Advertisements
Advertisements
प्रश्न
Find the slope of a line, correct of two decimals, whose inclination is 30°
उत्तर
Slope of line = m = tan θ
= tan 30°
= `1/sqrt 3`
APPEARS IN
संबंधित प्रश्न
Find the slope of the line passing through the points A(2, 3) and B(4, 7).
Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
The ordinate of a point lying on the line joining the points (6, 4) and (7, –5) is –23. Find the coordinates of that point.
Determine whether the given point is collinear.
L(1,2), M(5,3) , N(8,6)
Find the value of x so that the line passing through (3, 4) and (x, 5) makes an angle 135° with positive direction of X-axis.
Show that the points A(- 2, 5), B(2, – 3) and C(0, 1) are collinear.
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.
Determine whether the following points are collinear. A(–1, –1), B(0, 1), C(1, 3)
Given: Points A(–1, –1), B(0, 1) and C(1, 3)
Slope of line AB = `(square - square)/(square - square) = square/square` = 2
Slope of line BC = `(square - square)/(square - square) = square/square` = 2
Slope of line AB = Slope of line BC and B is the common point.
∴ Points A, B and C are collinear.
