InterviewSolution
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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. |
Let a,b, c and d be non-zero numbers. If the point of intersection of the lines `4ax + 2ay+c=0` and `5bx+2by +d=0` lies in the fourth quadrant and is equidistant from the two axes, thenA. `2bc-3ad=0`B. `2bc+3ad=0`C. `2ad-3bc=0`D. `3bc+2ad=0` |
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Answer» Let coordinate of the intersection point in fourth quadrant be `(alpha,-alpha)`. Since, `(alpha,-alpha)` lies on both lines `4ax+2ay+c=0` and `5bx+2by+d=0` `:. 4aa-2aa+c=0impliesalpha=(-c)/(2a)`……….`(i)` and `5balpha-2balpha+d=0impliesalpha=(-d)/(3b)`.........`(ii)` From Eqs. `(i)` and `(ii)`, we get `(-c)/(2a)=(-d)/(3b)implies3bc=2ad` `implies2ad-3bc=0` |
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| 2. |
Three lines `px+qy+r=0`, `qx+ry+p=0` and `rx+py+q=0` are concurrent , ifA. `p+q+r=0`B. `p^(2)+q^(2)+r^(2)=pr+rq`C. `p^(3)+q^(3)+r^(3)=3pqr`D. None of these |
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Answer» Given lines `px+qy+r=0`, `qx+ry+p=0` and `rx+py+q=0` are concurrent. `:. |{:(p,q,r),(q,r,p),(r,p,q):}|=0` Applying `R_(1) to R_(1)+R_(2)+R_(3)` and taking common from `R_(1)` `(p+q+r)|{:(1,1,1),(q,r,p),(r,p,q):}|=0` `implies(p+q+r)(p^(2)+q^(2)+r^(2)-pq-qr-pr)=0` `implies p^(3)+q^(3)+r^(3)-3pqr=0` Therefore, `(a)` and `(c )` are the answers. |
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| 3. |
If `px+qx+r=0` represent a family fo staright lines such that `3p+2q+4r=0` then (a) All lines are parallel (b) All lines are incosistance (c) All lines are concurrent at `((3)/(4),(1)/(2))` (d) All lines are concurrent at `(3,2)`A. Each lines passes through the origin.B. The lines are concurrent at the point `((3)/(4),(1)/(2))`C. The lines are all parallelD. The lines are not concurrent |
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Answer» Given, `px+qy+r=0` is the equation of line such that `3p+2q+4r=0` Consider, `3p+2q+4r=0` `implies(3p)/(4)+(2q)/(4)+r=0` (dividing the equation by `4`) `impliesp((3)/(4))+q((1)/(2))+r=0` `implies((3)/(4),(1)/(2))` satisfy `px+qy+r=0` So, the lines always passes through the point `((3)/(4),(1)/(2))`. |
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| 4. |
The equation of the line which bisects the obtuse angle between the line `x-2y+4=0` and `4x-3y+2=0` is |
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Answer» Given equations of lines are `x-2y+4=0` and `4x-3y+2=0` Here, `a_(1)a_(2)+b_(1)b_(2)=1(4)+(-2)(-3)=10 gt 0` For obtuse angle bisector, we take negative sign. `:. (x-2y+4)/(sqrt(5))=(4x-3y+2)/(5)` `impliessqrt(5)(x-2y+4)=-(4x-3y+2)` `implies(4+sqrt(5))x-(2sqrt(5)+3)y+(4sqrt(5)+2)=0` |
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| 5. |
The number of integer values of `m ,`for which the x-coordinate of the point of intersection of the lines, `3x+4y=9`and `y=x m+1`is also an integer is2 (b) 0 (c) 4(d) 1A. `2`B. `0`C. `4`D. `1` |
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Answer» On solving equtions `3x+4y=9` and `y=mx+1`, we get `x=(5)/(3+4m)` Now, for `x` to be an integer, `3+4m=+-5` or `+-1` The intergral values of `m` satisfying these conditions are `-2` and `-1` |
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| 6. |
For `a gt b gt c gt 0`, if the distance between `(1,1)` and the point of intersection of the line `ax+by-c=0` and `bx+ay+c=0` is less than `2sqrt2` then,(A) `a+b-cgt0` (B) `a-b+clt0` (C) `a-b+cgt0` (D) `a+b-clt0`A. `a+b-c gt 0`B. `a-b+c lt 0`C. `a-b+c gt 0`D. `a+b-c lt 0` |
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Answer» PLAN Application of inequality sum and difference, along with lengths of perpendicular. For this type of questions involving inequally we should always check all options. Situation analysis Check all the inequalities according to options and use lengths of perpendicular from the point `(x_(1)y_(1))` to `ax+by+c=0` i.e. `(|ax_(1)+by_(1)+c|)/(sqrt(a^(2)+b^(2)))` As `a gt b gt c gt 0` `a-c gt 0` and `b gt 0` `implies a+b-c gt 0`.......`(i)` `a-b gt 0` and `c gt 0`.....`(ii)` `:.` Option `(c )` are correct. Also, the point of intersection for `ax+by+c=0` and `bx+ay+c=0` i.e. `((-c)/(a+b),(-c)/(a+b))` The distance between `(1,1)` and `((-c)/(a+b),(-c)/(a+b))` i.e. less than `2sqrt(2)`, `impliessqrt((1+(c )/(a+b))^(2)+(1+(c )/(a+b))^(2)) lt 2sqrt(2)` `implies((a+b+c)/(a+b))sqrt(2) lt 2sqrt(2)` `impliesa+b+c lt 2a+2b` `impliesa+b-c gt 0` From Eqs. `(i)` and `(ii)` , option `(a)` is correct. |
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| 7. |
The coordinates of `A,B,C` are `(6,3)`, `(-3,5)`, `(4,-2)` respectively and `P` is any point `(x,y)` . Show that the ratio of the areas of the triangles `DeltaPBC` and `DeltaABC` is `|(x+y-2)/(7)|`. |
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Answer» `("Area of " DeltaPBC)/("Area of" DeltaABC)=((1)/(2)|x(5+2)+(-3)(-2-y)+4(y-5)|)/((1)/(2)|6(5+2)+(-3)(-2-3)+4(3-5)|)` `=(|7x+7y-14|)/(|42+15-8|)=(7|x+y-2|)/(49)=|(x+y-2)/(7)|` |
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| 8. |
The vertices of a triangle are `[at_(1)t_(2),a(t_(1)+t_(2))]`,`[at_(2)t_(3),a(t_(2)+t_(3))]`, `[at_(3)t_(1),a(t_(3)+t_(1))]`. Find the orthocentre of the triangle. |
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Answer» Let `ABC` be a triangle whose vertices are `A[at_(1)t_(2),a(t_(1)+t_(2))]`, `B[at_(2)t_(3),a(t_(2)+t_(3))]` and `C[at_(1)t_(3),a(t_(1)+t_(3))]`. Then, Slope of `BC=(a(t_(2)+t_(3))-a(t_(1)+t_(3)))/(at_(2)t_(3)-at_(1)t_(3))=(1)/(t_(3))` Slope of `AC=(a(t_(1)+t_(3))-a(t_(1)+t_(2)))/(at_(1)t_(3)-at_(1)t_(2))=(1)/(t_(1))` So, the equation of a line through A perpendicular to `BC` is `y-a(t_(1)+t_(2))=-t_(3)(x-at_(1)t_(2))`........`(i)` and the equation of a line through `B` perpendicular to `AC` is `y-a(t_(2)+t_(3))=-t_(1)(x-at_(2)t_(3))`.........`(ii)` The point of intersection of Eqs. `(i)` and `(ii)` , is the orthocentre. On subtracting Eq. `(ii)` from Eq. `(i)`, we get `x=-a`. On putting `x=-a` in Eq. `(i)` , we get `y=a(t_(1)+t_(2)+t_(3)+t_(1)t_(2)t_(3))` Hence, the coordinates of the orthocentree are `[-a,a(t_(1)+t_(2)+t_(3)+t_(1)t_(2)t_(3))]`. |
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| 9. |
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)))` |
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Answer» Let the vertices of triangle be `A(1,sqrt(3))`, `B(0,0)` and `C(2,0)`. Here , `AB=BC=CA=2`. Therefore it is an equilateral triangle . So, the incentre coincides with centroid. `:. I-=((0+1+2)/(3),(0+0+sqrt(3))/(3))` `I-=(1,1//sqrt(3))` |
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| 10. |
Suppose that the points `(h,k)`, `(1,2)` and `(-3,4)` lie on the line `L_(1)`. If a line `L_(2)` passing through the points `(h,k)` and `(4,3)` is perpendicular to `L_(1)`, then `k//h` equalsA. `-(1)/(7)`B. `(1)/(3)`C. `3`D. `0` |
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Answer» Given points are `P(1,2)`, `(-3,4)` and `(h,k)` are lies on line `L_(1)`, so slope of line `L_(1)` is `m_(1)=(4-2)/(-3-1)=(k-2)/(h-1)` `impliesm_(1)=(-1)/(2)=(k-2)/(h-1)`……..`(i)` `implies 2(k-2)=-1(h-1)` `implies2k-4=-h+1` `impliesh+2k=5`……..`(ii)` and slope of line `L_(2)` joining points `(h,k)` and `(4,3)`, is `m_(2)=(3-k)/(4-h)`........`(iii)` Since, lines `L_(1)` and `L_(2)` are perpendicular to each other. `impliesm_(1)m_(2)=-1` `implies(-(1)/(2))((3-k)/(4-h))=-1` [from Eqs. `(i)` and `(iii)`] `implies3-k=8-2h` `implies2h-k-5`.....`(iv)` On solving Eqs. `(ii)` and `(iv)`, we get `(h,k)=(3,1)` So, `(k)/(h)=(3)/(1)=3` |
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| 11. |
The points `(0,(8)/(3))`, `(1,3)` and `(82,30)` are vertices ofA. an obtuse angled triangleB. an acute angled triangleC. a right angled triangleD. None of the above |
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Answer» Since , vertices of a triangle are `(0,8//3)`, `(1,3)` and `(82,30)` Now, `(1)/(2)|{:(0,8//3,1),(1,3,1),(82,30,1):}|` `=(1)/(2)[-(8)/(3)(1-82)+1(30-246)]` `=(1)/(2)[216-216]=0` `:.` Points are collinear. |
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| 12. |
The straight line `x+y=0`, `3x+y-4=0`, `x+3y-4=0` form a triangle which isA. isosclesB. equilateralC. right angledD. None of the above |
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Answer» The points of intersection of three lines are `A(1,1)`, `B(2,-2)`, `C(-2,2)`. Now, `|AB|=sqrt(1+9)=sqrt(10)`, `|BC|=sqrt(16+16)=4sqrt(2)`, and `|CA|=sqrt(9+1)=sqrt(10)` `:.` Triangle is in isosceles. |
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| 13. |
If `P(1,2)Q(4,6),R(5,7),`and `S(a , b)`are the vertices of a parallelogram `P Q R S ,`then`a=2,b=4`(b) `a=3,b=4``a=2,b=3`(d) `a=1orb=-1`A. `a=2,b=4`B. `a=3,b=4`C. `a=2,b=3`D. `a=3,b=5` |
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Answer» `PQRS` is a parallelogram if and only if the mid point of the diagonals `PR` is same as that of the mid-point of `QS`. That is, if and only if `(1+5)/(2)=(4+a)/(2)` and `(2+7)/(2)=(6+b)/(2)` `impliesa=2` and `b=3`. |
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| 14. |
If `|{:(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):}|` , 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))` must be congruent. |
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Answer» Since, `|{:(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):}|` represents area of triangles are equal, which does not impies triangles are congrvent. Hence, given statement is false. |
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| 15. |
If the sum of the distances of a point from two perpendicular lines ina plane is 1, then its locus isa square(b) a circlea straight line (d) two intersecting linesA. squareB. circleC. straight lineD. two intersecting lines |
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Answer» By the given conditions, we can take two perpendicular lines as `x` and `y` axes. If `(h,k)` is any point on the locus, then `|h|+|k|=1`. Therefore, the locus is `|x|+|y|=1`. This consist of a square of side `1`. Hence, the required locus is a square. |
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| 16. |
Let the algebraic sum of the perpendicular distance from the points (2,0), (0,2), and (1, 1) to a variable straight line be zero. Then the linepasses through a fixed point whose coordinates are___ |
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Answer» Let the variable straight line be `ax+by+c=0`……`(i)` where, algebaric sum of perpendicular from `(2,0)`, `(0,2)` and `(1,1)` is zero. `:. (2a+0+c)/(sqrt(a^(2)+b^(2))+(0+2b+c)/(sqrt(a^(2)+b^(2))+(a+b+c)/(sqrt(a^(2)+b^(2))=0` `implies3a+3b+3c=0` `implies a+b+c=0`........`(ii)` From Eqs. `(i)` and `(ii)` `ax+by+c=0` always passes through a fixed point `(1,1)`. |
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| 17. |
A line cuts the x-axis at `A (7, 0)` and the y-axis at `B(0, - 5)` A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, find the locus of R |
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Answer» The equation of the line `AB` is `(x)/(7)+(y)/(-5)=1`……`(i)` `implies5x-7y=35` Equation of line perpendicular to `AB` is `7x+5y=lambda` ……..`(ii)` It meets `X`-axis at `P(lambda//7,0)` and `Y`-axis at `Q(0,lambda//5)`. The equations of lines `AQ` and `BP` are `(x)/(7)+(5y)/(lambda)=1` and `(7x)/(lambda)-(y)/(5)=1`, respectively. Let `R(h,k)` be their point of intersection of lines `AQ` and `BP`. Then, `(h)/(7)+(5k)/(lambda)=1` and `(7h)/(lambda)-(k)/(5)=1` `implies(1)/(5k)(1-(h)/(7))=(1)/(7h)(1+(k)/(5))` [on eliminating `lambda`] `impliesh(7-h)=k(5+k)` `impliesh^(2)+k^(2)-7h+5k=0` Hence, the locus of a point is `x^(2)+y^(2)-7x+5y=0`. |
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| 18. |
`y=10^x` is the reflection of `y=log_(10) x` in the line whose equation is |
| Answer» `y=10^(x)` is reflection of `y=log_(10)x` about `y=x`. | |
| 19. |
The points `(-a,-b)`, `(0,0)`. `(a,b)` and `(a^(2),ab))` areA. collinearB. vertices of a rectangleC. vertices of a parallelogramD. None of the above |
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Answer» The point `O(0,0)` is the mid-point of `A(-a,-b)` and `B(a,b)`. Therefore, `A,O,B` are collinear and equation of line `AOB` is `y=(b)/(a)x` Since, the fourth point `D(a^(2),ab)` satisfies the above equation. Hence, the four points are collinear. |
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| 20. |
The point (4, 1) undergoes the following three transformations successively: (a) Reflection about the line y = x (b) Translation through a distance 2 units along the positive direction of the x-axis. (c) Rotation through an angle `pi/4` about the origin in the anti clockwise direction. The final position of the point is given by the co-ordinates.A. `((1)/(sqrt(2)),(7)/(sqrt(2)))`B. `(-sqrt(2),7sqrt(2))`C. `(-(1)/(sqrt(2)),(7)/(sqrt(2)))`D. `(sqrt(2),7sqrt(2))` |
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Answer» Let `B,C,D` be the position of the point `A(4,1)` after the three operations `I,II` and `III` respectively. Then, `B` is `(1,4), C(1+2,4)` i.e. `(3,4)` . The point `D` is obtained from `C` by rotationg the coordinate axes through an angle `pi//4` in anti-clockwise direction. Therefore, the coordinates of `D` are given by `X=3cos.(pi)/(4)-4sin.(pi)/(4)=-(1)/(sqrt(2))` and `Y=3sin.(pi)/(4)+4cos.(pi)/(4)=(7)/(sqrt(2))` `:.` Coordinates of `D` are `(-(1)/(sqrt(2)),(7)/(sqrt(2)))`. |
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| 21. |
A line is perpendicular to the line `2x-3y+5=0` and passes through the points `(15, beta)` and (7, 17)` then value of `beta` will be equal to (A) `35/3` (B) `-35/3` (C) `5` (D) `-5`A. `(35)/(3)`B. `-5`C. `-(35)/(3)`D. `5` |
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Answer» Slope of the line `2x-3y+17=0` is `(2)/(3)=m_(1)`, (let) and the slope of line joining the points `(7,17)` and `(15,beta)` is `(beta-17)/(15-7)=(beta-17)/(8)=m_(2)`(let) According to the question, `m_(1)m_(2)=-1` `implies(2)/(3)xx(beta-17)/(8)=-1impliesbeta-17=-12impliesbeta=5`. |
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| 22. |
Tangent at `(1,e)` on the curve `y=xe^(x^2)`, also passes through the point (a) `((4)/(3),2e)` (b) `((5)/(3),e)` (c) `((4)/(3),3e)` (d) `((3)/(4),3e)`A. `((4)/(3),2e)`B. `(3,6e)`C. `(2,3e)`D. `((5)/(3),2e)` |
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Answer» Given equation of curve is `y=xe^(x^(2))` …..`(i)` Note that `(1,e)` lie on the curve, so the point of contact is `(1,e)`. Now, slope of tangent, at point `(1,e)` , to the curve `(i)` is `(dy)/(dx)|_(((1,e)))=(x(2x)e^(x^(2))+e^(x^(2)))_(((1,e)))` `=2e+e=3e` Now, equation of tangent is given by `(y-y_(1))=m(x-x_(1))` `y-e=3e(x-1)impliesy=3ex-2e` On checking all the options, the option `((4)/(3),2e)` satisfy the equation of tangent. |
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| 23. |
Let P = (-1, 0), Q = (0, 0) and R = (3, `3sqrt3`) be three points. The equation of the bisector of the angle PQRA. `(sqrt(3))/(2)x+y=0`B. `x+sqrt(3)y=0`C. `sqrt(3)x+y=0`D. `x+(sqrt(3))/(2)y=0` |
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Answer» The line segment `QR` makes an angle of `60^(@)` with the positive direction of `X`-axis. So, the bisector of the angle `PQR` will make an angle of `60^(@)` with the negative direction of `X`-axis it will therefore have angle of inclination of `120^(@)` and so, its equation is `y-0=tan120^(@)(x-0)` `impliesy=-sqrt(3)x` `impliesy+sqrt(3)x=0` |
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