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प्रश्न
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
उत्तर
Writing the given lines 4x + 3y = 9
And px – 6y + 3 = 0 in the form of y = mx + c
4x + 3y = 9
`=>` 3y = −4x + 9
`y = (-4)/3x + 9/3`
`=> y = (-4)/3x + 3` ...(i)
Here, m1 = slope = `(-4)/3`
And px – 6y + 3 = 0
`=>` –6y = –px – 3
`y = (-p)/(-6)x - 3/(-6)`
= `p/6x + 1/2`
`=> y = p/6x + 1/2` ...(ii)
Here, slope m2 = `p/6`
∵ Lines are parallel
∴ Their slopes are equal i.e., m1 = m2
∴ `p/6 = (-4)/3`
`=>` 3p = –24
`p = (-24)/3`
p = –8
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If A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of `square`ABCD, show that `square`ABCD is a parallelogram.
Solution:
Slope of line = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
∴ Slope of line AB = `(2 - 1)/(8 - 6) = square` .......(i)
∴ Slope of line BC = `(4 - 2)/(9 - 8) = square` .....(ii)
∴ Slope of line CD = `(3 - 4)/(7 - 9) = square` .....(iii)
∴ Slope of line DA = `(3 - 1)/(7 - 6) = square` .....(iv)
∴ Slope of line AB = `square` ......[From (i) and (iii)]
∴ line AB || line CD
∴ Slope of line BC = `square` ......[From (ii) and (iv)]
∴ line BC || line DA
Both the pairs of opposite sides of the quadrilateral are parallel.
∴ `square`ABCD is a parallelogram.
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