First slide

Equation of line in Straight lines

Question

The point (2t² + 2t + 4, t² + t + 1) lies on the line x + 2y = 1 for

Moderate

A

all real values of t
B

some real values of t
C

$t = \frac{- 4 + \sqrt{7}}{8}$
D

none of these

Solution

The point (2t² + 2t + 4, t² + t + 1) lies on the line x + 2y = 1
if (2t² + 2t + 4) + 2(t² + t + 1) = 1
i.e., 4t² + 4t + 5 = 0
Here, discriminent = 16 – 4 × 4 × 5 = –64 < 0.
$∴$ No real value of t is possible.
Hence, the given point cannot lie on the line.

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