Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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).
Show that the given points are collinear: (– 3, – 4), (7, 2) and (12, 5)
The line through the points (– 2, a) and (9, 3) has slope `-1/2` Find the value of a.
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
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.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (− 8, 8) is
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.
