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प्रश्न
उत्तर
Slope of line AB = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
⇒ 1 = `(2"p" + - 5)/("p" + 2)`
⇒ p + 2 = 2p - 4
⇒ 6 = p
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संबंधित प्रश्न
Show that the points P(a, b + c), Q(b, c + a) and R(c, a + b) are collinear.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Fill in the blank using correct alternative.
Out of the following, point ........ lies to the right of the origin on X– axis.
Find the slope of a line passing through the given pair of points (9,-2) and (-5,5)
Find the slope of a line parallel to the given line 5x + 2y = 11
Find the value of a line perpendicular to the given line 5x+2y-9 = 0
Find the slope of the line passing through the points A(6, -2) and B(–3, 4).
Find the slope of the line passing through the points A(4,7) and B(2,3).
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.
