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Find the value of ‘P’ for which lines (P - 1)x + y - 5 = 0 and (2 - P)x - 3y + 2 = 0 are parallel?

Find the value of ‘P’ for which lines (P - 1)x + y - 5 = 0 and (2 - P)x - 3y + 2 = 0 are parallel? Correct Answer 1 / 2

Given:

L1; (P - 1)x + y - 5 = 0

L2; (2 - P)x - 3y + 2 = 0

L1 / / L2

Formula used:

If two lines are parallel, then, the slope of both lines will be same.

Calculation:

L1; (P - 1)x + y - 5 = 0

⇒ y = - (P - 1)x + 5

⇒ y = (1 - P)x + 5

We know y = mx + c, where m = slope, comparing the above equation, we get:

∴ Slope = (1 - P)       ----(1)

L2; (2- P)x - 3y + 2 = 0

⇒ 3y = (2 - P)x + 2

⇒ y = (2 - P)x / 3 + 2 / 3

We know y = mx + c, where m = slope, comparing the above equation, we get:

∴ Slope = (2 - P) / 3       ----(2)

Now, (1 - P) = (2 - P) / 3

⇒ (3 - 3P) = (2 - P)

⇒ (3P - P) = (3 - 2)

⇒ 2P = 1

⇒ P = 1 / 2

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