Advertisements
Advertisements
प्रश्न
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem
A(1, – 4), B(2, – 3) and C(4, – 7)
Advertisements
उत्तर
The vertices are A(1, – 4), B(2, – 3) and C(4, – 7)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(-3 + 4)/(2 - 1) = 1/1` = 1
Slope of BC = `(-7 + 3)/(4 -2) = (-4)/2` = – 2
Slope of AC = `(-7 + 4)/(4 - 1) = - 3/3` = – 1
Slope of AB × Slope of AC = 1 × – 1 = – 1
∴ AB is ⊥r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:
Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
AB = `sqrt((2 - 1)^2 + (-3 + 4)^2`
= `sqrt(1^2 + 1^2)`
= `sqrt(2)`
BC = `sqrt((4 - 2)^2 + (- 7 + 3)^2`
= `sqrt((2)^2 + (- 4)^2`
= `sqrt(4 + 16)`
= `sqrt(20)`
AC = `sqrt((4 - 1)^2 + (-7 + 4)^2`
= `sqrt(3^2 + (- 3)^2`
= `sqrt(9 + 9)`
= `sqrt(18)`
BC2 = AB2 + AC2
`(sqrt(20))^2 = (sqrt(2))^2 + (sqrt(18))^2`
20 = 2 + 18
20 = 20
⇒ Pythagoras theorem verified
APPEARS IN
संबंधित प्रश्न
What is the slope of a line whose inclination with positive direction of x-axis is 0°
What is the inclination of a line whose slope is 1
Find the slope of a line joining the points
`(5, sqrt(5))` with the origin
Find the slope of a line joining the points
(sin θ, – cos θ) and (– sin θ, cos θ)
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 right angled triangle and check whether they satisfy Pythagoras theorem.
L(0, 5), M(9, 12) and N(3, 14)
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)
A quadrilateral has vertices at A(– 4, – 2), B(5, – 1), C(6, 5) and D(– 7, 6). Show that the mid-points of its sides form a parallelogram.
If slope of the line PQ is `1/sqrt(3)` then slope of the perpendicular bisector of PQ is
