Advertisements
Advertisements
प्रश्न
Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Advertisements
उत्तर
AB =`sqrt((-3 -1)^2 + (0 +3)^2) = sqrt(16+9) = sqrt(25)` = 5
BC =`sqrt((4 + 3)^2 + (1 +0)^2)= sqrt(49+1)= sqrt(50) = 5sqrt(2)`
CA =`sqrt((1 -4)^2 + (-3 - 1)^2) = sqrt(9 + 16) = sqrt(25)` = 5
∵ AB = CA
A, B, C are the vertices of an isosceless triangle.
AB2 + CA2 = 25 + 25 = 50
BC2 = `(5sqrt(2))^2` = 50
∴ AB2 + CA2 = BC2
Hence, A, B, C are the vertices of a right-angled triangle.
Hence, ΔABC is an isosceles right-angled triangle.
Area of ΔABC = `(1)/(2) xx "AB" xx "CA"`
= `(1)/(2) xx 5 xx 5`
= 12.5 sq.units
APPEARS IN
संबंधित प्रश्न
If the opposite vertices of a square are (1, – 1) and (3, 4), find the coordinates of the remaining angular points.
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Find the distance of the following points from the origin:
(i) A(5,- 12)
Find all possible values of x for which the distance between the points
A(x,-1) and B(5,3) is 5 units.
If P (x , y ) is equidistant from the points A (7,1) and B (3,5) find the relation between x and y
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Prove that the points (0 , -4) , (6 , 2) , (3 , 5) and (-3 , -1) are the vertices of a rectangle.
Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.
