Advertisements
Advertisements
Question
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
Advertisements
Solution
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7) are the vertices of a quadrilateral.
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(-4 + 4)/(9 - 3) = 0/6`= 0
Slope of BC = `(-7 + 4)/(5 -9)= (-3)/(-4) = 3/4`
Slope of CD = `(-7 + 7)/(7 - 5) = 0/2` = 0
Slope of AD = `(-7 + 4)/(7 - 3) = (-3)/4 = - 3/4`
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly, the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.
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).
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.
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)
The slope of the line joining (12, 3), (4, a) is `1/8`. The value of ‘a’ is
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
