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प्रश्न
उत्तर
Slope of line MN = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
= `(3 - 9)/(-2 - 4) = (-6)/(-6)`
= -1
Slope of line parallel to MN = Slope of MN = 1
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संबंधित प्रश्न
A(5, 4), B(−3, −2) and C(1, −8) are the vertices of a triangle ABC. Find:
- the slope of the altitude of AB,
- the slope of the median AD and
- the slope of the line parallel to AC.
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Angles made by the line with the positive direction of X–axis is given. Find the slope of these line.
60°
Find the slope of the lines passing through the given point.
E(–4, –2) , F (6, 3)
Find k, if B(k, –5), C (1, 2) and slope of the line is 7.
Determine whether the given point is collinear.
\[P\left( 1, 2 \right), Q\left( 2, \frac{8}{5} \right), R\left( 3, \frac{6}{5} \right)\]
Verify whether the following points are collinear or not:
A(1, –3), B(2, –5), C(–4, 7).
With out Pythagoras theorem, show that A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right angled.
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.
