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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.
| 251. |
The co-ordinates of the orthocentre of the triangle bounded by the lines, `4x -7y + 10=0;x+y=5` and `7x + 4y=15` isA. (-1,-2)B. (1,-2)C. (-1,2)D. (1,2) |
| Answer» Correct Answer - D | |
| 252. |
If the points `((a^3)/(a-1),(a^2-3)/(a-1)),((b^3)/(b-1),(b^3-3)/(b-1))a n d((c^3)/(c-1),(c^3-3)/(c-1))`where `a , b ,c`are different from 1 lieon the line `l x+m y+n=0``a+b+c=-m/l``a b+b c+c a+n/l=0``a b c=((3m+n))/l``a b c-(b c+c a+a b)+3(a+b+c)=0`A. `a+b+c = -(m)/(l)`B. `ab+bc+ca=(n)/(l)`C. `abc = ((m+n))/(l)`D. abc-(bc+ca+ab) +3(a+b+c)=0 |
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Answer» Correct Answer - A::B::D Since the given points lie on the line lx+my+n=0, a,b,c are the roots of the equation `l((t^(3))/(t-1)) + m((t^(3)-3)/(t-1))+n=0` `"or " l t^(3) +mt^(2) + nt-(3m+n) = 0 " " (1)` `"Hence, "a+b+c = -(m)/(l)` `ab+bc+ca = (n)/(l)" " (2)` `abc = (3m+n)/(l) " " (3)` So, from (1), (2), and (3), we get abc-(bc+ca+ab)+3(a+b+c)=0` |
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| 253. |
Let L be the line belonging to the family of straight lines (a+2b) x+(a-3b)y+a-8b =0, a, b `in `R, which is the farthest from the point (2, 2). If L is concurrent with the lines x-2y+1=0 and `3x-4y+ lambda=0`, then the value of `lambda` isA. 2B. 1C. -4D. 5 |
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Answer» Correct Answer - D Lines x+4y+7 =0 and x-2y+1 =0 intersect at (-3, -1) which must satisfy the line `3x-4y+lambda=0. "Then " -9+4+lambda=0 " or " lambda = 5.` |
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| 254. |
Let `l` be the line belonging to the family of straight lines `(a + 2b)x+ (a - 3b)y +a-8b = 0, a, b in R`, which is farthest from the point `(2, 2),` then area enclosed by the line `L` and the coordinate axes isA. x+4y+7=0B. 2x+3y+4=0C. 4x-y-6=0D. none of these |
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Answer» Correct Answer - A Given lines (x+y+1) +b(2x-3y-8) = 0 are concurrent at the point of intersection of the line x+y+1=0 and 2x-3y-8=0, which is (1, -2). Now, the line through A(1, -2), which is farthest from the point B(2, 2), is perpendicular to AB. Now, the slope of AB is 4. Then the required line is y+2 =-(1/4)(x-1) or x+4y=0. |
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| 255. |
Let `l` be the line belonging to the family of straight lines `(a + 2b)x+ (a - 3b)y +a-8b = 0, a, b in R`, which is farthest from the point `(2, 2),` then area enclosed by the line `L` and the coordinate axes isA. 4/3 sq. unitsB. 9/2 sq. unitsC. 49/8 sq. unitsD. none of these |
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Answer» Correct Answer - C Also, this line x+4y+7 =0 meets the axes C(-7,0) and D(0, -7/4). Then the area of triangle OCD is (1/2)|(-7)(-7/4)| or 49/8. |
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| 256. |
The limiting position of the point of intersection of the lines `3x+4y=1 and (1+c)x+3c^2 y=2` as c tends to 1, isA. (-5,4)B. (5 , -4)C. (4 , -5)D. none of these |
| Answer» Correct Answer - A | |
| 257. |
P(2,1) , Q (4,-1) , R (3,2) are the vertices of a triangle and if through P and R lines parallel to opposite sides are drawn to intersect in S , then the area of PQRS , isA. 6B. 4C. 8D. 12 |
| Answer» Correct Answer - B | |
| 258. |
Given three straight lines `2x+11 y-5=0,24 x+7y-20=0,`and `4x-3y-2=0`. Then,they form a triangleone line bisects the angle between the other twotwo of them are parallelA. `2p_(1) = p_(2)`B. `p_(1) = p_(2)`C. `p_(1) = 2p_(2)`D. none of these |
| Answer» Correct Answer - B | |
| 259. |
Given three straight lines `2x+11 y-5=0,24 x+7y-20=0,`and `4x-3y-2=0`. Then,they form a triangleone line bisects the angle between the other twotwo of them are parallelA. they from a triangleB. they are concurrentC. one line bisects the angle between the other twoD. two of them are parallel |
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Answer» Correct Answer - C For the two lines 24x+7y-20=0 and 4x-3y-2=0, the angle bisectors are given by `(24x+7y-20)/(25) =+-(4x-3y-2)/(5)` Taking positive sign, we get 2x+11y-5=0 |
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| 260. |
The centroid of the triangle formed by the pair of straight lines `12x^2- 20xy+7y^2=0` and the line `2x-3y + 4 = 0 `isA. `(-(7)/(3),(7)/(3))`B. `(-(8)/(3),(8)/(3))`C. `((8)/(3),(8)/(3))`D. `((4)/(3),(4)/(3))` |
| Answer» Correct Answer - C | |
| 261. |
Find the equation of the line passing through the point of intersection of the lines `4x+7y-3=0 and 2x-3y+1=0`, which has equal intercepts on the axes. |
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Answer» Let the point equation of the required line be `(x)/(a)+(y)/(a)=1` i.e. x+y=a …..(i) The equation of the given line are `4x+7y=3......(i)` `2x-3y=-1......(ii)` On solving (ii) and (iii), we get `x=(1)/(13), y=(5)/(13)` So, the point of intersection of the given line is `P((1)/(13),(5)/(13))` Putting `x=(1)/(13) and y=(5)/(13)` in (i), we get `a=((1)/(13)+(5)/(13))=(6)/(13)` Hence, the required equation is `x+y=(6)/(13)` i.e. 13x+13y-6=0 |
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| 262. |
Pind the distance between the parallel lines 8x+15y-36=0 and 8x+15y+32=0. |
| Answer» Correct Answer - 4 units | |
| 263. |
The line L given by `x/5 + y/b = 1` passes through the point (13,32).the line K is parallel to L and has the equation `x/c+y/3=1` then the distance between L and K isA. `17//sqrt(15)`B. `23//sqrt(17)`C. `23//sqrt(15)`D. `sqrt(17)` |
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Answer» Correct Answer - B Let `m_(1)` and `m_(2)` be the slopes of lines L and K respectively. Then, `m_(1)=-(b)/(5) and m_(2)=-(3)/(c)` It is given that the lines L and K are parallel. `m_(1)=m_(2)` `implies -(b)/(5)=-(3)/(c)implies bc=15` Line L passes through the point (13, 32). `therefore (13)/(5)=(32)/(b)=1implies b=-20` `therefore ` bc =15 `implies c=-(3)/(4)` Hence , the equations of lines L and K are `4x-y-20=0` and `4x-y+3=0` `therefore ` Distance between L and K `=(23)/(sqrt(16+1))=(23)/(sqrt(17))` |
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| 264. |
Find the incentre of a triangle formed by the lines `x "cos" (pi)/(9) + y "sin" (pi)/(9) = pi, x "cos" (8pi)/(9)+ y "sin" (8pi)/(9) = pi " and " x "cos" (13pi)/(9) + y "sin" ((13pi)/(9)) = pi.` |
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Answer» Correct Answer - (0,0) Incentre of the triangle is the point which is equidistant from all the sides of triangle. Clearly, distance of origin from all the three given lines is `pi`. Also, (0,0) is interior point of triangle which can be verified by drawing lines on the plane. So, incentre of the triangle is (0,0). |
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| 265. |
If two members of family `(2+lambda)x+(1+2lambda)y-3(1+lambda) = 0` and line x+y=0 make an equilateral triangle, the the incentre of triangle so formed isA. `((1)/(3), (1)/(3))`B. `((7)/(6), -(5)/(6))`C. `((5)/(6), (5)/(6))`D. `(-(3)/(2), -(3)/(2))` |
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Answer» Correct Answer - A Given equation of family of lines is `(2x+y-3)+lambda(x+2y-3) =0, lambda in R.` The family of lines is concurrent at point of intersection of 2x+y-3 = 0 and x+2y-3=0 which is A(1,1). Two members `L_(1)" and "L_(2)` of this family and line x+y=0 form an equilateral triangle. Let foot of perpendicular from A(1, 1) on the line x+y=0 be M(h,k). `(h-1)/(1) = (k-1)/(1) = (-(1+1))/(2)` `therefore M-=(h,k)-=(0,0)` Since triangle is equilateral, AM is median, altitude and angle bisector. `AM= sqrt((1-0)^(2)+(1-0)^(2)) = sqrt(2)` Incentre coincides with centroid G such that AG `=(2)/(3)sqrt(2)`. `"Slope of " AM is 1 = "tan" 45^(@)`. `therefore G-=(1-(2sqrt(2))/(3)"cos" 45^(@), 1-(2sqrt(2))/(3)"sin" 45^(@))` `-=((1)/(3),(1)/(3))` |
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| 266. |
A line passes through the point `(2,2)` and is perpendicular to the line `3x + y =3,` then its `y`-intercept isA. `(1)/(3)`B. `(2)/(3)`C. `1`D. `(4)/(3)` |
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Answer» Correct Answer - D Given line is `y=3-3x`. Then , slope of the required line `=(1)/(3)` `because` Equation of the required line is `y-2=(1)/(3)(x-2)` `rArr3y-6=x-2` `rArr x-3y+4=0` For y-intercept put `x=0`. `0-3y+4=0` `rArr y=(4)/(3)` |
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| 267. |
Given three straight lines `2x+11 y-5=0,24 x+7y-20=0,`and `4x-3y-2=0`. Then,they form a triangleone line bisects the angle between the other twotwo of them are parallelA. form a triangleB. are only concurrentC. are concurrent with one line bisecting the angle between the other twoD. none of these |
| Answer» Correct Answer - C | |
| 268. |
The line L given by `x/5+y/b=1`passes through the point (13,32). The line K is parallel to L and has the equation `x/c+y/3=1`Then the distance between L andK is(1) `sqrt(17)`(2) `(17)/(sqrt(15))`(3) `(23)/(sqrt(17))`(4) `(23)/(sqrt(15))` |
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Answer» `K || L` `L: x/5 + y/b = 1` `L(13,32) = 13/5 + 32/b = 1` `32/b = 1- 13/5` `32/b = -8/5` `b= -32 xx5/8 = -20` `x/5 - y/20 = 1` `y = 20 x/5 -20 =4x - 20` slope of L=`4` slope of K=`4` `x/c + y/3 = 1` `y = -x/c 3 + 3` `=> -3/c = 4` `c= -3/4` k: `y= 4x + 3` distance= `|(c_1 - c_2)/(sqrt(a^2 + b^2))| = |(-20-3)/(sqrt(1+16))|` `= 23/sqrt17` answer |
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| 269. |
The x-coordinate of the incentre of the trianglethat has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1,0) is(1) `2-sqrt(2)`(2) `1+sqrt(2)`(3) `1-sqrt(2)`(4) `2+sqrt(2)` |
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Answer» `0 = (a+x)/2` `x=-a` `1= (b+y)/2` `y= 2-b` now,` 1= (-a + 2 -a)/2` `2 = -2a + 2` `a= 0` & `4-2b = 0` `b=2` `AC= sqrt((0-2)^2 + (2-0)^2) = sqrt 8 ` `I_n = (ax_1 + bx_2 + c x_3)/(a+b+c)` where, a,b,c are length of the sides & `x_1, x_2 , x_3` are opposite vertex coordinates so,` (2 xx 2 + 2 xx 0 + sqrt8 xx 0)/(2 + 2 + sqrt8)` `= 4/(4 + 2 sqrt2) xx (4 - 2sqrt 2)/(4 - 2sqrt 2)` `= (16- 8 sqrt2)/(16- 8)` `= (16- 8 sqrt2)/8` `= (4-2sqrt2)/2 = 2 - sqrt2` option 1 is correct |
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| 270. |
LetA(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle withAC as its hypotenuse. If the area of the triangle is 1, then the set ofvalues which k can take is given by(1) `{1,""3}`(2) `{0,""2}`(3) `{-1,""3}`(4) `{-3,-2}` |
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Answer» `AC^2=AB^2+BC^2` `(h-2)^2+(k-1)^2=(h-1)^2+(k-1)^2+1` `h^2-4h+4+k^2-2k+1=h^2-2h+1+k^2-2k+1+1` `-2h=-5+3` `h=1` `A=1/2|(k-1)|xx1=1` `|k-1|=2` `k=1+_2` `=-1,3` option`3(-1,3)` |
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| 271. |
A line passes through the point `(2,2)` and is perpendicular to the line `3x + y =3,` then its `y`-intercept isA. `1//3`B. `2//3`C. 1D. `4//3` |
| Answer» Correct Answer - D | |
| 272. |
A line passes through (2,2) and is perpendicular to the line `3x+y=3`, the y-intercept is?A. `1//3`B. `2//3`C. 1D. `4//3` |
| Answer» Correct Answer - D | |
| 273. |
Let PS be the median of the triangle with vertices `P(2,""2),""Q(6,-1)""a n d""R(7,""3)`. The equation of the linepassing through `(1,-1)`and parallelto PS is(1) `4x-7y-11""=""0`(2) `2x+""9y+""7""=""0`(3) `4x+""7y+""3""=""0`(4) `2x-9y-11""=""0` |
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Answer» slope of line `PS= (2-1)/(2- 13/2)` `= -2/9` slope of line `AB= m= -2/9` point `(1,-1)` `y- y_1 = m(x- x_1)` `y + 1 = - 2/9 (x-1)` `9y + 9 = -2x+2` `2x + 9y + 7=0` option 2 is correct |
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| 274. |
A triangle ABC, right angled at A, has points A and B as (2, 3) and (0,-1), respectively. If C is BC=5, then the point C isA. (-4,2)B. (4,2)C. (3,-3)D. (0,-4) |
| Answer» Correct Answer - B | |
| 275. |
A line passes through (1,0). The slope of the line, for which its intercept between `y = x-2 and y=-x + 2` subtends a right angle at the origin isA. `pm(2)/(3)`B. `pm(3)/(2)`C. `pm1`D. none of these |
| Answer» Correct Answer - D | |
| 276. |
If `A` and `B` are two fixed points, then the locus of a point which moves in such a way that the angle `APB ` is a right angle isA. a circleB. an ellipseC. a parabolaD. none of these |
| Answer» Correct Answer - A | |
| 277. |
Find the equation of a line parallel to the y-axis at a distance of (i)6 units to its right (ii) 3 units to its left. |
| Answer» Correct Answer - (i)x-6=0 (ii)x+3=0 | |
| 278. |
Let PS be the median of the triangle with vertices `P(2,""2),""Q(6,-1)""a n d""R(7,""3)`. The equation of the linepassing through `(1,-1)`and parallelto PS is(1) `4x-7y-11""=""0`(2) `2x+""9y+""7""=""0`(3) `4x+""7y+""3""=""0`(4) `2x-9y-11""=""0`A. 4x-7y-1=0B. 2x+9y+7=0C. 4x+7y+3=0D. 2x-9y-11=0 |
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Answer» Correct Answer - B S is the midpoint of QR, `therefore S((13)/(2), 1)` Slope of PS `=-(2)/(9)` `therefore " Equation of required line is: " y+1 = -(2)/(9)(x-1)` `therefore 2x+9y+7=0` |
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| 279. |
If the coordinates of the vertices of triangle `A B C`are `(-1,6),(-3,-9)`and `(5,-8)`, respectively, then find the equation of the median through `Cdot` |
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Answer» Correct Answer - 13x+14y+47=0 The required equation fo median is `y+8 = (-(3//2)+8)/(-2-5)(x+5)` or 13x+14y+47=0 |
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| 280. |
Find the angle made by the line `x+sqrt3y-6=0` with the positive direction of the x-axis. |
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Answer» Let the required angle be `theta`. `"Now", x+sqrt3y-6=0 Rightarrow y=-(1)/(sqrt3)x+(6)/(sqrt3)` `therefore m=-(1)/(sqrt3) Leftrightarrow tan theta =-(1)/(sqrt3)=-tan 30^(@)=tan(180^(@)-30^(@))=tan 150^(@)` `therefore theta=150^(@)` Hence, the given line makes an angle of `150^(@)` with the positive direction of the x-axis. |
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| 281. |
Given that `P(3,1),Q(6. 5),`and `R(x , y)`are threepoints such that the angle `P R Q`is a rightangle and the area of ` R Q P`is 7, findthe number of such points `Rdot` |
| Answer» Correct Answer - C | |
| 282. |
If one diagonal of a square is along the line `x=2y` and one of its vertex is `(3,0)`, then its sides through the vertex are given by the equations -A. `y-3x+9=0,3y+x-3=0`B. `y+3x+9=0,3y+x-3=0`C. `y-3x+3=0,3y-x+3=0`D. `y-3x+3=0,3y+x+9=0` |
| Answer» Correct Answer - A | |
| 283. |
If A(1, 4, B(2, 3) and C(-1,-2) are the vertices of a `triangle ABC`, find the equation of (i) the median through A (ii) the altitude through A (iii) the perpendicular bisector of BC. |
| Answer» Correct Answer - (i)13x-y-9=0, (ii)3x-y+1=0 (iii)3x-y-4=0 | |
| 284. |
Reduce the equation `sqrt3y+y+2=0` to (i) slope-intercept form and final the slope and y-intercept. (ii) intercepts form and find the intercepts on the axes. |
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Answer» We have `(i)sqrt3x+y+2=0 Rightarrow y=-sqrt3x-2` This is of the form `y=mx+c," where "m=-sqrt3 and c=-2` `therefore y=-sqrt3-2` in the slope-intercept form. Here, slope=`-sqrt3` and y-intercept=-2. (ii) `sqrt3+y+2=0 Rightarrow sqrt3+y=-2` `Rightarrow ((sqrt3)/(-2))x+((1)/(-2))y=1` `Rightarrow (x)/(((-2)/(sqrt3)))+(y)/(-2)=1` This is of the form `(x)/(a)+(y)/(b)=1,` where `a=(-2)/(sqrt3) and b=-2` Thus, `Rightarrow (x)/(((-2)/(sqrt3)))+(y)/(-2)=1` is intercepts form. Here, x-intercept=`(-2)/(sqrt3)` and y-intercept=-2 |
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| 285. |
If the pair of straight lines `xy - x - y +1=0` & the line `ax+2y-3=0` are concurrent then `a =`A. `-1`B. 0C. 3D. 1 |
| Answer» Correct Answer - D | |
| 286. |
Find the equation of the perpendicular drawn from the origin to the line 4x-3y+5=0. Also, find the coordinates of the foot of the perpendicular. |
| Answer» `(3x+4y=0,(-(4)/(3),(3)/(5)))` | |
| 287. |
The coordinates of the foot of the perpendicular from the point `(2,3)`on the line `-y+3x+4=0`are given by`((37)/(10),-1/(10))`(b) `(-1/(10),(37)/(10))``((10)/(37),-10)`(d) `(2/3,-1/3)`A. (37,/10,-1/10)B. (-1/10,37/10)C. (10/37,-10)D. (2/3,-1/3) |
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Answer» Correct Answer - B The line passing through (2,3) and perpendicular to -y+3x+4=0 is `(y-3)/(x-2) =-(1)/(3)` or 3y+x-11 = 0 Therefore, the foot is x=-1/10, y=37/10. |
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| 288. |
The line, which is parallel to `X`-axis and crosses the curve `y = sqrtx` at an angle `45^@`, isA. x=1/4B. `y=1//4`C. `y=1//2`D. `y=1` |
| Answer» Correct Answer - C | |
| 289. |
The line `lx+my+n=0` intersects the curve `ax^2 + 2hxy + by^2 = 1` at the point P and Q. The circle on PQ as diameter passes through the origin. Then `n^2(a+ b)` equals (A) `l^2+m^2` (B) `2lm`(C) `l^2-m^2` (D) `4lm`A. `n^(2)(a+b)=l^(2)+m^(2)`B. `l^(2)(a+b)=n^(2)+m^(2)`C. `m^(2)(a+b)=l^(2)+n^(2)`D. none of these |
| Answer» Correct Answer - A | |
| 290. |
If lines `x+2y-1=0,a x+y+3=0,`and `b x-y+2=0`are concurrent, and `S`is the curve denoting the locus of `(a , b)`, then the least distance of `S`from the origin is`5/(sqrt(57))`(b) `5//sqrt(51)``5//sqrt(58)`(d) `5//sqrt(59)`A. `5//sqrt(57)`B. `5//sqrt(51)`C. `5//sqrt(58)`D. `5//sqrt(59)` |
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Answer» Correct Answer - C The lines are concurrent if `|{:(1,2,-1),(a,1,3),(b,-1,2):}|=0` or 7b-3a+5=0 The locus of (a,b) is 3x-7y=5. Least distance from (0,0) = Length of perpendicular from (0,0) `=(5)/(sqrt(58))` |
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| 291. |
Find the equations of the lines which pass through the origin and areinclined at an angle `tan^(-1)m`to the line `y=m x+cdot` |
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Answer» Correct Answer - `y =0, y = (2mx)/(1-m^(2))` Angle between both the lines is `"tan"^(-1)m+- "tan"^(-1)m = "tan"^(-1) (2m)/(1-m^(2)) or "tan"^(-1)0` Therefore, the equations of the lines are `y=0, y= (2mx)/(1-m^(2))` |
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| 292. |
The area enclosed within the curve `|x|+|y|=1` isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B | |
| 293. |
The equation of the bisector of the acute angle between the lines `2x-y+4=0`and `x-2y=1`is`x-y+5=0``x-y+1=0``x-y=5`(d) none oftheseA. `x + y+ 5=0`B. `x-y+ 1=0`C. ` x- y - 5=0`D. none of these |
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Answer» Correct Answer - B The equations of the lines are `2x - y + 5 =0` We have , `2x -1+(-1) xx 2 le 0 ` i.e, `a_(1)a_(2) +b_(1)b_(2) lt 0` Therefore, the equation of the bisector of acute angle is `(2x-y+4)/(sqrt(1+4)) = (-x+2y+1)/(sqrt((-1)^(2)+2^(2)))` `rArr 2x - y + 4 = - x + 2y + 1` `rArr 3x -3y + 3 = 0 rArr x -y + 1 =0` |
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| 294. |
The equation of the bisector of the obtuse angle between the lines `x-2y+4=0` and `4x-3y+2=0` is |
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Answer» Equation of bisectors `(x-2y+4)/sqrt5=(4x--3y+2)/sqrt(4^2+3^2)` `sqrt5x-2sqrt5y+4sqrt5=4x-3y+2` `y(3-2sqrt5)+x(sqrt5-4)+4sqrt5-2=0` `m(beta_1)=(4-sqrt5)/(3-2sqrt5)` this will be negative So, this is obtuse. `(x-2y+4)/sqrt5=-(4x--3y+2)/sqrt(4^2+3^2)` In this case it will be positive So, this is acue. . |
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| 295. |
The distance between the lines `4x + 3y = 11` and `8x + 6y = 15` isA. `7//2`B. 4C. `7//10`D. none of these |
| Answer» Correct Answer - C | |
| 296. |
The points on `x+y=4` that lie at a unit distance from the line `4x+3y-10=` areA. (3,1) and (-7,11)B. (-3,7) and (2,2)C. (-3,7) and (-7,11)D. none of these |
| Answer» Correct Answer - A | |
| 297. |
Let `A3(0,4) and Bs(21,0) in R`. Let the perpendicular bisector of AB at M meet the y-axis at R. Then the locus of midpoint P of MR is `y=x^2 + 21`A. `x^(2) + y^(2) = (1)/(4) `B. `(y-2)^(2) - x^(2) = 4 `C. ` y + x^(2) = 2`D. `3x^(2) + y^(2) = 8` |
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Answer» Correct Answer - C The equation of MR is `y - 2 = (t)/(2) (x - t) implies tx - 2y + 4 -t^(2) = 0` It cuts y-axis at `( 0 , 2 - (t^(2))/(2))` Let P , (h , k) be the mid-point of MR . Then , ` h = (0 + t)/(2) , k = ( 2 - (t^(2))/(2) + 2)/(2)` `implies t = 2h , k = 2 - (t^(2))/(4) implies k = 2 - h^(2) implies h^(2) + k = 2` Hence , the locus of (h , k) is `x^(2) + y = 2`. |
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| 298. |
The foot of the perpendicular on the line `3x+y=lambda`drawn from the origin is `Cdot`If the line cuts the `x`and the y-axis at `Aa n dB`, respectively, then `B C: C A`is`1:3`(b) `3:1`(c) `1:9`(d) `9:1`A. `1:3`B. `3:1`C. `1:9`D. `9:1` |
| Answer» Correct Answer - D | |
| 299. |
If the straight line `a x+c y=2b ,`where `a , b , c >0,`makes a triangle of area 2 sq. units with the coordinate axes, then`a , b , c`are in GPa, -b; c are in GP`a ,2b ,c`are in GP (d) `a ,-2b ,c`are in GPA. a,b,c are in GPB. a,-b, c are in GPC. a,2b,c are in GPD. a,-2b, c are in GP |
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Answer» Correct Answer - A::B The area of the triangle is given by `(1)/(2) xx (2b)/(a) xx (2b)/(c) = (2b^(2))/(ac) = 2` `"or " b^(2) = ac` Therefore, a,b,c are in GP. So, a,-b, c are alos in GP. |
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| 300. |
Find the equation of the bisector of the obtuse angle between the lines`3x-4y+7=0`and `12 x+5y-2=0.` |
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Answer» Correct Answer - 21x+77y-101=0 Firstly, make the constant terms `(c_(1), c_(2))` positive. 3x-4y+7 = 0 and -12x-5y+2=0 `therefore a_(1)a_(2) + b_(1)b_(2) = (3)(-12) + (-4)(-5)` =-36+20=-16 Hence, "-" sign gives the obtuse bisector. Therefore, the obtuse bisector is `((3x-4y+7))/(sqrt((3)^(2) + (-4)^(2))) = ((-12x-5y+2))/(sqrt((-12)^(2) + (-5)^(2)))` or 13(3x-4y+7) = -5(-12x-5y+2) or 21x+77y-101 =0 |
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