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Question

Mathematics Question on Straight lines

If the point (1,a)(1, a) lies between the straight lines x+y=1x + y = 1 and 2(x+y)=32(x+y) = 3 then a lies in interval

A

(32,)\left(\frac{3}{2},\infty\right)

B

(1,32)\left(1,\frac{3}{2}\right)

C

(,0)\left(-\infty, 0\right)

D

(0,12)\left(0,\frac{1}{2}\right)

Answer

(0,12)\left(0,\frac{1}{2}\right)

Explanation

Solution

Since, (1,a)(1, a) lies between x+y=1x + y = 1 and 2(x+y)=32(x+y) = 3 \therefore Put x=1x = 1 in 2(x+y)=32\left(x+y\right) = 3. We get the range of y. Thus, 2(1+y)=3y=321=122\left(1+y\right) = 3 \Rightarrow y = \frac{3}{2}-1 = \frac{1}{2} Thus ????ies in (0,12)\left(0, \frac{1}{2}\right)