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Find the value of p for which the lines, px+3y+5=0px + 3y + 5 = 0 and 8x+2y−3=08x + 2y - 3 = 0 are parallel.

Solution

✅ Correct Option: 1

Two lines are parallel when they have the same slope.

Line 1: px+3y+5=0px + 3y + 5 = 0

Line 2: 8x+2y−3=08x + 2y - 3 = 0


Converting Line 2 to slope-intercept form:

8x+2y−3=08x + 2y - 3 = 0

2y=−8x+32y = -8x + 3

y=−4x+32y = -4x + \dfrac{3}{2}

Slope of Line 2 is −4-4.


Converting Line 1 to slope-intercept form:

px+3y+5=0px + 3y + 5 = 0

3y=−px−53y = -px - 5

y=−p3x−53y = -\dfrac{p}{3}x - \dfrac{5}{3}

Slope of Line 1 is −p3-\dfrac{p}{3}.


For the lines to be parallel, their slopes must be equal:

−p3=−4-\dfrac{p}{3} = -4

p=12p = 12

Therefore, p=12p = 12.

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