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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.
| 51. |
Find the points of interesting of the lines `4x+3y=5 and x=2y-7 |
| Answer» Correct Answer - (-1,3) | |
| 52. |
The nearest point on the line `3x + 4y = 25` from the origin isA. (-4,5)B. (3 , -4)C. (3,4)D. ( 3, 5) |
| Answer» Correct Answer - B | |
| 53. |
The perpendicular distance of a line from the origin is 5 units and its slope is -1. Find the equation of the line. |
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Answer» Correct Answer - `x+y+5sqrt2=0 or x+y=5sqrt2=0` Let the required equation be y=-x+c `therefore (|0+0-c|)/(sqrt(1^(2)+1^(2)))=5 Rightarrow |c| =5sqrt2 Rightarrow c=5sqrt2 or c=-5sqrt2` |
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| 54. |
The distance between the parallel lines given by `(x+7y)^2+4sqrt(2)(x+7y)-42=0` isA. `4//5`B. `4sqrt2`C. 2D. `10 sqrt2` |
| Answer» Correct Answer - A | |
| 55. |
A vertex of a square is at the origin and its one side lies along the line 3x-4y-10=0. Find the area of the square. |
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Answer» Correct Answer - 4 sq.units Side of the square=length of perp. From of perp. From (0,0) on 3x-4y-10=0 `(|3xx0-4xx10|)/(sqrt(3^(2)(-4)^(2)))=2"units"` Area of the square=`(2xx2)` sq units=4 sq units |
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| 56. |
Find the distance between the parallel lines p(x+y)+g=0 and p(x+y)-r=0. |
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Answer» Correct Answer - `(|q+r|)/(sqrt2p` Putting x=0 in px+py+q=0, we get y=(-q/p) Required distance=length of perp. From `P(0,(-q)/(p))` on px+py-r=0 `(|pxx0+pxx((-q)/(p))-r|)/(sqrt(p^(2))+p^(2))=(|-q-r|)/(sqrt2p)=(|q+r|)/(sqrt2p)` |
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| 57. |
Find the distance between the parallel lines y=mx+c and y=mx+d. |
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Answer» Correct Answer - `(|d-c|)/(sqrt(1+m^(2)))` Putting x=0 in y=mx+c, we get y=c. Thus, P(0,c) is a point on y=mx+c Required distance=length of perp. From P(0,c) on y=mx+d `(|mxx0-c+d|)/(sqrt(1+m^(2)))=(|d-c|)/(sqrt1+m^(2))` |
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| 58. |
if `P` is the length of perpendicular from origin to the line `x/a+y/b=1` then prove that `1/(a^2)+1/(b^2)=1/(p^2)` |
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Answer» The given line is `bx+ay-ab = 0 " "(1)` It is given that p is the length of the perpendicular from the origin to (I), that is, `p = (|b(0) + a(0) -ab|)/(sqrt(b^(2) + a^(2)))` `=(ab)/(sqrt(a^(2) + b^(2)))` ` " or " p^(2) = (a^(2)b^(2))/(a^(2) + b^(2))` ` " or " (1)/(p^(2)) = (a^(2)+ b^(2))/(a^(2)b^(2)) = (1)/(a^(2)) + (1)/(b^(2))` |
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| 59. |
Prove that the line 5x-2y-1=0 is midpoint to the line `5x-2y-9=0 and 5x-2y+7=0` |
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Answer» Converting each of the given equations to the form y=mx+C, We get `5x-2y-1=0 Rightarrow y=(5)/(2)x -(1)/(2).......(i)` `5x-2y-9=0 Rightarrow y=(5)/(2)x -(9)/(2).......(ii)` `5x-2y+7=0 Rightarrow y=(5)/(2)x +(7)/(2).......(iii)` Clearly, the slope of (i) is equal to the slope of each of (ii) and (iii), So, the line (i) is parallel to each of the given line (ii) and (iii). Let the given line be `y=mx+C, y=mx+C_(1) and y=mx+C_(2)` respectively. Then, `m=(5)/(2),C=-(1)/(2)=-(1)/(2),C_(1)=-(9)/(2) and C_(2)=(7)/(2)` Let `d_(1) and d_(2)` be the distance of (i) from (ii) and (iii) respectively. `"Then", d_(1)=(|C_(1)-C|)/(sqrt(1+m^(2)))=(-(9)/(2)+(1)/(2))/(sqrt(1+(25)/(4)))=|-4|=(4xx2)/(sqrt29)=(8)/(sqrt29)"units"` `and d_(2)=(|C_(2)-C|)/(sqrt(1+m^(2)))=((7)/(2)+(1)/(2))/(sqrt(1+(25)/(4)))=(4xx(2)/(sqrt29))=(8)/(sqrt29)"units"` Thus, `d_(1)=d_(2)` This shows that (i) is equidistant from (ii) and (iii). Hence, 5x-2y-1=0 is mid-parallel to the lines 5x-2y-9=0 and 5x-2y+7=0 |
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| 60. |
Find the coordinates of a point on `x+y+3=0,`whose distance from `x+2y+2=0`is `sqrt(5)dot` |
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Answer» Let the required point be P(a,b). Then, `a+b+3=0` `"Also",(|a+2b+2|)/(sqrt(1^(2)+2^(2)))=sqrt5=|a+2b+2|=5` `Rightarrow (a+2b+2=5) or (a+2b+2=-5)` `Rightarrow a+2b=3 or a+2b=-7` `"Thus", a+2b=3......(ii) and a+2b=-7......(iii)` On solving (i) and (ii), we get a=-9 and b=6. On solving (i) and (iii), we get a=1 and b=-4 Hence, the required points are A(-9,6) and B(1,-4). |
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| 61. |
Find the equation of the line midway between the parallel lines `9x+6y-7=0` and 3x+2y+6=0` |
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Answer» Converting each of the given equation to the form `y=mx+C`, We get, `9x+6y-7=0 Rightarrow y=(-3)/(2)x+(7)/(6)....(i)` `3x+2y+6=0 Rightarrow y=(-3)/(2)x-3....(ii)` Clearly, the slope of each one of the given line is `(-3)/(2)` Let the given lines by `y=mx+C_(1) and y=mx+C_(2)`. Then, `m=(-3)/(2), C_(1)=(7)/(6) and C_(2)=-3` Let the required line. Then, L is parallel to each one of (i) and (ii) and equidistant from each one of them. `therefore "slope of L"=(-3)/(2)` Let the equation of L be y`=(-3)/(2)x+C......(iii)` Then, distance between (i) and (iii) must be equal to the distance between (ii) and (iii). `therefore (|C_(1)-C|)/(sqrt(1+m^(2)))=(|C_(2)-C|)/(sqrt(1+m^(2)))=|C_(1)-C|=|C_(2)-C|` `Rightarrow |(7)/(6)-C|=|-3-C| Rightarrow |(7)/(6)-C|=|3+c|` `Rightarrow (7)/(6)-C=3+C Rightarrow 2C=|(-11)/(6)Rightarrow C=(-11)/(2)` `therefore "equation of L is y"=(-3)/(2)x-(11)/(12),i.e. 18+12y+11=0` Hence, the line `18x+12y+11=0` is midway between the parallel line `9x+6y-7=0 and 3x+2y+6=0`. |
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| 62. |
The distance between the parallel lnes `y=2x+4 and 6x-3y-5` is (A) `1` (B) `17/sqrt(3)` (C) `7sqrt(5)/15` (D) `3sqrt(5)/15`A. `17//sqrt(3)`B. 1C. `3//sqrt(5)`D. `17sqrt(15)//15` |
| Answer» Correct Answer - D | |
| 63. |
The acute angle between the lines y=3 and `y=sqrt(3x)+9` is: |
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Answer» Here, equation of Line 1 (`L_1`) is, `y =3` Equation of Line 2 (`L_2`) is , `y =sqrt3x+9` The slope of `L_1,(m_1) = 0` Slope of `L_2,(m_2) = sqrt3` So, for acute angle `theta` between them, `tan theta = (m_1+m_2)/(1-m_1m_2) = (0+sqrt3)/(1-0) = sqrt3` `=>theta = pi/3` |
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| 64. |
The point P(1,1) is translated parallal to the line 2x=y in the first quadrant through a unit distance.The coordinates of the new position of P are:A. `(1 pm (2)/(sqrt5) , 1 pm (1)/(sqrt5))`B. `(1 pm (1)/(sqrt5) , 1 pm (2)/(sqrt5))`C. `((1)/(sqrt5) , (2)/(sqrt5))`D. `((2)/(sqrt5) , (1)/(sqrt5))` |
| Answer» Correct Answer - B | |
| 65. |
The point P(1,1) is translated parallal to the line 2x=y in the first quadrant through a unit distance.The coordinates of the new position of P are: |
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Answer» `x=1pmcostheta` `x=1pm1/sqrt5` `y=1pmsintheta` `y=2pm2/sqrt5`. |
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| 66. |
If a circle Passes through point (1,2) and orthogonally cuts the circle `x^2 + y^2 = 4`, Then the locus of the center is: |
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Answer» `(C_1P)^2+(C_2P)^2=(C_1C_2)^2` `h^2+k^2=4+r^2-(1)` `(h-1)^2+(k-2)^2=r^2` subtracting equation 2 from 1 `h^2+k^2-(h-1)^2-(k-1)^2=4+r^2-r^2` `h^2+k^2-h^2+2h-1-k^2+4k-4=4` 2h+4=9 2x+4y=9. |
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| 67. |
Distance of a point (2,5) from the line 3x+y+4=0 measured parallal to the line 3x-4y+8=0 is:a. 15/2b. 9/2c. 5d. none |
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Answer» `3x-4y+8=0` m=3/4 `(k-5)/(h-2)=3/4` `4k-20=3h-6` 3h-4k=-14-(1) 3h+k=-4-(2) Subtracting equation 2 from equation1 k=2 h=-2 PQ=`sqrt((2+2)^2+(5-2)^2)` PQ=`sqrt(4^2+3^2)` PQ=`sqrt25` PQ=5cm |
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| 68. |
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) Orthocenter |
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Answer» G is rational I may not be rational `x^2+y^2+2gx+2fy+c=0` g and f are rational, circumcenter of ration. `(y-y_2)/(x-x_2)=(y_3-y_2)/(x_3-x_2)-(1)` `(y-y_1)/(x-x_1)=(y_3-y_2)/(x_3-x_2)-(2)` From 1 and 2 orthocenter is a rational. |
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| 69. |
The incenter of the triangle formed by the axes and the line `x/a+y/b=1` |
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Answer» `(x_1,y_1)=(a,0)` `(x_2,y_2)=(0,b)` `(x_3,y_3)=(0,0)` `l=sqrt(0^2+b^2)=|b|` `m=sqrt(a^2+0^2)=|a|` `n=sqrt(a^2+b^2)` `I=(a|b|+b|a|+o)/(|a|+|b|+sqrt(a^2+b^2)),(b|a|+0+0)/(|a|+|b|+sqrt(a^2=b^2)}` `I=(a|b|)/(|a|+|b|+sqrt(a^2+b^2)),(|a|b)/(|a|+|b||sqrt(a^2+b^2)}` When a>0,b>o,|a|=a,|b|=b. option c is correct. |
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| 70. |
A line passing through the point P(1,2) meets the line `x+y=7` at the distance of 3 units from P. Then the slope of this line satisfies the equation :A. `8x^(2)-9x+1=0`B. `7x^(2)-18x+7=0`C. `16x^(2)-39x+16=0`D. `7x^(2)-6x-7=0` |
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Answer» Correct Answer - B The equation of a line through P(1,2) is `(x-1)/(cos theta)=(y-2)/(sin theta)` The coordinates of point of this line at a distance of 3 units from P(1,2) are given by `(x-1)/(cos theta)=(y-2)/(sin theta)=pm3` . Letthe coordinates of the points be `(1 pm 3 cos theta, 2 pm sin theta)` . These points lie on `x+y=7`. `(1 pm 3 cos theta)+(2 pm 3 sintheta)=7` `implies pm3(cos theta + sin theta)=4` `implies 9(1+sin 2theta)=16` `implies (18tan theta)/(1+tan^(2)theta)=7` `implies 7 tan^(2)theta-18 tan theta +7=0` `implies tan theta ` is a root of `7x^(2)-18x+7=0` |
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| 71. |
A circle S passes through the point (0, 1) and is orthogonal to the circles `(x -1)^2 + y^2 = 16` and `x^2 + y^2 = 1`. Then(A) radius of S is 8(B) radius of S is 7(C) center of S is (-7,1)(D) center of S is (-8,1) |
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Answer» `delta:x^2+y^2+2ga+2fy+c=0` `delta_1-delta_2=0` `x^2+y^2-2x-15=0` `x^2+y^2-1=0` `delta_1-delta_2=-2x-14=0` `delta:x^2+y^2+14x+2fy+c=0` `0+1+0+2f+c=0` `1+2f+c=0-(1)` `2gg_1+2ff_1=c+c_1` `2(7)(-1)+0=c-15` `c=1` `f=-1` centre=(-7,1) option b and c is correct. |
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| 72. |
The perpendicular drawn from origin to the line `y=mx+c` meets the line at point `(-1,-2)`, `(c,m)=?`A. `((5)/(2),(1)/(2))`B. `((1)/(2),(5)/(2))`C. `(-(1)/(2),(-5)/(2))`D. None of these |
| Answer» Correct Answer - A | |
| 73. |
Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length `2sqrt7` on y-axis is (are) |
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Answer» `(x-a)^2+(y-b)^2=r^2` `y=(x-a)^2=r^2-b^2` `x=pm3` `b=r` `a=pm3` `(x-a)^2+(y-b)^2=b^2` `(-a)^2+(y-b)^2=b^2` `9+y^2+b^2-2yb=b^2` `y^2-2yb+e=0` `-(2sqrt7)^2+(2b)^2=4*9` `4b^2=36+28=64` `b^2=16` `b=pm4` Putting this value in above equation. `x^2+y^2pm6xpm8=0`. |
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| 74. |
The equation of the straight line passing through the point `(3,2)` and perpendicular to the line `y=x` isA. `x-y=5`B. `x+y=5`C. `x+y=1`D. `x-y=1` |
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Answer» Correct Answer - B Since, line passses through the point (3,2)and perpendicular to the line y=x. `because` slope `(m)=-1` [since, line is perpendicular to the line y =x] `therefore` Equation of line which passes through (3,2) is `y-2=-1(x-3)` `rArr y-2=-x+3` `rArrx+y=5` |
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| 75. |
The equation of the line passing through the point `(1,2)` and perpendicular to the line `x+y+1=0` isA. `y-x+1=0`B. `y-x-1=0`C. `y-x+2=0`D. `y-x-2=0` |
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Answer» Correct Answer - B Given point is (1,2)and slope of the required line is 1. `because x+y+1=0rArry=-x-1rArrm_1=-11` `therefore ` slope of the line `=(-1)/(-1)=1` `therefore` Equation of required line is `y-2=1(x-1)` `rArry-2=x-1` `rArry-x-1=0` |
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| 76. |
Find the tangent of the angle between the lines whose intercepts n theaxes are respectively `a ,-b` and `b ,-a`A. `(a^2-b^2)/(ab)`B. `(b^2-a^2)/(2)`C. `(b^2-a^2)/(2ab)`D. None of these |
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Answer» Correct Answer - C Sinc, intercepts on the axes are a,-b then equation of the line is `(x)/(a)-(y)/(b)=1` `rArr(y)/(b)-(x)/(a)=1` `rArr y=(bx)/(a)-b` So, the lope of this line i.e., `m_1=(b)/(a)` Also, for intercepts on the axes as b and -a, then equation of the line is `(x)/(b)-(y)/(a)=1` `rArr(y)/(a)-(x)/(b)-1rArry=(a)/(b)x-a` and slope of this line i.e,`m_2=(a)/(b)` `tan theta =(b/a-a/b)/(1+a/b.b/a)=((b^2-a^2)/(ab))/(2)=(b^2-a^2)/(2ab)` |
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| 77. |
Find the distance between the parallel line 15x+8y-34=0 and 15x+8y+31=0. |
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Answer» Converting each of the given equations to the form `y=mx+C`, we get `15x+8y-34=0 Rightarrow y=(-15)/(8)x+(17)/(4).......(i)` `15x+8y+31=0 Rightarrow y=(-15)/(8)x-(31)/(8).......(ii)` Clearly, the slope of the given line are equal and so they are parallel. The given line are of the form `y=mx+C_(1) and y=mx+C_(2)"Where ",m=(-15)/(8), C_(1)=(17)/(4) and C_(2)=(-31)/(8)` `therefore` distance the given lines. `=(|C_(2)-C_(1)|)/(sqrt(1+m^(2))),"where m"=(-15)/(8),C_(1)=(17)/(4) and C_(2)=(-31)/(8)` `=(|(-31)/(8)-(17)/(4)|)/(sqrt(1+((-15)/(8))^(2)))=(|-(65)/(8)|)/(sqrt(1+(225)/(64)))=(((65)/(8)))/(sqrt(1+(289)/(64)))=((65)/(8)xx(8)/(17))=(65)/(17)"units"` Hence, the distance between the given line is `(65)/(17)` units. |
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| 78. |
The points A(2,3), B(4,-1) and C(-1,2) are the vertices of A ABC. Find the length of perpendicular from C on AB and hence find the area of `triangle ABC`. |
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Answer» Correct Answer - `(7)/(sqrt5)` units,7 sq. units `"Equation of AB is"=(y-3)/(x-2)=(-4)/(2) Rightarrow 2x+y-7=0` Length of the perpendicular from C(1,-2) to 2x+y-7=0 is equal to `(|2xx(-1)+2-7|)/(sqrt(2^(2)+1^(2)))=(7)/(sqrt5)"units"` `ar(triangle ABC)=(1)/(2)xxABxx(7)/(sqrt5)=((1)/(2)xxsqrt20xx(7)/(sqrt5))"sq units. =7 sq. units"` |
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| 79. |
Using slopes, find the value of x for which the points A(5, 1), B(1,-1) and C(x-4) are collinear. |
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Answer» Correct Answer - X=11 For collinearity, we must have, slope of AB=slope of BC |
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| 80. |
What are the points on the yaxis whose distance from the line `x/3+y/4=1`is 4 units. |
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Answer» Correct Answer - (8,0) and (-2,0) Let the required point be P(x,o) Given line is 4x+3y-12=0 Now, `(|4x+3xx0-12|)/(sqrt(4^(2)+3^(2)))=4 Leftrightarrow |x-3|=5 Rightarrow x-3=5 or x-3=-5 ` `Rightarrow x=8 or x=-2` |
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| 81. |
Find the slope of the line, which makes an angle of `30o`with the positive directionof yaxis measured anticlockwise. |
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Answer» Correct Answer - `-sqrt3` The given line makes an angle of `(90^(@)+30^(@))=120^(@)` with the positive direction of the x-axis. Hence, `m=tan 120^(@)=-sqrt3` |
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| 82. |
Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114. Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`. For all positions of pont P, angle APB is maximum when point P isA. (0, 12)B. (0, 15)C. (0, 18)D. (0, 21) |
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Answer» Correct Answer - C 0 |
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| 83. |
Match the following lists: |
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Answer» Correct Answer - `a to p, s;" b" to q,s; " c" to p,r; " d" to s` a. The given lines are concurrent. So, `|{:(3, 1, -4),(1, -2, -6),(lambda, 4, lambda^(2)):}| = 0` `"or " lambda^(2) +2lambda-8 =0` `"or " lambda=2, -4` b. The points are collinear. Hence, `|{:(lambda+1, 1, 1),(2lambda+1, 3, 1),(2lambda+2, 2lambda, 1):}| = 0` `"or " 2lambda^(2) - 3lambda-2 =0 " or " lambda = 2, -(1)/(2)` c. The point of intersection of x-y=1 and 3x+y-5=0 is (1,2). It lines on the line `x+y-1-|(lambda)/(2)| = 0 " "therefore lambda = +-4` d. The midpoint of (1, -2) and (3, 4) will satisfy `y-x-1+lambda=0` `therefore lambda = 2` |
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| 84. |
Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114. Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`. The maximum value of angle APB isA. `(pi)/(3)`B. `(pi)/(2)`C. `(2pi)/(3)`D. `(3pi)/(3)` |
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Answer» Correct Answer - B 0 |
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| 85. |
Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114. Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`. For all positions of pont Q, and AQB is maximum when point Q isA. (0, 54)B. (0, 58)C. (0, 60)D. (0, 1) |
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Answer» Correct Answer - B 0 |
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| 86. |
The equation to the line bisecting the join of (3,-4) and (5,2) and having intercepts on the x-axis and y-axis in the ratio of 2:1 is: |
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Answer» `x=(3+5)/2=4 and y=(2-4)/2=-1` Point P(4,-1) x intercept=2a and y intercept=a general formula `x/a+y/b=1` `x/(2a)+y/a=1-(1)` Putting point P `4/(2a)-1/a=1` a=1 putting this value in equation 1 `x/2+y/1=1` x+2y-2=0. |
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| 87. |
A stick of length `l` slides with its ends on two mutually perpendicular lines. Find the locus of the middle point of the stick. |
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Answer» Let the stick intercepts x-axis at point `(a,0)` and y-axis at point `(0,b)` and `(alpha, beta)` are midpoints of the stick. then, `(a+0)/2 = alpha => a = 2alpha` `(b+0)/2 = beta=> b= 2beta` Then, `l^2 = (a-0)^2+(0-b)^2` Here, `l` is the length of stick. `=>l^2 = a^2+b^2 ` `=>l^2 = (2alpha)^2+(2beta)^2` `=>l^2 = 4alpha^2+4beta^2` `alpha^2+beta^2 = (l/2)^2` If we replace `(alpha,beta)` by `(x,y)`, then , our equation becomes, `x^2+y^2 = (l/2)^2` So, locus will be a circle with center at origin and radius `l/2`. |
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| 88. |
Find the area of the triangle formed by the lines `y-x=0,x+y=0`and `x-k=0`. |
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Answer» Equation of line 1,`L_1->y-x = 0=> x= y->(1)` Equation of line 2,`L_2->x+y = 0=> x = -y->(2)` Equation of line 3,`L_3->x-k = 0=> x = k->(3)` Now, if we draw these lines, it will form a triangle with vertices, `(0,0),(k,k),(k,-k)`. Please refer to video for the diagram. Now, from the diagram, we can see that, Base of triangle `=2k` Altitude of triangle ` = k` So, Area of triangle = 1/2*base*height `:.` Area `= 1/2*2k*k = k^2` square units |
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| 89. |
Two points P(a,0) and Q(-a,0) are given, R is a variable on one side of the line PQ such that `/_RPQ-/_RQP` is a positive constant `2alpha`. FInd the locus of point R. |
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Answer» `/_RPQ=phi` `/_RQP=theta` `phi-theta=2alpha` `tan(phi-theta)=2alpha` `(tanphi-tantheta)/(1+tanthetatanphi)=tan2alpha` `tantheta=y/(x+a),tanphi=y/(a-x)` `(y/(x+a)+y/(x-a))/(1-y^2/((x-a)(x-b)))=tan2alpha` `(2xy)/(a^2-x^2+y^2)=tan2alpha`. |
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| 90. |
Find the coordinates of the center of the circle inscribed in a triangle whose vertices are (-36,7) , (20,7) and (0,-8). |
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Answer» Let `ABC` is the triangle with the vertices `A(-36,7), B(20,7) and C(0,-8)`. Then, `AB = sqrt((20-(-36))^2 + (7-7)^2) = 56` `BC = sqrt((0-(-20))^2+((-8)-7)^2 ) )= sqrt(400+225) = sqrt(625) = 25` `AC = sqrt((-36-0)^2 + (7-(-8))^2)) = sqrt(1521) = 39` Let, `O(O_x,O_y)` is the center of the circle inscribed in triangle `ABc`. Then, `O_x = (A_xa+B_xb+C_xc)/P and O_y = (A_ya+B_yb+C_yc)/P` Here, `P = AB+BC+AC = 56+25+39 = 10` and `a = 25,b = 39, c = 56` So, `O_x = (-36**25+20**39+0**56)/120 = -120/120 = -1` `O_y = (7**25+7**39+(-8)**56)/120 = 0/120 = 0` So, coordinates of the center of the circle are `O(-1,0)`. |
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| 91. |
If the slope of the line passing through the points (2,5) and (x,3) is 2. find the value of x. |
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Answer» Let A(2,5) and B(x,3) be the given points. Then slope of AB=`(3-5)/(x-2)=(-2)/((x-2))` `therefore (-2)/((x-2))=2 Rightarrow 2x-4=-2` `Leftrightarrow 2x=2 Leftrightarrow x=1` Hence, x=1. |
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| 92. |
Find the value of x so that the inclination of the line joining the points (x,-3) and (2,5) is `135^(@)` |
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Answer» Let A(x,-3) and B(2,5) be the given points. Then slope of AB=`(5-(-3))/(2-x)=(8)/((2-x))` But slope of AB`=tan135^(@)=tan(180^(@)-45^(@))=-tan45^(@)=-1` `therefore (8)/((2-x))=-1 Leftrightarrow x-2=8 Leftrightarrow x=10` Hence, x=10 |
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| 93. |
What is the inclination of a line whose slope isA. zeroB. positiveC. negative?D. not defined? |
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Answer» Let `theta` be the inclination of the given line. Then, m=`tan theta` (i) `m=0 Rightarrow tan theta=0` `Rightarrow theta=0^(@)" "[therefore 0^(@) le theta lt 180^(@)]` (ii) `m gt 0 Rightarrow tan theta gt 0` `Rightarrow theta` lies between `0^(@) and 90^(@)` `Rightarrow theta` is acute. (iii) `m lt 0 Rightarrow tan theta lt 0` `Rightarrow theta` lies between `90^(@) and 180^(@)` `Rightarrow theta` is obtuse. We know that a vertical line is the only line whose slope is not defined. And, the inclination of a vertical line is `90^(@)`. Hence, the inclination of a line whose slope is not defined, is `90^(@)`. |
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| 94. |
Find the equation of a line whose inclination with the x-axis is `150^(@)` and which passes through the point (3, -5). |
| Answer» Correct Answer - `x+sqrt3y+(-3+5sqrt3)=0` | |
| 95. |
Find the equation of a line whose inclination with the x-axis is `30^(@)` and which passes through the point (0, 5). |
| Answer» Correct Answer - `x-sqrt3y+5sqrt3=0` | |
| 96. |
Find the coordinates of one vertex of an equilateral triangle with centroid at the origin and the opposite side `x +y-2 =0`.A. (-1,1)B. (2,2)C. (-2,-2)D. none of these |
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Answer» Since the triangle is equilateral . Therefore , centroid coincides with the circumcentre and orthocentre . The equation of the perpendicular bisector of x + y - 2= 0 and passing through the centroid (0,0) is x - y = 0 . The vertex of the triangle must lie on x- y = 0 and the origin and the vertex must lies on the same side of x + y - 2 = 0. So we can choose (-2,-2) as the vertex . |
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| 97. |
Find the points on `y-a xi s`whose perpendicular distance from the line `4x-3y-12=0`is `3.` |
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Answer» Correct Answer - (0,1) and (0,9) Let the required point be `P(0, alpha)`. It is given that the length of the perpendicular from `P(0, alpha)` on 4x-3y-12=0 is 3. Therefore, `|(4(0)-3alpha-12)/(sqrt(4^(2)+(-3)^(2)))|=3` `"or " |3alpha+12| = 15` `"or " |alpha+4| = 5` `"or " alpha+4 =+-5` `"or " alpha = 1, -9` Hence, the required points are (0,1) and (0,-9). |
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| 98. |
The straight line passing through `P(x_1,y_1)` and making an angle `alpha` with x-axis intersects `Ax + By + C=0` in Q then PQ= |
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Answer» Correct Answer - `(|Ax_(1) + By_(1) + C|)/(|A "cos" alpha +B "sin" alpha|)` Line is passing through the point P`(X_(1), Y_(1))` and making an angle `alpha` with x-axis. One of the points on the line at distance r from the point P is given `(x_(1)+r "cos" alpha, y_(1) +r "sin" alpha)`. If this point id Q, then it lies on the line. So, A`(x_(1) +r "cos" alpha) +B(y_(1) +r "sin" alpha) +C=0`. `therefore r = (|Ax_(1) +By_(1)+C|)/(|A"cos" alpha + B"sin"alpha|)` |
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| 99. |
Given the four lines with the equations `x+2y -3 = 0,3x+4y-7=0, 2x +3y - 4 = 0, 4x+5y -6 =0, then |
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Answer» Here, we are given 4 lines `x+2y-3=0` `3x+4y-7=0` `2x+3y-4=0` `4x+5y-6=0` If, we subtract twice of third equation from fourth equation, we get `y=2` and putting it in any of the equation, we get `x=-1`. If we put these value in first line equation, `-1+2(2)-3=0` that is ` true`. If we put these value in second line equation, `2(-1)+3(2)-4=0` that is not true. That means three lines are intersecting at point(-1,2), but the fourth one is not that concludes They are not concurrent and they are not forming quardilateral.So, option `C` None of these is the correct option. |
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| 100. |
Given four lines whose equations are `x +2y -3=0, 2x+3y-4=0, 3x + 4y -7 = 0 and 4x + 5y-6=0`, then the lines areA. concurrentB. sides of a squareC. sides of a rhombusD. none of these |
| Answer» Correct Answer - D | |