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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
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 |
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Answer» Correct Answer - A We have `(1)/(2)|{:(x_1,,y_1,,1),(x_2,,y_2,,1),(x_3,,y_3,,1):}|=|{:(a_1,,b_1,,1),(a_2,,b_2,,1),(a_3,,b_3,,1):}|or (1)/(2)|{:(x_1,,y_1,,1),(x_2,,y_2,,1),(x_3,,y_3,,1):}|=(1)/(2)|{:(a_1,,b_1,,1),(a_2,,b_2,,1),(a_3,,b_3,,1):}|` Hence, the area of triangle with vertices `(x_1,y_1),(x_2,y_2),(x_3,x_3)` is the same as the area of triangle with vertices `(a_1,b_1),(a_2,b_2),(a_3,b_3)`. Hence, the two triangles are equal in area. |
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| 2. |
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. |
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Answer» `P_1` is midpoint of `A_1A_2`. `therefore" "P_1-=((x_1+x_2)/(2),(y_1+y_2)/(2))` `P_2` divides `P_1A_3` in `1:2`. `therefore" "P_2-=((2((x_1+x_2)/2)+x_3)/(2+1),(2((y_1+y_2)/2)+y_3)/(2+1))` `-=((x_1+x_2+x_3)/(3),(y_1+y_2+y_3)/(3))` Now, `P_3` divides `P_2A_4` in ` 1:3` `therefore" "P_3-=((3.((x_1+x_2+x_3)/3)+x_4)/(3+1),(3.((y_1+y_2+y_3)/3)+y_4)/(3+1))` `-=((x_1+x_2+x_3+x_4)/(4),(y_1+y_2+y_3+y_4)/(4))` Proceeding in this manner, we get `P_n-=((x_1+x_2+x_3+....x_n)/(n),(y_1+y_2+y_3+....y_n)/(n))`. |
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| 3. |
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 |
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Answer» Correct Answer - D Arranging the lines in desceding order of slope, we have `m_1=5,m_2=3,and m_3=-1` `therefore tan A =(m_1-m_2)/(1+m_1m_2)=(2)/(1+15)=(1)/(8)` `tan B=(m_1-m_3)/(1+m_2m_3)=(3+1)/(1-3)=-2` `tan C=(m_3-m_1)/(1+m_3m_1)=(-1-5)/(1-5)=(3)/(2)` `Sigma tan^(2)A=(1)/(64)+4+(9)/(4)=(1+256+144)/(64)=(401)/(64)` or `p+q=465` |
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| 4. |
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` |
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Answer» Correct Answer - C Let `(e^t+e^(-t),e^(t)-e^(-t))-=(h,k)` or `h=e^t+e^(-t) and k=e^(t)-e^(-t)` Squareing and subtracing we get `h^2-k^2=(e^t+e^(-t))^2-(e^(t)-e^(-t))^2=4` Therefore,the locus is `x^2-y^2=4`. |
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| 5. |
Find the locus of the point `(t^2-t+1,t^2+t+1),t in Rdot` |
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Answer» Let `(h,h)-=(t^2-t+1,t^2+t+1)` or `h=t^2-t+1` and `k=t^2+t+1` or `k-h=2t` or `t=(k-h)/(2)` or `h=((k-h)/(2))^(2)-((k-h)/(2))+1` The required locus is `x=((x-y)/(2))^2-((y-x)/(2))+1` |
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| 6. |
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` |
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Answer» Correct Answer - C Here, `x=(a)/(2)(2+(1)/(t))` .....(1) `y=(a)/(2)(1-(1)/(t))`......(2) Adding (i) and (ii) we get `x+y=a` which is the required locus. |
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| 7. |
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` |
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Answer» Correct Answer - A Let the vertices be `O(0,0),A(alpha,0)`, and `B(alpha_1,beta_1)`, where `0le alphale1 ` and `1lealpha_(1)^(2)+beta_(1)^(2) le4` So, the area of `DeltaOAB` is maximum where `alpha=1` and `(alpha_1,beta_1)` is (2,0) In this case, `a=1,b=2`, and `c=sqrt5`,which satisfies `2lecle3`. Therefore, the maximum area is 1. |
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| 8. |
Convert `rsintheta=rcostheta+4`into itsequivalent Cartesian equation. |
| Answer» Given `r sin theta =rcos theta +4`. Therefore, `y=x+4`. | |
| 9. |
Convert `r=cos e cthetae^(rcostheta)`into itsequivalent Cartesian equation. |
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Answer» `r=cosecthetae^(rcostheta)=(1)/(sintheta)e^(rcostheta)` or `rsintheta=e^(rcostheta)` or `y=e^(x)` |
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| 10. |
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` |
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Answer» Correct Answer - B Given that a,b,c are in A.P. `rArra-b=b-c` Now, `(a,x),(b,y)and (c,z)` are collinear. `rArr(x-y)/(a-b)=(y-z)/(b-c)` `rArr (x-y)/(y-z)=(a-b)/(b-c)=1` `rArrx-y=y-z` So, x,y,z are in A.P. Thus x,y,z are in A.P and also in G.P. `therefore x=y=z`. |
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| 11. |
If the point `(2,3),(1,1),a n d(x ,3x)`arecollinear, then find the value of `x ,`using slopemethod. |
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Answer» Given Points `A(2,3),B(1,1)`, and `C(x,3x)` are collinear. Then slope of AB=Slope of BC or `(1-3)/(1-2)=(3x-1)/(x-1)` or `3x-1 =2x-2` or `x=-1` |
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| 12. |
Find the locus of a point such that the sum of itsdistance from the points (0, 2) and `(0,-2)`is 6. |
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Answer» Let P(h,k) be any point on the locus and let A(0,2) and `B(0,-2)` be the given points. By the given condition. We get `PA+PB=6` or `sqrt((h-0)^2+(k-2)^2)+sqrt((h-0)^2+(k+2)^2)=6` or `sqrt(h^2+(k-2)^2)=6-sqrt((h-0)^2+(k+2)^2)` or `h^2+(k-2)^2=36-12sqrt(h^2+(k+2)^2)+h^2+(k+2)^2` or `-8k-36=-12sqrt(h^2+(k+2)^2)` `(2k+9)=3sqrt(h^2+(k+2)^2)` or `(2k+9)^2=9{h^2+(k+2)^2}` or `4k^2+36k+81=9h^2+9k^2+36k+36` or `9h^2+5k^2=45` Hence the locus of (h,k) is `9x^2+5y^2=45`. |
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| 13. |
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 |
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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)` |
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| 14. |
For what value of `k`are thepoints `(k ,2-2k),(-k+1,2k)a n d(-4-k ,6-2k)`collinear? |
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Answer» Let the three given points be `A-=(x_1,y_1)-=(k,2-2k),B-=(x_2,y_2)-=(-k+1,2k)`, and `C-=(x_3,y_3)-=(-4-k,6-2k)`. If the given points are collinear, then `Delta=0`, i.e., `x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0` or `k(2k-6+2k)+(-k+1)(6-2k-2+2k)+ (-4-k)(2-2k) =0` or `k(4k-6)-4(k-1) +(4+k)(4k-2)=0` or `4k^2 -6k-4k+4+4k^2+14k-8=0` or `8k^2+4k-4=0` or `2k^2+k-1=0` or `(2k^2-1)(k+1)=0` i.e., `k=(1)/(2)` or `-1` But for `t=1//2`, the first two points are coincident. Hence ,`K=-1`. |
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| 15. |
If `a ,b ,c`are the `p t h ,q t h ,r t h`terms, respectively, of an `H P`, show that the points `(b c ,p),(c a ,q),`and `(a b ,r)`are collinear. |
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Answer» Let the first term be A and the common difference be D of the corresponding AP. Given pth term of `HP=a` `therefore` pth term `AP=(1)/(a)` `therefore A + (p-1)D=(1)/(a)` Similarly, `A+(q-1)D=(1)/(b)` `A+(r-1)D=(1)/(c)` Subtracting,we get `(p-q)D=(1)/(a)=(1)/(b)` and `(q-r)D=(1)/(b)-(1)/(c)` Dividing, we get `(p-q)/(q-r) =(bc-ac)/(ac-ab)` and Slope of `BC =(r-q)/(ab-ca)=(q-p)/(ca-bc)` [Using (1)] `therefore` Slope of BC =Slope of AB Therefore, A, B, and C are collinear. |
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| 16. |
if `A(cosalpha,sinalpha),B(sinalpha,-cosalpha),C(1,2)` are the vertices of `DeltABC`,then as `alpha` Find the locsus of its centroid. |
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Answer» Let (h,k) be the triangle . Then, `h=(cosalpha+sinalpha+1)/(3)` and `k=(sinalpha-cosalpha+2)/(3)` or `3h-1=cosalpha+sinalpha` and `3k-2=sinalpha-cosalpha` Squareing and adding, we get ` (3h-1)^2+(3k-2)^2=2` or `9(h^2+k^2)-6h-12k+3=0` or `3(h^2+k^2)-2h-4k+1=0` Therefore the locus of the centroid is `3(x^2+y^2)-2x-4y+1=0` . |
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| 17. |
If the roots of the equation `(x_(1)^(2)-a^2)m^2-2x_1y_1m+y_(1)^(2)+b^2=0(agtb)` are the slopes of two perpendicular lies intersecting at `P(x_1,y_1)`, then the locus of P isA. `x^2+y^2=a^2+b^2`B. `x^2+y^2=a^2-b^2`C. `x^2-y^2=a^2+b^2`D. `x^2-y^2=a^2-b^2` |
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Answer» Correct Answer - B Equation `(x_1^(2)-a^2)m^2-2x_1y_1m+y_1^(2)+b^2=0` has roots `m_1` and `m_2` `therefore m_1m_2=(y_(1)^(2)+b^2)/(x_(1)^(2)-a^2)=-1` (Given) `therefore x^2+y^2=a^2-b^2` |
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| 18. |
If the point `(x ,-1),(3, y),(-2,3),a n d(-3,-2)`taken inorder are the vertices of a parallelogram, then find the values of `xa n dydot` |
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Answer» Correct Answer - `x=2,y=4` The midpoints of the diagonals must be the same. Therefore, ` (x-2)/(2)=(-3+3)/(2)` or `x=2` and ` (-1+3)/(2)=(-2+y)/(2)` or `y=4` |
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| 19. |
The line joining the points `(x ,2x)a n d(3,5)`makes anobtuse angle with the positive direction of the x-axis. Then find the valuesof `xdot` |
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Answer» Correct Answer - `xin(5//2,3)` The slope joining points `A(x,2x)` and `B(3,5)` is `(2x-5)//(x-3)` if AB makes an angle `theta` with the positive direction of the x-axis, then `tan theta=(2x-5)/(x-3)` Since, `theta` is obtuse, `tanthetalt0` or `(2x-5)/(x-3)lt0` or `x in ((5)/(2),3)` |
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| 20. |
If line `3x-ay-1=0` is parallel to the line `(a+2)x-y+3=0` then find the values of a. |
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Answer» The slope of the line `3x-ay-1=0` is `3//a`. The slope of the line `(a+2)x-y+3=0` is `a+2`. Since the lines are parallel, we have `a+2(3)/(a)` or `a^2+2a-3=0` or `(a-1)(a+3)=0` or `a=1or a=-3`. |
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| 21. |
Triangle ABC lies in the cartesian plane and has an area of 70 sq. units. The coordinates of B and C are `(12,19)`, and `(23,20)` respectively. The line containing the median to the side BC has slope `-5`. Find the possible coordinates of point A. |
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Answer» Correct Answer - `(15,32)`or `(20,7)` Let the coordinates of point `A be (h,k)`. Midpoint of BC is `D(35//2,39//2)`. Slope of AD is `((39)/(2)-k)/((35)/(2)-h)=-5` `rArr 39-2k=-5(35-3h)` `rArr39-2k=-175+10h` `rArr5h+k=107` Also, area of `DeltaABC` is 70 sq.units. `therefore|{:(h,,h,,1),(12,,19,,1),(23,,20,,1):}|=+-140` `rArr11k-h=337` (1) or `11k-h=57` (2) Solving (1) and (2), we get `h=15,k=32 ` Solving (1) and (3), we get `h=20,k=7` So, possible coordinates of A are `(15,32) or (20,7)`. |
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| 22. |
Let `A(6,4)a n dB(2,12)`be two given point. Find the slope of a lineperpendicular to `A Bdot` |
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Answer» Let m be the slope of AB. Then, `m=(12-4)/(2-6)=(8)/(-4) =-2` So, the slope of a line perpendicular to AB is `(1)/(m)=(1)/(2)` |
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| 23. |
Convert the following points from polar coordinates to the corresponding Cartesian coordinates. (i) `(2,pi//3)` (ii) `(0.pi//2)` (iii) `(-sqrt(2),pi//4)` |
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Answer» (i) `(2,pi//3)-=(r,theta)` We have ` x=rcostheta=2"cos"(pi)/(3)=2xx(1)/(2)=1` and `y=rsintheta=2"sin"(pi)/(3)=2xxsqrt(3)/(2)=sqrt(3)` Therefore, the point is `(1,sqrt(3)` in the Cartesian coordinates. (iii) `x=-sqrt(2) "cos"(pi)/(4)=-sqrt(2)((1)/(sqrt2))=-1` `y=-sqrt(2)"sin"(pi)/(4) =-sqrt(2)((1)/(sqrt2))=-1` So, the equivalent Cartesian coordinates for the given polar coordinates are `(-1,-1)`. |
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| 24. |
Express the polar equation `r-2costheta`inrectangular coordinates. |
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Answer» We use the formulas `r^2=x^2+y^2` and `x=rcostheta`. ltbr Given `r=2costheta`. Therefore, `r^2=2rcostheta` or `x^2+y^2=2x` or `x^2x+y^2=0` or `x^2-2x+1+y^2=1` or `(x-1)^2+y^2=1`. |
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| 25. |
Find the minimum distance of any point on the line`3x+4y-10=0`from theorigin using polar coordinates. |
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Answer» Correct Answer - 2 units Let the polar coordinates of any point on the curve be `P(r,theta)`. Then its cartesian coordinates are `P(rcostheta,rsintheta)`. P lies on the line. Therefore, `3rcostheta+4rsintheta-10=0` or `r=(10)/(3costheta+4sintheta)` Now, `(3costheta+4sintheta)_(max)=5` `therefore r_(min)=(10)/(5)=2` |
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| 26. |
If `(x , y)`and `(x ,y)`are the coordinates of the same point referred to two sets ofrectangular axes with the same origin and it `u x+v y ,`where `u`and `v`are independent of `xa n dy`, becomes `V X+U Y ,`show that `u^2+v^2=U^2+V^2dot` |
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Answer» Let the axes rotate at angle `theta`. If `(x,y)` is the point with respect to the old axes and (x,y) are the coordinates with respect to the new axes, then `{(x,=,Xcostheta-Ysintheta,,),(y,=,Xsintheta+Ycostheta,,):}` Then `ux+vy=u(Xcostheta-Ysintheta)+v(Xsintheta + Y costheta` `=(ucostheta +vsintheta )X+(-usintheta +vcos theta )Y` But given new exression is `VX+UY`. Then, `VX+UY=(ucostheta +vsintheta )X+(-usintheta +vcos theta) Y` On comparing the coefficients of X and Y, we get `ucostheta +vsintheta=V` (1) and `-usintheta +vcostheta=U` (2) Squareing and adding (1) and (2) , we get `u^2+v^2=U^2+V^2` |
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| 27. |
If the points `A(0,0),B(cosalpha,sinalpha)`, and `C(cosbeta,sinbeta)` are the vertices of a right- angled triangle, thenA. `sin.(alpha-beta)/(2)=(1)/(sqrt2)`B. `cos.(alpha-beta)/(2)=(1)/(sqrt2)`C. `cos.(alpha-beta)/(2)=-(1)/(sqrt2)`D. `sin.(alpha-beta)/(2)=-(1)/(sqrt2)` |
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Answer» Correct Answer - A::C::D Since `AB=AC=1`, the triangle is right -angled at point A. we have `tanalphatanbeta=-1` or `cos(alpha-beta)=0or alpha-beta=+-(pi)/(2)` |
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| 28. |
Given the equation `4x^2+2sqrt(3)x y+2y^2=1`. Through what angle should the axes be rotated sothat the term `x y`is removedfrom the transformed equation. |
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Answer» Correct Answer - `pi//6,2pi//3` Comparing the given equation with `ax^2+2hxy+by^2` we get `a=4,h=sqrt3,b=2`. Let `theta` be the angle through which the axes are to be rotated. Then, `tan 2theta=(2h)/(a b)` or `tan2theta=(2sqrt3)/(4-2)=sqrt(3)=tan.(pi)/(3)` or `2theta=(pi)/(3),pi+(pi)/(3)` or `theta=(pi)/(6),(2pi)/(3)` |
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| 29. |
In each of the following, check how the points A,B and C are situated. (i) `A(-2,2),B(8,-2),C(-4,-3)` (ii) `A(-a,-b),B(a,b),C(a^2,ab),agt1`(iii) `A(4,0),B(-1,-1),C(3,5)` |
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Answer» Correct Answer - (i) ABC is triangle right angled at A (ii) collinear points (iii) ABC is right angled isoceles triangle (i) `A(-2,2), B((8,-2),C(-4,-3)` `AB=sqrt((8-(-2))^2+(-2-2)^(2))` `=sqrt(100+16)=2sqrt(29)` `BC=sqrt((8-(-4))^2+(-2-(-3))^(2))` `=sqrt(144+1)=sqrt(5)=sqrt(29)` `CA=sqrt((-2-(-4))^2+(2-(-3))^(2))` `=sqrt(4+25)=sqrt(29)` Thus, `AB^2+CA^2=BC^2` So, triangle is right angled at A. (ii) `A(-a,-b),B(a,b),C(a^2,ab)` `AB=sqrt((2a)^2+(2b)^(2))=2sqrt(a^2+b^(2))` `BC=sqrt((a^2-a)^2+b^2(a-1)^(2))=(a-1)sqrt(a^2+b^(2)` `AC=sqrt((a^2+a)^2+b^2(a+1)^(2))=(a+1)sqrt(a^2+b^2)` Thus, `AB+BC=AC`. So, points are collinear. (iii) `A(4,0),B(-1,-1),C(3,5)`. `AB=sqrt((-1-4)^2+b^2(-1-0)^(2))=sqrt(25+1)=sqrt(26)` `BC=sqrt((3+1)^2+b^2(5+1)^(2))=sqrt(16+36)=sqrt(52)` `CA=sqrt((4-3)^2+b^2(0-5)^(2))=sqrt(1+25)=sqrt(26)` Thus, `AB=CA and BC62=AB^2+CA62` So, triangle ABC is right angled isoceles. |
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| 30. |
Show that the distance between the points `P(acosalpha, asinalpha)` and `Q(a sin beta,a sinbeta)` is `2a sin(a-b)/(2)` |
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Answer» Correct Answer - NA `PQ=sqrt(a^2(cosalpha-cosbeta)^2=a^2(sinalpha-beta)^2)` `a=sqrt(sin^2alpha+cos^2alpha+cos^2beta+sin^2beta-2cosalphacosbeta-2sinalphasinbeta)` `a=sqrt(2(1-cos(alpha-beta))` `a=sqrt(2xx2sin^(2).(alpha-beta)/(2))` `=2a"sin"(alpha-beta)/(2)` |
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| 31. |
`P`and `Q`are points on the line joining `A(-2,5)`and `B(3,1)`such that `A P=P Q=Q B`. Then, the distance of the midpoint of `P Q`from the origin is3 (b) `(sqrt(37))/2`(b) 4(d) 3.5A. 3B. `sqrt(37//2)`C. 4D. `3.5` |
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Answer» Correct Answer - B The midpoint of PQ is the midpoint of AB which is `R(1//2,3)`. Therefore, `OR=sqrt((1)/(4)+9)=sqrt(37)/(2)` . |
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| 32. |
The vertices of a triangle are `A(x_1,x_1tantheta_1),B(x_2, x_2tantheta_2),`and `C(x_3, x_3tantheta_3)dot`If thecircumcenter of ` A B C`coincideswith the origin and `H(a , b)`is theorthocentre, show that`a/b=(costheta_1+costheta_2+costheta_3)/(sintheta_1+sintheta_2+sintheta_3)` |
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Answer» Correct Answer - NA Since the circumcenter is at the origin, the orthocenter is `x_1+x_2+x_3,x_1tantheta_1+x_2tantheta_2+x_3tantheta_3)` `therefore a=x_1 +x_2+x_3` and `b=x_1tantheta_1+x_2tantheta_2+x_3tantheta_3` Also `x_1^(2)+x_1^(2) tantheta_1^(2) =x_2^(2)+x_2^(2)tan theta_2^(2)=x_3^(2)=x_3^(2)tan theta_3^(2)` or `x_1sec theta_1=x_2sec theta_2=x_3sec theta_3=lambda` (say) Now, `(a)/(b) =(x_1+x_2+x_3)/(x_1tantheta_1+x_2tan theta_2+x_3tantheta_3)` `=(lambdacostheta_1+lambdatheta_2+lambdatheta_3)/(lambdacostheta_1thantheta_1+ lambdacostheta_2tantheta_2+lambdacostheta_2tan theta_3)` `=(costheta_1+costheta_2+costheta_3)/(sintheta_1+sintheta+sintheta_3)` . |
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| 33. |
In a `triangle ABC` the sides `BC=5, CA=4` and `AB=3`. If `A(0,0)` and the internal bisector of angle A meets BC in D `(12/7,12/7)` then incenter of `triangle ABC` isA. `(2,2)`B. `(3,2)`C. `(2,3)`D. `(1,1)` |
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Answer» Correct Answer - D We have `a=5,b=4,c=3` Incentre I divides AD int the ratio `b+c:a`. Therefore,I divides AD in the ratio `7:5` Hence, coordinates of I are (1,1). |
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| 34. |
Find the incentre of the triangle with vertices `(1, sqrt3), (0, 0)` and `(2, 0)` |
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Answer» Correct Answer - `(1,(1)/(sqrt3))` Here `AB=BC=CA=2` So, it is an equilateral triangle and the incentre coincides with centroid. Therefore, centroid `((0+1+2)/(3),(0+0+sqrt3)/(3))-=(1,(1)/sqrt3)` |
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| 35. |
The vertices of a triangle are `(p q ,1/(p q)),(p q)),(q r ,1/(q r)),`and `(r q ,1/(r p)),`where `p ,q`and `r`are the roots of the equation `y^3=3y^2+6y+1=0`. The coordinates of its centroid are`(1,2)`(b) `2,-1)``(1,-1)`(d) `2,3)`A. `(1,2)`B. `(2,-1)`C. `(1,-1)`D. `(2,3)` |
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Answer» Correct Answer - B p,q,r, are the roots of equation `y^3-3y^2+6y+1=0`. So, `p+q+r=3,pq+qr+rp=6`, and `pqr=-1`. Now, the centroid of the triangle is `((pq_+qr+rp)/(3),((1)/(pq)+(1)/(qr)+(1)/(rp))/(3))` i.e., `((pq_+qr+rp)/(3),(p_+q+r)/(3pqr))-=((6)/(3),(3)/(6))or (2,-1)` |
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| 36. |
If the middle points of the sides of a triangleare `(-2,3),(4,-3),a n d(4,5)`, then find the centroid of the triangle. |
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Answer» Correct Answer - `(2,5//3)` As we know that the centroid of triangle ABC and that of the triangle formed by joining the middle points of the sides of triangle ABC are the same. So,the required centroid is `((4+4-2)/(3),(5-3+3)/(3))=(2,(1)/(sqrt3))` |
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| 37. |
If two vertices of a triangle are `(-2,3)`and `(5,-1)`the orthocentre lies at the origin, and the centroid on the line `x+y=7`, then the third vertex lies at`(7,4)`(b) `8,14)``(12 ,21)`(d) none of theseA. (7,4)B. (8,14)C. (12,21)D. none of these |
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Answer» Correct Answer - D Given O(0,0) is the orhtocenter . Let A(h,k) be the third vertex, and `B(-2,3)` and `C(5,-1)` the other two vertices. Then the slope of the line through A and O is k/h. while the line through B and C has the slope `-4//7`. By the property of the orthocenter, these two lines must be perpendiuclar. So, we have `(k/(h))(-(4)/7)=-1or (k)/(h)=(7)/(4)`....(1) Also, ` (5-2+h)/(3)+(-1+3+k)/(3)=7` ....(ii) `h+k=16` |
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| 38. |
Statement 1: If in a triangle, orthocentre, circumcentre and centroid are rational points, then its vertices must also be rational points. Statement : 2 If the vertices of a triangle are rational points, then the centroid, circumcentre and orthocentre are also rational points.A. Statement 1 is true, Statement 2 is true and Statement 2 is correct explanation for Statement 1.B. Statement 1 is true , Statement 2 is true and Statement 2 is not the correct exlpanation for Statement 1.C. Statement 1 is true, Statement 2 is false.D. Statement 1 is false, Statement 2 is true. |
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Answer» Correct Answer - D Statement 2 is obvisouly correct. For statement 1. let the circumcentre be at (0,0) and the vertices of the triangle be `(x_1,y_1),(x_1,y_1),(x_2,y_2)` and `(x_3,y_3)`. Then centroid is `((x_1+x_2+x_3)/(3),(y_1+y_2+y_3)/(3))`, and orthocentre of the triangles becomes `(x_1+x_2+x_3,y_1_y_2+y_3)`. This implies that if the centroid then othocentre is also rational but `(x_1+x_2+x_3)` can be rational even if `x_1,x_2,x_3` are not all rational. for example , `A(1,0), B(-1//2,sqrt3//2)` and `C(-1//2,-sqrt3//2)`, where G,H and C are at (0,0) i,e., rational points. |
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| 39. |
Let A (2,-3) and B(-2,1) be vertices of a triangle ABC. If the centroid of this triangle moves on line 2x + 3y = 1, then the locus of the vertex C is the line : |
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Answer» Correct Answer - `2x+3y=9` Let C be `(alpha,beta)`. The centroid is `((2-2+alpha)/(3),(-3+1+beta)/(3)),i.e.,((alpha)/(3),(beta-2)/(3))` This lies on `2x+3y=1`. Therefore,we get `2(alpha/(3))+3(beta-2/(3))=1` or `2alpha+3beta=9` |
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| 40. |
In an acute triangle `A B C`, if the coordinates of orthocentre `H`are `(4,b)`, of centroid `G`are `(b ,2b-8)`, and of circumcenter `S`are `(-4,8)`, then `b`cannot be`4`(b) `8`(c)12 (d) `-12`But no common value of `b`is possible.A. 4B. 8C. 12D. `-12` |
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Answer» Correct Answer - A::B::C::D As H (orhtocenter), G (centroid), and C (circumcenter) are collinear we have `|{:(4,,b,,1),(b,,2b-8,,1),(-4,,8,,1):}|=0` or `|{:(4,,b,,1),(b-4,,b-8,,0),(-(b+4),,16-2b,,0):}|=0` or `(b-4)(16-2b)+(b+4)(b-4)=0` or `2(b-4)(8-b)+(b+4)+(b-8)=0` or `(8-b)[2b-8)-(b+4)=0` or `(8-b)(b-12)=0` Hence `b=8 or 12`, which is wrong because collinearity does not explain centroid, orthocenter, and circumcenter. Now, H.G, and C are collinear and `HG//GC=2`. Therefore, `(-8+4)/(3)=b or b=(-4)/(3)` and `(16+b)/(3)=2b-8 or b=8` But no common value of b is possible. |
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| 41. |
Locus of the point of intersection of the lines `xcosalpha+ysinalpha=a` and `xsinalpha-ycosalpha=b` where `alpha` is variable. |
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Answer» Correct Answer - `x^2+y^2=a^2+b^2` Let (h,k) be the point of intersection of `x cosalpha+ysinalpha=a` and `xsinalpha-ycosalpha=b`. Then, `hcosalpha+ksinalpha=a` (1) `hsinalpha-kcosalpha=b` (2) Squareing and adding (1) and (2),we get `(hcosalpha+ksinalpha)^2+(hsinalpha-kcosalpha)^2=a^2+b^2` or `h^2+k^2=a^2+b^2` Hence, the locus of (h,k) is `x^2+k^2=a^2+b^2` |
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| 42. |
If `(2,-3), (6,-5)` and `(-2,1)` are three consecutive vetcies of a rohombus, then its area isA. 24B. 36C. 18D. 48 |
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Answer» Correct Answer - D We have `A(2,-3),B(6,5),C(-2,1)` `2xxArea` of `DeltaABC` `=2xx24=48` |
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| 43. |
ABCD is a rectangle with `A(-1,2),B(3,7)` and `AB:BC=4:3`. If P is the centre of the rectangle, then the distance of P from each corner is equal toA. `sqrt(14)/(2)`B. `3sqrt(41)/(4)`C. `2sqrt(41)/(3)`D. `5sqrt(41)/(8)` |
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Answer» Correct Answer - D Since ABCD is a rectangle, `(AC)^2=(AB)^2+(BC)^2` `=(AB)^2[1+(9)/(16)]=(25)/(16)(AB)^2` and `PA=(AC)/(2)=(5AB)/(8)` `(5)/(8)sqrt((-1-3)^2+(2-7)^2)=5(sqrt41)/(8)` |
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| 44. |
The sum of the squares of the distances of amoving point from two fixed points (a,0) and `(-a ,0)`is equal toa constant quantity `2c^2dot`Find theequation to its locus. |
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Answer» Let `P(h,k)` be any position of the moveing point and let `A(a,0), B(-a,0)` be the given points. Then, we have `PA^2+PB^2=2c^2` (Given) or `(h-a)^2+(k-0)^2+(h+a)^2+(k-0)^2=2c^2` `2h^(2)2k^2+2a^2=2c^2` or `h^2+k^2=c^2-a^2` Hence, the equation to locus `(h,k)` is `x^2+y^2 =c^2-a^2`. |
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| 45. |
ABCD is a square Points `E(4,3)` and `F(2,5)` lie on AB and CD, respectively,such that EF divides the square in two equal parts. If the coordinates of A are `(7,3)`,then the coordinates of other vertices can beA. `(7,2)`B. `(7,5)`C. `(-1,3)`D. `(-1,5)` |
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Answer» Correct Answer - D Since EF divides the squre in two equal parts, it must pass through the center of the square and center is midpoint of EF. So, center of the sqaure is `(3,4)`. |
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| 46. |
The area of triangle `A B C`is `20c m^2dot`The coordinates of vertex `A`are `-5,0)`and those of `B`are `(3,0)dot`The vertex `C`lies on the line `x-y=2`. The coordinates of `C`are`(5,3)`(b) `(-3,-5)``(-5,-7)`(d) `(7,5)`A. (5,3)B. (-3,-5)C. (-5,-7)D. (7,5) |
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Answer» Correct Answer - B Let any point on the line `x-y=2` be `C(h,h-2)`. The given area of `DeltaABC` is. `|(1)/(2)||{:(h,,h-2,,1),(-5,,0,,1),(3,,0,,1):}|=20` or ` |8(h-2)|=40` or `h-2=+-5` or `h=7,-3` |
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