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प्रश्न
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
उत्तर
y = 3x + 7
Slope of this line = 3
2y + px = 3
2y = −px + 3
`y = -(px)/2 + 3/2`
Slope of this line = `-p/2`
Since, the lines are perpendicular to each other, the product of their slopes is –1.
∴ `(3)(-p/2) = -1`
`(-3p)/2 = -1`
`p = 2/3`
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संबंधित प्रश्न
Find the slope of the line perpendicular to AB if : A = (0, −5) and B = (−2, 4)
Find the slope and the inclination of the line AB if : A = `(-1, 2sqrt(3))` and B = `(-2, sqrt(3))`
Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Find the slope of the lines passing through the given point.
P (–3, 1) , Q (5, –2)
Find the slope of a line, correct of two decimals, whose inclination is 45°
Find the slope of a line passing through the given pair of points (3,7) and (5,13)
Find the slope of a line passing through the following pair of points
(5pq,p2q) and (5qr,qr2)
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.
