`A B C` is a variable triangle such that `A` is (1, 2), and `Ba n dC` on the line `y=x+lambda(lambda` is a variable). Then the locus of the orthocentr

Monjur Alahi

`A B C` is a variable triangle such that `A` is (1, 2), and `Ba n dC` on the line `y=x+lambda(lambda` is a variable). Then the locus of the orthocentre of `"Delta"A B C` is `x+y=0` (b) `x-y=0` `x^2+y^2=4` (d) `x+y=3`
A. x+y=0
B. x-y=0
C. `x^(2) + y^(2)=4`
D. x+y=3


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - D
As the altitude from A is fixed and the orthocenter lies on the altitude, x+y=3 is the required locus.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
A variable line `x/a + y/b = 1` moves in such a way that the harmonic mean of a and b is 8. Then the least area of triangle made by the line with the

Correct Answer - C Given line is `(x)/(a)+(y)/(b)=1` It meets the axes at point A(a,0) and B(0,b). Area of triangle AOB, `Delta = (1)/(2)` ab Given that 8 is H.M. of...

2 Answers 4 views
If two members of family `(2+lambda)x+(1+2lambda)y-3(1+lambda) = 0` and line x+y=0 make an equilateral triangle, the the incentre of triangle so forme

Correct Answer - A Given equation of family of lines is `(2x+y-3)+lambda(x+2y-3) =0, lambda in R.` The family of lines is concurrent at point of intersection of 2x+y-3 = 0 and...

2 Answers 3 views
If `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C`, where A, B and C are matrices then find matrices B a

We have `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C` Putting `lambda=0`, we get `C=[(1,-2),(1,1)]` Putting `lambda=1`, we get `A+B+C=[(0, -1),(3,0)]` ...(1) Putting `lambda=-1`, we get `A-B+C=[(4,-3),(-3,0)]` Subtracting (2) from (1), we get `2B[(0, -1),(3,0)]-[(4,-3),(-3,0)]=[(-4,2),(6,0)]` `:. B=[(-2,...

2 Answers 1 views
If the sum of the slopes of the normal from a point `P` to the hyperbola `x y=c^2` is equal to `lambda(lambda in R^+)` , then the locus of point `P` i

Correct Answer - `x^(2)=lambdac^(2)` Equation of hyperbola is `xy = c^(2)`. Differentiating w.r.t. x, we get `(dy)/(dx)=-(y)/(x)` Thus, slope of normal at any point on the hyperbola is `-(dx)/(dy)=(x)/(y)`. If point...

2 Answers 2 views
If the coordinates of a variable point `P` are `(acostheta,bsintheta),` where `theta` is a variable quantity, then find the locus of `Pdot`

Let `P -=(x,y)`. According to the question, `x=acostheta`(1) `y=bsintheta` (2) Sqaureing and adding (1) and (2), we get `(x^2)/(a^2)+(y^2)/(b^2)=cos^2theta+sin^2theta` or `(x^2)/(a^2)+(y^2)/(b^2)=1`

2 Answers 1 views
Vertices of a variable triangle are `(3,4); (5costheta, 5sintheta)` and `(5sintheta,-5costheta)` where `theta` is a parameter then the locus of its ci

Correct Answer - D Distance of all the points from (0,0) are 5 units. That means the circumcenter of the triangle formed by the given points is (0,0). If `G-=(h,k)` is...

2 Answers 2 views
If l and m are variable real number such that `5l^(2)+6m^(2)-4lm+3l=0`, then the variable line lx+my=1 always touches a fixed parabola, whose axes is

Correct Answer - B (2) Any parabola whose axes whose axes is parallel to the x-axis will be of the from `(y-a)^(2)=4b(x-c)` (1) Now, lx+my=1 can be rewritten as `y-a=-(l)/(m)(x-c)+(1-am-lc)/(m)` (2)...

2 Answers 1 views
In triangle `A B C` , base `B C` and area of triangle are fixed. The locus of the centroid of triangle `A B C` is a straight line that is parallel to

Correct Answer - A `Delta = (1)/(2) (BC) h`, where h is the distance of vertex A from `Delta_(GBC) = (Delta)/(3) = ((BC)h)/(6)`, where G is the centroid `rArr h =...

2 Answers 1 views
Find all values of `lambda` for which the `(lambda-1)x+(3lambda+1)y+2lambdaz=0(lambda-1)x+(4lambda-2)y+(lambda+3)z=0 2x+(3lambda+1)y+3(lambda-1)z=0` p

Correct Answer - D `(d)` `|{:(lambda-1,3lamda+1,2lambda),(lambda-1,4lambda-2,lambda+3),(2,3lambda+1,3(lambda-1)):}|=0` `implieslambda=0` or `3` If `lambda=0`, the equations become `-x+y=0`, `-x-2y+3z=0` and `2x+y-3z=0`, `:. (x)/(6-3)=(y)/(6-3)=(z)/(-1+4)`

2 Answers 1 views
A spectroscopic instrument can resolve two nearly wavelength `lambda and lambda = Delta lambda if lambda//Delta lambda` is smaller than `8000` This is

Correct Answer - [38]

2 Answers 1 views
X