Home
Class 11
MATHS
Let the base of a triangle lie along the...

Let the base of a triangle lie along the line `x = a` and be of length `a.` The area of this triangle is `a2`, if the vertex lies on the line

A

straight line

B

circle

C

ellipse

D

parabola

Text Solution

Verified by Experts

Let the base of the triangle be the line segment joining B(0, 0) and C(a, 0) and let the vertex be A(h, k), where a is fixed. Also, let
`(AB)/(AC)=lambda, lambda !=1`
`rArr AB^(2)=lambda^(2)AC^(2)`
`rArrh^(2)+k^(2)=lambda^(2){(h-a)^(2)+k^(2)}`
`rArr h^(2)(1-lambda^(2))+k^(2)(1-lambda^(2))+2a lambda^(2)h-a^(2)lambda^(2)=0`
So, A(h, k) lies on
`x^(2)+y^(2)+2a(lambda^(2))/(1-lambda^(2))x-(a^(2)lambda^(2))/(1-lambda^(2))=0`, which is a circle.
Promotional Banner

Similar Questions

Explore conceptually related problems

Let the base of a triangle lie along the line x = a and be of length 2a. The area of this triangles is a^(2) , if the vertex lies on the line

Let the base of a triangle lie along the line x = a and be of length 2a. The area of this triangles is a^(2) , if the vertex lies on the line

Let the base of a triangle lie along the line x = a and be of length a. The area of this triangles is a^(2) , if the vertex lies on the line

Let the base of a triangle lie along the line x = a and be of length a. The area of this triangles is a^(2) , if the vertex lies on the line

The base of a triangle lies along the line x=a and is of length a. The area of the triangle is a^2. The locus of the third vertex is

The base of a triangle lies along x= 1and is of length 1. The area of triangle is 1. The locus of vertex is

The base of a triangle lies along x=5 and is of length 1 .The area of triangle is 1 . The locus of vertex is

The base of an equilateral triangle is along the line given by 3x+4y=9 . If a vertex of the triangle is (1,2) then length of a side of the triangle is

One side of length 3a of a triangle of area a^2 square units lies on the line x = a. Then one of the lines on which the third vertex lies, is :