Advertisements
Advertisements
प्रश्न
The points (– 4, 0), (4, 0), (0, 3) are the vertices of a ______.
विकल्प
right triangle
isosceles triangle
equilateral triangle
scalene triangle
उत्तर
The points (– 4, 0), (4, 0), (0, 3) are the vertices of a isosceles triangle.
Explanation:
Let A(– 4, 0), B(4, 0), C(0, 3) are the given vertices.
Now, distance between A(– 4, 0) and B(4, 0),
AB = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
AB = `sqrt([4 - (-4)]^2 + (0 - 0)^2`
= `sqrt((4 + 4)^2`
= `sqrt(8^2)`
= 8
Distance between B(4, 0) and C(0, 3),
BC = `sqrt((0 - 4)^2 + (3 - 0)^2`
= `sqrt(16 + 9)`
= `sqrt(25)`
= 5
Distance between A(– 4, 0) and C(0, 3),
AC = `sqrt([0 - (-4)]^2 + (3 - 0)^2`
= `sqrt(16 + 9)`
= `sqrt(25)`
= 5
∵ BC = AC
Hence, ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.
संबंधित प्रश्न
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Name the type of quadrilateral formed, if any, by the following point, and give reasons for your answer:
(−3, 5), (3, 1), (0, 3), (−1, −4)
The value of 'a' for which of the following points A(a, 3), B (2, 1) and C(5, a) a collinear. Hence find the equation of the line.
Find all possible values of x for which the distance between the points
A(x,-1) and B(5,3) is 5 units.
Find the distance of the following point from the origin :
(13 , 0)
A point A is at a distance of `sqrt(10)` unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.
Find distance between point A(7, 5) and B(2, 5)
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.