Let `A-=(3,-4),B-=(1,2)`. Let `P-=(2k-1+1)` be a variable point such that `PA+PB` is the minimum. Then k isA. `7//9`B. 0C. `7//8`D. none of these

Let `A-=(3,-4),B-=(1,2)`. Let `P-=(2k-1+1)` be a variable point such t

1.	Let `A-=(3,-4),B-=(1,2)`. Let `P-=(2k-1+1)` be a variable point such that `PA+PB` is the minimum. Then k isA. `7//9`B. 0C. `7//8`D. none of these
Answer» Correct Answer - C We know that `PA+PBgeAB` (by triangle inequality). So, `PA+PB` is the minimum if `PA+PB=AB` , i.e., A,P,B are collinear. Therefore, `\|{:(3,,-4,,1),(1,,2,,1),(2k-1,,2k+1,,1):}\|=0` or `3(2-2k-1)+4(1-2k+1)+1(2k+1)+1(2k+1-4k+2)=0` or `3-6k+8-8k+3-2k=0` or `14-16k=0` `therefore k=(7)/(8)`

Discussion

No Comment Found

Related InterviewSolutions

If`|x_1y_1 1x_2y_2 1x_3y_3 1|=|a_1b_1 1a_2b_2 1a_3b_3 1|`then the two triangles with vertices `(x_1, y_1),(x_2,y_2),(x_3,y_3)`and `(a_1,b_1),(a_2,b_2),(a_3,b_3)`areequal to area(b) similarcongruent(d) none of theseA. equal in areaB. similarC. congruentD. none of these
Let `A_1,A_2,A_3,...,A_n` are n Points in a plane whose coordinates are `(x_1,y_1),(x_2,y_2),....,(x_n,y_n)` respectively. `A_1A_2` is bisected at the point `P_1,P_1A_3` is divided in the ratio `1:2` at `P_2,P_2A_4` is divided in the ratio `1:3` at `P_3,P_3A_5` is divided in the ratio `1:4` at `P_4` and the so on until all n points are exhausted. find the coordinates of the final point so obtained.
Consider three lines as follows.`L_1:5x-y+4=0``L_2:3x-y+5=0``L_3: x+y+8=0`If these lines enclose a triangle `A B C`and the sum of the squares of the tangent to the interior angles can beexpressed in the form `p/q ,`where `pa n dq`are relatively prime numbers, then the value of `p+q`is500 (b) 450(c) 230 (d)565A. 500B. 450C. 230D. 465
The locus of the moving point whose coordinates are given by `(e^t+e^(-t),e^t-e^(-t))`where `t`is a parameter, is`x y=1`(b) `x+y=2``x^2-y^2=4`(d) `x^2-y^2=2`A. `xy=1`B. `x+y=2`C. `x^2-y^2=4`D. `x^2-y^2=2`
Find the locus of the point `(t^2-t+1,t^2+t+1),t in Rdot`
The locus of a point represent by `x=(a)/(2)((t+1)/(t)),y=(a)/(2)((t-1)/(t))`, where `t=in R-{0}`, isA. `x^2+y^2=a^2`B. `x^2-y^2=a^2`C. `x+y=a`D. `x-y=a`
The maximum area of the triangle whose sides `a ,b`and `5sintheta),`and `(5sintheta,-5costheta),`where `theta in Rdot`The locus of its orthocentre is`(x+y-1)^2+(x-y-7)^2=100``(x+y-7)^2+(x-y-1)^2=100``(x+y-7)^2+(x+y-1)^2=100``(x+y-7)^2+(x-y+1)^2=100`A. 1B. `1//2`C. 2D. `3//2`
Convert `rsintheta=rcostheta+4`into itsequivalent Cartesian equation.
Convert `r=cos e cthetae^(rcostheta)`into itsequivalent Cartesian equation.
Let a,b,c be in A.P and x,y,z be in G.P.. Then the points `(a,x),(b,y)` and `(c,z)` will be collinear ifA. `x^2=y`B. `x=y=z`C. `y^2=z`D. `x=z^2`

Let `A-=(3,-4),B-=(1,2)`. Let `P-=(2k-1+1)` be a variable point such that `PA+PB` is the minimum. Then k isA. `7//9`B. 0C. `7//8`D. none of these

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment