InterviewSolution
Saved Bookmarks
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. |
Find the incentre of the triangle with vertices `(1, sqrt3), (0, 0)` and `(2, 0)`A. `(1,sqrt(3)//2)`B. `(2//3,1,sqrt(3))`C. `(2//3,sqrt(3)//2)`D. `(1//1,sqrt(3))` |
|
Answer» Correct Answer - D Let the vertices of the triangle be `A(1,sqrt(3)),(B(0,0) and C(2,0)`. We observe that `AB=2=BC=AC` So, the triangle is equilateral. Consequently, its incetnre coincides with the whose coordinates are `((1+0+2)/(3),(sqrt(3)+0+0)/(3))=(1-(1)/(sqrt(3)))` |
|
| 2. |
The coordinates of the point where origin is shifted is (-1,2) so that the equation`2x^(2)+y^(2)-4x+4y=0` become?A. `X^(2)+2Y^(2)=6`B. `2X^(2)+y^(2)=6`C. `2X^(2)+Y^(2)=4`D. `X^(2)+2Y^(2)=4` |
|
Answer» Correct Answer - B Substituting `x=X+1,y=Y-2` in the equation `2x^(2)+y^(2)-4x+4y=-0`, we get `2(X+1)^(2)+(Y-2)^(2)-4(X+1)+4(Y-2)=0` `rArr 2X^(2)+Y^(2)-6=0rArr2X^(2)+Y^(2)=6` |
|
| 3. |
Statement-1 : The points A(-2,2), B(2,-2) and C(1,1) are the vertices of an obtuse angled isoscles triangle. Statement-2: Every abtuse angle tirangle is isosceles.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True |
|
Answer» Correct Answer - C We have , `AB=4sqrt(2),BC=sqrt(10) and AC=sqrt(10)` Since AB is the largest side. So, c is the greatest angle. Now, `cosC=(BC^(2)+CA^(2)-AB^(2))/(2BC.CA)` `rArr cos C=(20-32)/(20)=-(3)/(5)lt0` `rArr C` is an obtus angle `rArr Delta ABC` is an obtuse angled iosceels triangle. So, statement-1 is true. Statement-2 is not ture as the triangle with angles `120^(@),40^(@),20^(@)` is obtuse angled but not isosceles. |
|
| 4. |
If the vertices of a triangle `PQR` are rational points, then which of the following points of this triangle may not be rational -(a) Centroid (b) Incenter(c) Circumcenter (d) OrthocenterA. centroidB. incentreC. circumentreD. orthoentre |
|
Answer» Correct Answer - B Let `A(x_(1),y_(1)), B(x_(2),y_(2)) and C(x_(3),y_(3))` be the vertices of `Delta ABC` such that `x_(1),x_(2),x_(3),y_(1),y_(2),y_(3)` are rational numbers, Then, `(x_(1)+x_(2)+x_(3))/(3) and (y_(1)+y_(2)+y_(3))/(3)` are alo rational numbers. So, the coordinates of the centroid are always rational numbers. Let P{ bhe the circumcentre of `Delta ABC.` then, `PA=AB=PC` `rArr PA^(2)=PB^(2)=PC^(2)` `rArr PA^(2)=PB^(2) and PB^(2)=PC^(2)` Theses two relations two linear equations in terms of the coordiants of point P such that the coefficients are rational numbers. So, the coordinates of P are also rational numbers. Since the centroid G divides the segment joining circumcentre P and orthocentre H in the ratio of `1:2` Therefore, co-ordinates of H will have rational values. The coordinates of the incentre are `((ax_(1)+bx_(2)+cx_(3))/(a+b+c),(ay_(1)+by_(2)+cy+_(3))/(a+b+c))` where `a=Bc,b=CA and c=AB` Clearly, `a=sqrt((x_(2)-x_(3))^(2)+(y_(2)-y_(3))^(2))` `b=sqrt((x_(3)-x_(1))^(2)+(y_(2)-y_(3))^(2))` may not be ratioal numbers. Therefore, coordinats of the incentre may not be rational numbers. |
|
| 5. |
21. If a vertex of a triangle is (1, 1) and the midpoints of two sides of the triangle through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle isA. `((1)/(3),(7)/(3))`B. `(2,(7)/(3))`C. `(-(1)/(3),(7)/(3))`D. `(-1,(7)/(3))` |
|
Answer» Correct Answer - B Let ABC to the given triangle whose vertices are `A(1,1) B(x_(1),B(x_(1),y_(2))`. Let D(-1,2), and E(3,2) be the mid -points of AB and AC respecvtive. Then, `(x_(1)|1)/(2)=-1,(y_(1)|1)/(2)=2,(x_(2)|1)/(2)=3 and (y_(2)+1)/(2)=2` `rArr x_(1)=-3,_(2),y_(1)=3,x_(2)=5 and y_(2)=3` Thus the coordinates of the vertces of `DeltaABC` are (1,1) B(-3,3) and C(5,3) `:.` Coordinates of the centorid are `((1-3+5)/(3),(1+3+3)/(3))-=(1,(7)/(3))` |
|
| 6. |
By shifting origin to `(-1,2)` the equation `y^2+8x - 4y + 12 = 0` changes as `Y^2 = 4aX` then `a=`A. 1B. 2C. `-2`D. `-1` |
|
Answer» Correct Answer - B By shifting the origin at (1,2) we have `x=X+1,y=Y+2` Substituting these values in `y^(2)-8x-4y+12=0` we have `(Y+2)^(2)-8h(X+1)-4(Y+2)+12=0` `rArr Y^(2)-8X=0 rArr Y^(2)=8X` |
|
| 7. |
The area of the quadrilateral whose vertices are `(1,2)(6,2),(5,3) and (3,4)`, isA. `(3)/(2)` sq. unirtsB. `(11)/(2)` sq. unirtsC. `(1)/(2)` sq. unirtsD. none of these |
|
Answer» Correct Answer - B We have, `Delta_(1)(1)/(2)|{:(1,6),(2,2):}|=-5,Delta_(2)=(1)/(2)|{:(6,5),(2,3):}|=4` `Delta_(3)(1)/(2)|{:(5,3),(3,4):}|=(11)/(5)and Delta_(4)=(1)/(2)|{:(3,1),(4,2):}|=1` `:.` Area of quadrilateral `=|-5+4+(11)/(4)+1|=(11)/(2)` sq. units |
|
| 8. |
If points `(a^2, 0),(0, b^2) and (1, 1)` are collinear, thenA. `(1)/(a^(2))+(1)/(b^(2))=1`B. `(1)/(a)+(1)/(b)=1`C. `a^(2)+b^(2)=1`D. none of these |
|
Answer» Correct Answer - A Given points are cooliner. `:.|{:(a^(2),0,1),(0,b^(2),1),(1,1,1):}|=0` `rArr a^(2)(b^(2)-a)+(0-b^(2))=0rArra^(2)b^(2)=a^(2)+b^(2)rArr (1)/(a^(2))+(1)/(b^(2))=1` |
|
| 9. |
If the coordinates of two points `A`and `B`are (3, 4)and `(5,-2)`, respectively, find the coordinates of any point `P`if `P A=P Bdot`Area of ` P A B`is 10 sq.units.A. (2,7)B. (7,2)C. (1,0)D. (0,1) |
|
Answer» Correct Answer - B Let the coodinates of P be (x,y). Then, `PA=PB` `rArr PA^(2)=PB^(2)` `rArr (x-3)^(2)+(y-4)^(2)=(x-5)^(2)+(y+2)^(2)` `x-3y-1=0" "....(i)` Now, Area of `DeltaPAB=10` `rArr(1)/(2)|{:(x,y,1),(2,4,1),(5,-2,1):}|=+-10` `rArr 6x+2y-26=+-20 " "....(ii)` Solving (i) with, equations in (ii), we obtain `x=7,y=2 and x=1,y=0` But, P does lie on any of the coordinarte axes. Thus, the coordinates of P are (7,2) |
|
| 10. |
The angles A, B and C of a `DeltaABC` are in A.P. If `AB = 6, BC =7`,then `AC =`A. 5B. 7C. 8D. none of these |
| Answer» Correct Answer - D | |
| 11. |
If the points A(1,1), B(-1,-1) and `C(-sqrt(3),sqrt(3))` are the vertices of a triangle isA. right-angledB. isoscles but not right angledC. equilateralD. name of these |
|
Answer» Correct Answer - D We have, `AB=2sqrt(2),BCsqrt((3-1)^(2)+(sqrt(3)+1)^(2))=2sqrt(2)` `and,AC=sqrt((3-1)^(2)+(sqrt(3)+1)^(2))=2sqrt(2)` `:. AB=BC=CA` So, `DeltaABC` is equilateral. |
|
| 12. |
If the coordinates of `t`ideas ABand AC of a ` A B C`are (3,5)and (-3,-3) respectively, then write the length of side BC.A. 10B. 15C. 20D. 30 |
|
Answer» Correct Answer - C We have , `BC=2DE=2sqrt((3+3)^(2)+(5+3)^(2))=20` |
|
| 13. |
If O is the origin and `P(x_(1),y_(1)), Q(x_(2),y_(2))` are two points then `POxxOQ sin angle POQ=`A. `x_(1)y_(2)+x_(1)y_(2)`B. `x_(1)y_(2)+x_(2)y_(1)`C. `|x_(1)y_(2)-x_(2)y_(1)|`D. none of these |
|
Answer» Correct Answer - C We have, Area of `DeltaOPQ`= Absoulte value of `(1)/(2)|{:(0,0,0),(x_(1),y_(1),1),(x_(2),y_(2),1):}|` `rArr"Area of" DeltaOPQ=(1)/(2)|x_(1)y_(2)-x_(2)y_(1)|` Also, Area of `DeltaOPQ=(1)/(2)OPxxOQxx sin anglePOQ` `:.(1)/(2) OPxxOQ sin anglePOQ=(1)/(2)|x_(1)y_(2)-x_(2)y_(1|` `rArr OPxxOQ sin angle POQ =|x_(1)y_(2)-x_(2)y_(2)|` |
|
| 14. |
If the distance between the points `P (a cos 48^@, 0) and Q(0, a cos 12^@)` is `d,` then `d^2-a^2=`A. `(a^(2))/(4)(sqrt(5)-1)`B. `(a^(2))/(4)(sqrt(5)+1)`C. `(a^(2))/(8)(sqrt(5)-1)`D. `(a^(2))/(8)(sqrt(5)+1)` |
| Answer» Correct Answer - D | |
| 15. |
If points `O(0,0) A(3,sqrt(3)) and B(3,a)` are the vertices of an equilaterla triangle then a=A. 2B. `-3`C. `-4`D. none of these |
|
Answer» Correct Answer - D It id given that `DeltaOAB` is equilateral. `:. OA=OBrArr 9+3=9+a^(2)rArr a=+-sqrt(3)` |
|
| 16. |
If O is the origin and `P(x_(1),y_(1)), Q(x_(2),y_(2))` are two points then `POxxOQ sin angle POQ=`A. `x_(1)x_(2)+y_(1)+y_(2)`B. `x_(1)y_(2)+x_(2)+y_(1)`C. `|x_(1)y_(2)-x_(2)y_(1)|`D. none of these |
| Answer» Correct Answer - C | |
| 17. |
If thecentroid of the triangle formed by the points `(a , b), (b , c)`and `(c , a)`is at theorigin, then `a^3+b^3+c^3=``a b c`(b) 0 (c) `a+b+c`(d) `3 a b c` |
| Answer» Correct Answer - C | |
| 18. |
If O is the origin P(2,3) and Q(4,5) are two, points, then `OP*OQ cos anglePOQ`=A. 8B. 15C. 22D. 23 |
| Answer» Correct Answer - D | |
| 19. |
If two vertices of an equilaterla triangle have integral coordinates, then the third vetex will haveA. integral coodinates which are rtionalB. coordinates which are rationalC. at least one coordinate irrationalD. coordinates which are irrational |
|
Answer» Correct Answer - C Let `A(x_(1),x_(2))B(x_(2),y_(2)) and C(x_(3),y_(3))` be the vertice of an equlateral triangle ABC such that `x_(1),x_(2) and y_(1),y_(2)` are integers. If we assume that none of the coordinates of the vertex C are irrational, then we find that `Delta`= Area of `DeltaABC` `rArr Delta=(1)/(2)|{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=A` rational number But, `Delta=(sqrt(3))/(4) ("Side")^(2)=(sqrt(3))/(4)xxA` rational number `rArr Delta`= an irrational number Thus, we arrive at a conradinte. Therefore , our supposition is wrong Hence, at least at least one coordinate of C is irrational. |
|
| 20. |
If the coordinates of two vertices of an equilateral triangle are `(2,4) and (2,6)`, then the coordinates of its third vertex areA. `(sqrt(3),5)`B. `(2sqrt(3),5)`C. `(2+sqrt(3),5)`D. (2,5) |
| Answer» Correct Answer - C | |
| 21. |
Statement-1: The orthocentre of the triangle having its verticews at A(2,0), B(4,0) and C(4,6) is at the point o(4,0) Statement-2 : Orthocentre of a right triangle is at the vertex forming a right angleA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True |
|
Answer» Correct Answer - A Statement-2 is ture. Using statement-2, we find that the orthocentre of `DeltaABC` is at the point (4,6) Clearly, statement-2 is a correct explanation for statement-1 |
|
| 22. |
Let `k`be an integer such that the triangle with vertices`(k ,-3k),(5, k)`and `(-k ,2)`has area `28s qdot`units. Then the orthocentre of this triangle is at the point :`(1,-3/4)`(2) `(2,1/2)`(3) `(2,-1/2)`(4) `(1,3/4)`A. (2,-1/2)B. (1,3/4)C. (1,-3/4)D. (2,1/2) |
|
Answer» Correct Answer - D Let ABC be the triangle the coordinates of whose vertices are A(h,3-k), B(5,k) and C(-k,2). It is given that Area of `DeltaABC=28` sq. units `rArr(1)/(2)|{:(k,-3k,1),(5,k,1),(-k,2,1):}|=+-28` `rArr(1)/(2)|{:(k,-3k,1),(5-k,5k,0),(-2k,2+3k,0):}|=+-28` `rArr (5-k)(2+3k)+8k^(2)=+-56` `rArr 5k^(2)+13+66=0 or 5 k^(2)+13 k` `rArr 5k^(2)+13k-46=0` `rArr (k-2)(5+3k)=0 rArr k=2` Hence, the coordinates of vertices are A(2,-6), B(5,2) and C(-2,2) . The equation of altitudes through vertices A and C are x=2 adn `3x+8y-10=0` respectively. These two altitudes instersect at (2,1/2). Hence, the coordinates of the orthocentre are (2,1/2) |
|
| 23. |
If P and Q are two points whose coordinates are `(a t^2,2a t)a n d(a/(t^2),(2a)/t)`respectively and S is the point (a,0). Show that `1/(S P)+1/(s Q)`is independent of t.A. aB. 4aC. 2aD. `(2)/(a)` |
|
Answer» Correct Answer - C We have, `SP=sqrt(("at"^(2)-a^(2))+("at"-0)^(2))=(t^(2)+t)` Replacing t by `-(1)/(t)` in SP, we get `SQ=(2)/(t^(2))(1+t^(2))` `:. (1)/(SP)+(1)/(SQ)=(1)/(a(t^(2)+1))+(t^(2))/(a(t^(2)+1))` `rArr (1)/(SP)+(1)/(SQ)=(1)/(a)` `rArr (2SPSQ)/(SP+SQ)=2arArr HM` of SP and SQ in 2a |
|
| 24. |
If area of the triangle formed by `(0, 0), (a^(x^2), 0), (0, a^(6x))` is `1/(2a^5)` sq. units then `x=`A. 1,5B. `-1,5`C. `1,-5`D. `-1,-5` |
|
Answer» Correct Answer - D We have, Area of `DeltaOAB=(1)/(2a^(5))` sq. units `rArr(1)/(2)xxa^(x^(2))xxa^(6x)=(1)/(2)a^(-5)` `rArr a^(x^(2)+6x)=a^(-5)rArr x^(2)+6x+5=0rArr x=-1,-5` |
|
| 25. |
The transformed equation of `x^(2)+6xy+8y^(2)=10` when the axes are rotated through an angled `pi//4` isA. `15x^(2)-14xy+3y^(2)=20`B. `15x^(2)+14xy-3y^(2)=20`C. `15x^(2)+14xy+3y^(2)=20`D. `15x^(2)-14xy-3y^(2)=20` |
| Answer» Correct Answer - C | |
| 26. |
By rotating the coordinates axes through `30^(@)` in anticlockwise sense the eqution `x^(2)+2sqrt(3)xy-y^(2)=2a^(2)` change toA. `X^(2)-Y^(2)=3a^(2)`B. `X^(2)-Y^(2)=a^(2)`C. `X^(2)-Y^(2)=2a^(2)`D. none of these |
| Answer» Correct Answer - B | |
| 27. |
If (x,y) and (X,Y) be the coordinates of the same point referred to two sets of rectangular axes with the same origin and if ax+by becomes pX+qY, where a,b are independent of x,y, thenA. `a^(2)-b^(2)=p^(2)-q^(2)`B. `a^(2)+b^(2)=p^(2)+q^(2)`C. `a^(2)+p^(2)=b^(2)+q^(2)`D. `a^(2)b^(2)=p^(2)q^(2)` |
|
Answer» Correct Answer - B Suppose one set of rectangular axes is obtained by rotating the coordinate axes of the other set by an angle `theta` in anticlocwise sense. Let (x,y) be the coordinates of the point with respect to odd axes and (X,Y) be the coordinates with respect to the axes. then, `x=X cos theta-Y=X sin theta cos theta`, `:. ax+by=a(Xcos theta-Y sin theta) +b ( X sin theta+Y cos theta)` `rArr ax+by=(a cos theta+bsintheta) X+(b cos theta-sin theta)Y` It is given that `ax+by` becomes pX+qY `:. P=acos theta+b sin theta and q=b cos theta-a sin theta` `rArr p^(2)+q^(2)=(a cos theta+b sin theta)^(2)+ ( b sin theta- a sin theta)` `rArr p^(2)+q^(2)=a^(2)+b^(2)` |
|
| 28. |
Consider three points `P = (-sin (beta-alpha), -cos beta)`, `Q = (cos(beta-alpha), sin beta)`, and `R = ((cos (beta - alpha + theta), sin (beta - theta))`, where `0< alpha, beta, theta < pi/4` ThenA. P lies on the line segmennt RQB. Q lies on the line segmet PRC. R lies on the line segment QPD. P,Q,R are non-colinear |
|
Answer» Correct Answer - D Let the coordinates of P and Q be `(x_(1),y_(1)) and (x_(2),y_(2))` respectively. Then `x_(1)=sin (beta-alpha), y_(1)=- cos beta, x_(2)= cos (beta-alpha),y_(2)= sin beta`. The coordinates of R are `(cos (beta-alpha+theta), sin (beta-theta))` , `or , (cos(beta-alpha)cos theta-sin theta(beta-alpha)sin theta, sin beta cos theta-cos beta sin theta)` `or (x_(1) sin theta+x_(2)cos, theta, y_(1)sin theta+y_(2) theta)` We observe that the poins having coordinates `((x_(1) sin theta+x_(2)cos theta)/(sin theta+ cos theta),(y_(1)sin theta+y_(2) cos theta)/(sin theta+cos theta))` divides PQ internally in the ratio `cos theta: sin theta` . So, P,Q S are coolinerr points. But P,Q R are non-Collinear. |
|
| 29. |
Line joining `A(bcosalpha,bsinalpha)` and `B(acosbeta,asinbeta)`, is produced to the point `M(x,y)` such that `AM:MB=b:a` then `xcos((alpha+beta)/2)+ysin((alpha+beta)/2) =`A. (-1)B. 0C. 1D. `a^(2)+b^(2)` |
|
Answer» Correct Answer - B Clearly , M divides AB expternally in the ratio `b:a`. `:. x=(b(acos beta)-a(b cos alpha))/(b-a) and,y=(b(a sin beta)-a(b sin alpha))/(b-a)` `rArr(x)/(y)=(cos beta-cos alpha)/(sin beta- sin alpha)` `rArr (x)/(y)(2 "sin" (alpha+beta)/(2) "sin"(alpha-beta)/(2))/(-2 "sin" (alpha+beta)/(2) "sin"(alpha-beta)/(2))` `rArr x "cos" ((alpha+beta)/(2)) + y "sin"((alpha+beta)/(2))=0` |
|