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)
What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment joining (4, 2) and (–6, 4).
If the three points (3, – 1), (a, 3) and (1, – 3) are collinear, find the value of a
If the points A(2, 2), B(– 2, – 3), C(1, – 3) and D(x, y) form a parallelogram then find the value of x and y.
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
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
