Advertisements
Advertisements
प्रश्न
Find the slope of a line joining the points
`(5, sqrt(5))` with the origin
Advertisements
उत्तर
The given points is `(5, sqrt(5))` and (0, 0)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
= `(0 - sqrt(5))/(0 - 5)`
= `sqrt(5)/5`
= `1/sqrt(5)`
APPEARS IN
संबंधित प्रश्न
What is the slope of a line whose inclination with positive direction of x-axis is 90°
Find the slope of a line joining the points
(sin θ, – cos θ) and (– sin θ, cos θ)
If the three points (3, – 1), (a, 3) and (1, – 3) are collinear, find the value of a
The line through the points (– 2, a) and (9, 3) has slope `-1/2` Find the value of a.
The line through the points (– 2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Show that the given points form a parallelogram:
A(2.5, 3.5), B(10, – 4), C(2.5, – 2.5) and D(– 5, 5)
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
The slope of the line joining (12, 3), (4, a) is `1/8`. The value of ‘a’ is
If slope of the line PQ is `1/sqrt(3)` then slope of the perpendicular bisector of PQ is
Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram
