Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर
The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of LM = `(12 - 5)/(9 - 0) = 7/9`

Slope of MN = `(14 - 12)/(3 - 9) = 2/(-6) = -1/3`
Slope of LN = `(14 - 5)/(3 - 0) = 9/3` = 3
Slope of MN × Slope of LN = `-1/3 xx 3` = –1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle.
Verification:
Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
LN = `sqrt((3 - 0)^2 + (14 - 5)^2`
= `sqrt(3^2 + 9^2)`
= `sqrt(9 + 81)`
= `sqrt(90)`
MN = `sqrt((9 - 3)^2 + (12 - 14)^2`
= `sqrt(6^2+ (- 2)^2`
= `sqrt(36 + 4)`
= `sqrt(40)`
LM = `sqrt((9 - 0)^2 + (12 - 5)^2`
= `sqrt(9^2 + 7^2)`
= `sqrt(81 + 49)`
= `sqrt(130)`
LM2 = LN2 + MN2
130 = 90 + 40
130 = 130
⇒ Pythagoras theorem is 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 0
What is the inclination of a line whose slope is 1
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 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 which is perpendicular to a line joining the points (0, 0) and (− 8, 8) is
If slope of the line PQ is `1/sqrt(3)` then slope of the perpendicular bisector of PQ is
