinates of vertex S are (a = 2, b = 4) which makes option (a) the correct answer.

Step by step solution

01

Compute slopes for given vertices

First, we need to compute the slopes of the opposite sides of the parallelogram. We will compute the slope of PQ and the slope of R(a, b). The slope of PQ is given by: \(m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6-2}{4-1} = \frac{4}{3}\) Now let's compute the slope of R(a, b), using the given point R(5,7) and a generic point S with coordinates (a, b): \(m_{RS} = \frac{b-7}{a-5}\)
02

Compare the slopes

Since the opposite sides of a parallelogram have equal slopes, we will set the slope of PQ equal to the slope of R(a, b) and find the relationship between a and b: \(\frac{4}{3} = \frac{b-7}{a-5}\)
03

Test the given options

Now we will test each given option for the coordinates of S(a, b) against the relationship found in Step 2: (a) a=2, b=4: \(\frac{4}{3} = \frac{4-7}{2-5}\) → \(\frac{4}{3} = \frac{-3}{-3}\) which is true. (b) a=3, b=4: \(\frac{4}{3} = \frac{4-7}{3-5}\) → \(\frac{4}{3} = \frac{-3}{-2}\) which is false. (c) a=2, b=3: \(\frac{4}{3} = \frac{3-7}{2-5}\) → \(\frac{4}{3} = \frac{-4}{-3}\) which is false. (d) a=2, b=5: \(\frac{4}{3} = \frac{5-7}{2-5}\) → \(\frac{4}{3} = \frac{-2}{-3}\) which is false. From the above analysis, only option (a) holds true.
04

Conclusion

The coordinates of vertex S are (a=2, b=4). Option (a) is the correct answer.

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Most popular questions from this chapter

The equations of the two lines each passing through \((5,6)\) and each making an acute angle of \(45^{\circ}\) with the line \(2 \mathrm{x}-\mathrm{y}+1=0\) is (a) \(3 x+y-21=0, x-3 y+13=0\) (b) \(3 x+y+21=0, x+3 y+13=0\) (c) \(y=2 x, y=3 x\) (d) \(3 \mathrm{x}+\mathrm{y}-21=0, \mathrm{x}-3 \mathrm{y}-13=0\)

If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(\mathrm{x}+2 \mathrm{y}=1\) and \(2 \mathrm{x}+4 \mathrm{y}=15\), then the range of \(\mathrm{t}\) is \(\ldots \ldots\) (a) \(0<\mathrm{t}<[5 /(6 \sqrt{2})]\) (b) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<0\) (c) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<[(5 \sqrt{2}) / 6]\) (d) None of these

\(\mathrm{A}(1,0)\) and \(\mathrm{B}(-1,0)\), then the locus of points satisfying \(\mathrm{AQ}-\mathrm{BQ}=\pm 1\) is \(\ldots \ldots\) (a) \(12 \mathrm{x}^{2}+4 \mathrm{y}^{2}=3\) (b) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=3\) (c) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=-3\) (d) \(12 x^{2}+4 y^{2}=-3\)

If \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) and \(\mathrm{P}\left(\mathrm{tx}_{2}+(1-\mathrm{t}) \mathrm{x}_{1}, \mathrm{t}_{2}+(1-\mathrm{t}) \mathrm{y}_{1}\right)\) where \(t<0\), then P divides \(\underline{A B}\) from \(A\) in the ratio \(\ldots \ldots\) (a) \(1-\mathrm{t}\) (b) \([(\mathrm{t}-1) / \mathrm{t}]\) (c) \([t /(1-t)]\) (d) \(t-1\)

The in centre of a triangle whose vertices \(\mathrm{A}(2,4), \mathrm{B}(2,6)\) and \(\mathrm{C}(2+\sqrt{3}, 5)\) is.... (a) \([2+(1 / \sqrt{3}), 5]\) (b) \([1+\\{1 /(2 \sqrt{3})\\},(5 / 2)]\) (c) \((2,5)\) (d) None of these
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