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.
| 101. |
For any real values of a,b,c such that 3a, +2b+4c=0, line ax+by+c=0 passes through the fixed point whose coordinates areA. (3,2)B. (2,4)C. (3,4)D. (3/4,1/2) |
| Answer» Correct Answer - D | |
| 102. |
the lines `(p+2q)x+(p-3q)y=p-q` for different values of `p&q` passes trough the fixed point is:A. `((3)/(2),(5)/(2))`B. `((2)/(5),(2)/(5))`C. `((3)/(5),(3)/(5))`D. `((2)/(5),(3)/(5))` |
| Answer» Correct Answer - D | |
| 103. |
The straight line `y = x - 2` rotates about a point where it cuts x-axis and become perpendicular on the straight line `ax + by + c = 0` then its equation isA. `ax +by +20 = 0`B. `ax - by - 2a = 0`C. `bx +ay - 2b = 0`D. `ay - bx +2b = 0` |
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Answer» Correct Answer - D Slopes of the line in the new position is bla, since it is perpendicular to the line `ax +by +c = 0` and it cuts the x-axis . Hence the required line passes through (2,0) and its slope is bla. The required is `y - 0 = (b)/(a) (x-2)` or `ay = bx - 2b` |
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| 104. |
The perpendicular distance of the line `px+qy+r=0` from the point (2,-1) is 4 unit , state which of the following conditions is true ?A. `q+r-2p=4(p^(2)+q^(2))`B. `q+r-2p=4sqrt(p+q)`C. `(q+r-2p)^(2)=4(p^(2)+q^(2))`D. `(2p-q+r)^(2)=16(p^(2)+q^(2))` |
| Answer» Correct Answer - D | |
| 105. |
The slope of the x- axis or of a line parallel to x-axis isA. `-1`B. 0C. undefinedD. 1 |
| Answer» Correct Answer - B | |
| 106. |
The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular ifA. b=`a sqrt(2)`B. a=`b sqrt(2)`C. b=`-a sqrt(2)`D. a=`5b sqrt(2)` |
| Answer» Correct Answer - B | |
| 107. |
· The co-ordinates of foot of the perpendicular from the point `(2, 4) `on the line `x+y = 1` are:A. `((1)/(2),(3)/(2))`B. `(-(1)/(2),(3)/(2))`C. `((4)/(3),(1)/(2))`D. `((3)/(4),-(1)/(2))` |
| Answer» Correct Answer - B | |
| 108. |
If A is (1,-2), B (3,k), C(-3,1) and D (k,4) where lines AB and CD are parallel then : K=A. `-2//7`B. `2//7`C. `-7//2`D. `7//2` |
| Answer» Correct Answer - A | |
| 109. |
If A is (-4,9) and B is a point on the Y-axis such that the slope of the line AB is (-1), then : B`-=`A. (0,1)B. (0,3)C. (5,0)D. (0,5) |
| Answer» Correct Answer - D | |
| 110. |
If A is (5,-3) and B is a point on the X-axis such that the slope of line AB is (-2), then : B=A. (7,2)B. (7/2,0)C. (0,7/2)D. (2/7,0) |
| Answer» Correct Answer - B | |
| 111. |
If the triangle whose vertices are A(4,3), B(6,-2) and C(k,-3) is right -angled at A, then : K=A. 3B. 8C. -11D. -5 |
| Answer» Correct Answer - C | |
| 112. |
If A is (`sqrt(3),1)` and B is `(sqrt(3),-1)`, then :m `angel`AOB=A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - C | |
| 113. |
The orthocentre of the triangle formed by the lines `xy=0 and x+y=1` , isA. (-2-1)B. (-2, 1)C. (0,0)D. none of these |
| Answer» Correct Answer - C | |
| 114. |
The algebraic sum of distances of the line `ax + by + 2 = 0` from `(1,2), (2,1) and (3,5)` is zero and the lines `bx - ay + 4 = 0` and `3x + 4y + 5=0` cut the coordinate axes at concyclic points. Then(a) `a+b=-2/7`(b) area of triangle formed by the line `ax+by+2=0` with coordinate axes is `14/5`(c) line `ax+by+3=0` always passes through the point `(-1,1)`(d) max `{a,b}=5/7`A. `a +b =- (2)/(7)`B. area of the triangle formed by the line `ax +by +2 = 0` with coordinate axes is `(14)/(5)`C. line `ax +by +3 = 0` always passes through the point `(-1,1)`D. max `{a,b} = (5)/(7)` |
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Answer» Correct Answer - C Line `ax +by +2 = 0` always passes through the point `(2,(8)/(3))` `:. 6a +8b +6 = 0` or `3a +4b +3 = 0` Now `bx - ay +4 = 0` and `3x +4y +5 = 0` meet axis in concyclic points. So, `m_(1)m_(2) =1` `((b)/(a)).(-(3)/(4)) =1` `rArr 4a +3b = 0` (ii) Solving (i) and (ii), we get `a = 9//7, b =- 12//7` `rArr` Line `ax +by +3 = 0` always passes through the point `(-1,1)`. |
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| 115. |
If the line kx+4y=6 passes through the point of intersection of the two lines 2x+3y=4 and 3x+4y=5, then : k=A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B | |
| 116. |
If the line ky=x+1 passes through the point on intersection of the two lines 2x-3y+5=0 and 3x+2y+1=0, then :k=A. -1B. 0C. 1D. none of these |
| Answer» Correct Answer - B | |
| 117. |
The straight lines`5x-9y-12=0andmx+10y=2` are perpendicular to each other , state which of the following is the value of m:A. 18B. `-9`C. 9D. `-18` |
| Answer» Correct Answer - A | |
| 118. |
If graph of `xy = 1` is reflected in `y = 2x` to give the graph `12x^(2) +rxy +sy^(2) +t = 0`, thenA. `r = 7`B. `s =- 12`C. `t = 25`D. `r +s =- 19` |
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Answer» Correct Answer - B::C::D Let image of point `A(alpha,beta)` about `y = 2x` is `B (a,b)` `rArr` mid point of AB i.e. `M -= ((alpha +a)/(2),(beta+b)/(2))` lie on `y -2x =0` `rArr ((beta +b)/(2)) =2 ((alpha +a)/(2)) rArr beta +b = 2alpha +2a `(i) and slope of `AB =- (1)/(2)` `rArr (beta-b)/(alpha-a) = -(1)/(2) rArr beta -b =- (1)/(2) alpha +(a)/(2)` (ii) Subtracting (i) and (ii), `rArr 2b = (5)/(2) alpha +(3)/(2) alpha rArr a = (4b-3a)/(5)` `rArr beta =(8b -6a)/(5) +2a - b = (3b+4a)/(5)` `rArr (alpha, beta)` lies on `xy =1` `rArr ((4b-3a)/(5))((3b+4a)/(5)) =1` `rArr 12b^(2) +7ab -12 a^(2) =25` `rArr 12 x^(2) - 7xy =12y^(2) +25 =0` |
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| 119. |
The set of real values of k for which the lines `x + 3y +1=0, kx + 2y-2=0` and `2x-y + 3 = 0` form a triangle isA. `R-{-4,(2)/(3)}`B. `R-{-4,(-6)/(5),(2)/(3)}`C. `R-{(-2)/(3),4}`D. R |
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Answer» Correct Answer - B Lines form triangle. Therefore, `x +3y +1 = 0` is not parallel to `kx +2y -2 = 0` `:. (-k)/(2) ne (-1)/(3) rArr k ne (2)/(3)` Also line `2x - y+3 =0` is not parallel to `kx +2y -2 = 0` `:. (-k)/(2) ne 2 rArr k ne -4` Further lines must not be concurrent `:. |{:(1,3,1),(k,2,-2),(2,-1,3):}|ne0rArrkne(-6)/(5)` |
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| 120. |
Pair of lines through `(1, 1)` and making equal angle with `3x - 4y=1` and `12x +9y=1` intersect x-axis at `P_1` and `P_2` , then `P_1,P_2` may beA. `((8)/(7),0)` and `((9)/(7),0)`B. `((6)/(7),0)` and `(8,0)`C. `((8)/(7),0)` and `((1)/(8),0)`D. `(8,0)` and `((1)/(8),0)` |
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Answer» Correct Answer - B Lines making equal angle with given two lines are always parallel to angle bisectors Equation of angle bisectors `(3x-4y-1)/(5) =(12x+9y-1)/(15)` So bisectors have slope `7,-(1)/(7)` Equation of required lines `y - 1 = 7(x-1)` and `y -1 =-(1)/(7) (x-1)` which intersect x-axis at `((6)/(7),0)` and (8,0). |
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| 121. |
Locus of the points which are at equal distance from `3x + 4y-11 = 0` and `12x + 5y + 2= 0` and which is near the origin is:A. `21x - 77y +153 = 0`B. `99x +77y - 133 = 0`C. `7x - 11y = 19`D. None of these |
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Answer» Correct Answer - B Points which at equal distance from `3x +4y -11 = 0` and `12x +5y +2 = 0` lie on the angle bisector of the lines. Now given two lines are on opposite sides with respect to origin. Since required line is near to the origin, we have to find the bisector of the angle containing the origin, which is `(2x+4y -11)/(5) =- ((12x +5y+2)/(13))` or `99x +77y - 133 =0` |
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