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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.
| 301. |
Find the equation of the bisector of the obtuse angle between the lines`3x-4y+7=0`and `12 x+5y-2=0.`A. 99x - 27y - 81 = 0B. 11x - 3y + 9 = 0C. 21x + 77y - 101 = 0D. 21x + 77y + 101 = 0 |
| Answer» Correct Answer - B | |
| 302. |
The equation of the bisector of that angle between the lines `x+2y-11=0,3x-6y-5=0`which contains the point `(1,-3)`is`(3x=19`(b) `3y=7``3x=19a n d3y=7`(d) None of theseA. 3x=19B. `3y=7`C. `3x=19 and 3y=7`D. none of these |
| Answer» Correct Answer - A | |
| 303. |
Find the range of `(alpha,2+alpha)`and `((3alpha)/2,a^2)`lie on the opposite sides of the line `2x+3y=6.` |
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Answer» Correct Answer - `alpha in (-oo, -2) uu (0,1)` We have L(x,y) = 2x+3y-6 ` therefore L(alpha, 2+alpha) = 5alpha` `" and " L((3)/(2) alpha, alpha^(2)) = 3alpha + 3alpha^(2)-6` Given points lie on the opposite sides of the line if `5alpha (3alpha + 3alpha^(2)-6) lt 0` `rArr alpha (alpha + 2)(alpha-1) lt 0` `rArr alpha in (-oo, -2) uu (0,1)` |
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| 304. |
The bisector of the acute angle formed between the lines 4x - 3y + 7 = 0 and 3x - 4y + 14 = 0 has the equationA. x + y + 3 = 0B. x - y - 3 = 0C. x - y + 3 =0D. 3x - y - 7 = 0 |
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Answer» Correct Answer - C The equations of given straight lines , by making constant terms positive , are 4x - 3y + 7 = 0 and 3x - 4y + 14 = 0 `because 4 xx 3 + (-3) (-4) = 24 gt 0 ` i.e, `a_(1) a_(2) + b_(1)b_(2) gt 0 ` So , the bisector of the acute angle is given by `(4x - 3y + 7)/(sqrt(4^(2) + (-3)^(2))) = -(3x - 4y + 14)/(sqrt(3^(2) (-4)^(2))) implies x - y + 3 = 0 ` |
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| 305. |
If P`(sin theta, 1//sqrt(2)) and Q(1//sqrt(2), cos theta), -pi le theta le pi` are two points on the same side of the line x-y=0, then `theta` belongs to the intervalA. `(-pi//4,pi//4)uu(pi//4,3pi//4)`B. `(-pi//4,pi//4)`C. `(pi//4,pi//4)`D. none of these |
| Answer» Correct Answer - A | |
| 306. |
The equation `(b-c)x+(c-a)y+(a-b)=0` and `(b^3-c^3)x+(c^3-a^3)y+a^3-b^3=0` will represent the same line ifA. a+b=-cB. c+a=-bC. b+c=-aD. a+b+c=0 |
| Answer» Correct Answer - D | |
| 307. |
Equation of the bisector of angle B of the triangle ABC is `y = x`. If A is `(2,6)` and B is `(1, 1)`; equation of side BC is (A) `2x + y - 3 = 0` (B) `x-5y+ 4 =0` (C) `x- 6y + 5 = 0` (D) none of theseA. 2x + y - 3 = 0B. x - 5y + 4 = 0C. x - 6y + 5 = 0D. none of these |
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Answer» The reflection or image of A (2,6) in the line mirror y =x will lie on BC. The reflection of A is P (6,2) . So , the equation of BC is `y -1 = (2-1)/(6-1) (x - 1) implies x - 5y + 4 = 0 ` |
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| 308. |
The equations of the sides of a triangle are `x-3y=0,4x+3y=5,3x+y=0.` The line `3x-4y=0` passes through (A) Incentre (B) Centroid (C) Orthocentre (D) CircumcentreA. the incentreB. the centroidC. the circumcentreD. the orthocentre , of the triangle |
| Answer» Correct Answer - D | |
| 309. |
Find the valueof `a`for whichthe lines `2x+y-1=0``2x+y-1=0``a x+3y-3=0``3x+2y-2=0`are concurrent.A. -3B. -1C. 1D. 4 |
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Answer» Correct Answer - A::B::C::D First and third line intersect at (0,1). This point also satisfies the second line for all real values of a. |
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| 310. |
Find the equation of the line, which makes intercepts `3` and 2 on the x and y axes respectively. |
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Answer» `xcostheta +4sintheta = p` let p=normal distance from origin `theta`=angle normal makes with +ve x direction given `P= 4` `theta=15^@` since we know that `cos2theta=2costheta^2-1` `take theta=15^@` `cos30^@=2cos^2 15-1` `sqrt 3/2+1=2cos^2 15` `cos15=sqrt((sqrt 3/4+2)` now `(sqrt 3+1)^2 =3+2sqrt 3+1` `=sqrt((sqrt 3 +1)^2/2 -2+2)/4 ` `cos15^@=sqrt((sqrt3+1)^2)/8` `=(sqrt3 +1)/(2sqrt2)` `sin15=sqrt1-cos^2 15` `=sqrt(1-(sqrt3+2)/4)` `=sqrt(4-(sqrt3+))/4` `sin15=sqrt(2-15)/4` `(sqrt3-1)^2=1+3-2sqrt3` `-sqrt3=((sqrt3-1)^2-4)/2` `sin15=sqrt(2+((sqrt3-1)^2/2)-2)/4` `sin15=sqrt(sqrt3-1)^2)/8` `=(sqrt3-1)/(2sqrt2)` `xcostheta+4sintheta=P` `x((sqrt3+1)/(2sqrt2))+4((sqrt3-1)/(2sqrt2)) =4` `x((sqrt3+1))+4((sqrt3-1))=8sqrt2` |
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| 311. |
The equation of the bisector of that angle between the lines x + y = 3 and 2x - y = 2 which contains the point (1,1) isA. `(sqrt5 - 2sqrt2) x + (sqrt5 + sqrt2) y - 3sqrt5 + 2 sqrt2 = 0`B. `(sqrt5 + 2sqrt2) x + (sqrt5 - sqrt2)y - 3sqrt5 - 2 sqrt2 = 0`C. 3x = 10D. none of these |
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Answer» Correct Answer - A First we re-write the equations of the two lines in such a way that the values of the expressions on the left hand sides of the equality for x = 1 , y = 1 become positive . Re-writing the given equations , we obtain -x-y + 3 = 0 and - 2x + y + 2 = 0. Now , we obtain the bisector of the angle containing point (1,1) for positive sign. The required bisector is given by `(-x-y + 3)/(sqrt((-1)^(2) + (-1)^(2))) = + (-2 x + y + 2)/(sqrt(-2)^(2) + 1^(2))` `implies (sqrt5 - 2 sqrt2 ) x + (sqrt5 + sqrt2)y - 3sqrt5 + 2sqrt2 = 0` . |
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| 312. |
The range of values of `theta` in the interval `(0, pi)` such that the points (3,5) and `(sin theta , cos theta)` lie on the same side of the line `x+y-1=0`, isA. `0 lt theta lt (pi)/(4)`B. `0 lt theta lt (pi)/(2)`C. `0 lt theta lt pi`D. `(pi)/(4) lt theta lt (3pi)/(4)` |
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Answer» Correct Answer - B We have line x+y-1=0. Let L(x,y) = x+y-1. `therefore L(3,5) = 3+5-1=7 gt 0` For points (3,5) and `"sin" theta, "cos" theta)` to lie on the same side line x+y-1=0, we must have `L("cos" theta, "sin" theta) gt 0` `"or " "sin" theta + "cos" theta-1 gt 0` `"sin" ((pi)/(4)+ theta) gt (1)/(sqrt(2))` `(pi)/(4) lt (pi)/(4) + theta lt (3pi)/(4)` `0 lt theta lt (pi)/(2)` |
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| 313. |
Find the set of positive values of `b`for which the origin and the point (1, 1) lie on the same side of thestraight line, `a^2x+ab y+1=0,AAa in Rdot` |
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Answer» Points (0,0) and (1,1) lie on same side of the line `a^(2) x +aby + 1= 0.` From (0,0). `a^(2)(0) + ab(0) + 1 gt 0` Therefore, for (1,1), we must have `a^(2) +ab + 1 gt 0 " " AA a in R`. `therefore D lt 0` `therefore b^(2)-4 lt 0` ` " or " b in (-2,2)` ` "But " b gt 0` `therefore b in (0,2)` |
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| 314. |
The range of values of `theta` in the interval `(0, pi)` such that the points (3,5) and `(sin theta , cos theta)` lie on the same side of the line `x+y-1=0`, isA. `(0, pi//2)`B. `(0,pi//4)`C. `(pi//4, pi//2)`D. none of these |
| Answer» Correct Answer - A | |
| 315. |
Two vertices of a triangle are `(5,-1)` and `(-2, 3)`. If the orthocentre of the triangle is the origin, then the third vertex isA. `(4,7)`B. `-4,-7)`C. `(2,-3)`D. `(5,-1)` |
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Answer» Correct Answer - B Let O (0,0) be the orthocentre , `A(x_(1),y_(1))` be the third vertex, and B(-2,b) and C (5,-1) be other two vertices of the given triangle . Then, `OA bot Bc rArr (y_(1)-0)/(x_(1)-0) xx (-1-3)/(5+2) = - 1 rArr 7 x_(1)= 4y_(1)" ".....(i)` and, `OB bot AC rArr (3)/(-2)xx(-1-y_(1))/(5-x_(1)) = - 1 rArr 2x_(1) - 3y_(1)= 13" ".....(ii)` Sloving (i)and (ii), we get `x_(1) = - 4, y_(1) = - 7` Thus, the orthocentre is (-4,-7) |
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| 316. |
Find the equation of the perpendicular bisector of the line segmentjoining the points `A(2,3)`and `B(6,-5)dot` |
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Answer» The slope of AB is given by `m = (-5-3)/(6-2) = -2` Therefore, the slope of the line perpendicular to AB is ` -(1)/(m) = (1)/(2)` Let P be the midpoint of AB. Then, the coordinates of P are `((2+6)/(2) , (3-5)/(2))`, i.e., (4,-1) Thus, the required line passes through P(4,-1) and has slope 1/2. So, its equation is `y+1 = (1)/(2)(x-4) " " [("Using " y-y_(1) = m(x_(1)-x_(1))]` or x-2y-6 = 0 |
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| 317. |
The equation of perpendicular bisector of the line segment joining the point (1,2) and (-2,0) is: |
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Answer» Let `AB` is the line segment joining the point `(1,2) and (-2,0)` and `C` is the midpoint og `AB`. The, coordinates of `C` will be `(-1/2,1)`. Slope of `AB, m_(AB) = (0-2)/(-2-1) = 2/3` Now, let `CL` is the perpendicular bisector of `AB`. Then, `m_(CL)**m_(AB) = -1` (Here, `m_(CL)` is slope of `CL`) `m_(CL)**2/3 = -1=> m_(CL) = -3/2` Now, we know the slope and coordinates of point C. So, the equation will be, `y-1 = -3/2(x+1/2) = > 4y-4 = -6x -3` `=>6x+4y-1 = 0` |
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| 318. |
Find the distance of the point `(3, - 5)`from the line `3x - 4y - 26 =0`. |
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Answer» the point given is `P(3,-5)` line given is `3x - 9y-26=0` as we know , `D= |(Ax_1+By_1+C)/(sqrt(A^2+B^2))|` `D= |(3*3+(_4)(-5) -26)/sqrt(3^2 + (-4)^2)|` `|(9+20-26)/(sqrt(9+16))|` `D= |3/5|` `=3/5` answer |
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| 319. |
If the area of the parallelogram formed by the lines 2x - 3y + a = 0 , 3x - 2y - a = 0 , 2x - 3y + 3a = 0 and 3x - 2y - 2a = 0 is 10 square units , then a =A. `pm1`B. `pm` 10C. `pm` 5D. none of these |
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Answer» Correct Answer - C We know that the area of the parallelogram formed by the lines `a_(1)x + b_(1)y + c_(1) = 0 , a_(2)x + b_(2)y+ c_(2) = 0 , a_(1) x + b_(1) y + d_(1) = 0` and `a_(2)x + b_(2)y + d_(2) = 0 ,` is `|((c_(1) - d_(1)) (c_(2) - d_(2)))/ ({:(a_(1), b_(1)), (a_(2) , b_(2)):}) |` `therefore |((3a-a) xx(-2a + a))/ ({:(2, -3), (3 , -2):}) | = 10 implies (2a^(2))/(5) = 10 implies a pm 15` |
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| 320. |
If the lines `x+2ay+a=0, x+3by+b=0` and `x+4cy+c=0` are concurrent then `a,b,c` are inA. A.P.B. G.P.C. H.P.D. none of these |
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Answer» Correct Answer - C Given lines will be concurrent , if `|{:(1,2a,a),(1,3b,b),(1,4c,c):}|=0` `implies -bc+2ac-ab=0 implies b =(2ac)/(a+c) implies a,b,c` are in H.P. |
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| 321. |
The equation of the line AB is `y = x`. If A and B lie on the same side of the line mirror `2x-y = 1`, then the equation of the image of AB is(a) `x+y-2=0`(b) `8x+y-9=0`(c) `7x-y-6=0`(d) `None of theseA. x + y - 2 = 0B. 8x + y - 9 = 0C. 7x - y - 6 = 0D. none of these |
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Answer» The required line is the line passing through the intersection of two lines and the image of a point on the line y = x in the line mirror 2x - y =1 Given lines intersect at (1,1) The image of (0,0) in the mirror 2x - y = 1 is given by `(x - 0)/(2) = (y-0)/(-1) -2 ((2 xx 0 -0-0 -1)/(4+ 1))` `implies x = (4)/(5) , y = (-2)/(5)` Thus , the required line passing through (1,1) and `((4)/(5) , (-2)/(5))` So , its equation is `y - 1 = ((-2)/(5) - 1)/((4)/(5) - 1) (x - 1)` `implies y - 1 = 7 (x -1) implies 7x - y - 6 = 0` |
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| 322. |
A line cutting off intercept `-3` from the `Y-`axis and the tangent at angle to the `X-`axis is `(3)/(5)`, its equation isA. `5y-3x+15=0`B. `3y-5x+15=0`C. `5y-3x-15=0`D. None of the above |
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Answer» Correct Answer - A Given that, c=-3and `m=(3)/(5)` `therefore` Equation of the line is `y=mx+c` `y=(3)/(5)x-3` `rArr5y=3x-15` `rArr5y-3x+15=0` |
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| 323. |
Two vertices of a triangle are (3,-2) and (-2,3) and its orthocentre is (-6,1) . Then the third vertex of this triangle can not lie on the lineA. 6 x + y = 0B. 4x + y = 2C. 5x + y = 2D. 3x + y = 3 |
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Answer» Correct Answer - C Let A(h,k) be the third vertex of `Delta`ABC whose two vertices are B(3,-2) and C(-2,3) and orthocentre at P(-6,1) . Then AP is perpendicular to BC and BP is perpendicular to AC . `therefore (k-1)/(h + 6) xx (3 +2 )/(-2 -3) = -1 ` and `(1 + 2)/(-6-3) xx (k-3)/(h + 2) = -1` `implies h - k + 7 = 0` and `3h - k + 9 = 0` `implies h = -1 , k = 6` Clearly , these values of h and k do not satisfy 5x + y = 2 . |
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| 324. |
If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points i. A(1, 3) ii. B(2,5) |
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Answer» Origin is shifted to (2, 3) = (h, k) Let the new co-ordinates be (X, Y). x = X + handy = Y + k x = X + 2 and y = Y + 3 …(i) i. Given, A(x, y) = A( 1, 3) x = X + 2 andy = Y + 3 …[From(i)] ∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0 ∴ The new co-ordinates of point A are (- 1,0). ii. Given, B(x, y) = B(2, 5) x = X + 2 and y = Y + 3 …[From(i)] ∴ 2 = X + 2 and 5 = Y + 3 ∴ X = 0 and Y = 2 ∴ The new co-ordinates of point B are (0, 2). |
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| 325. |
The equation of straight line equally inclined to the axes and equidistant from the point `(1, -2)` and `(3,4)` is:A. x + y + 1 = 0B. x + y + 2 = 0C. x - y - 2 = 0D. x -y - 1 = 0 |
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Answer» Correct Answer - D The slope of a line equally inclined with the co-ordinate axes is `pm` 1 . So , let its equation be `x pm y + lambda = 0 " " … (i)` CASEI When ` x + y + lambda = 0 ` is the line : It is equidistant from (1,-2) and (3,4) `therefore |(1-2 + lambda)/(sqrt2)| = |(3 + 4 + lambda)/(sqrt2)|` `implies | lambda -1| = |lambda + 7| implies lambda - 1 = - (lambda + 7) implies lambda = -3` So , the required line is x + y - 3 = 0. CASEII When ` x - y + lambda = 0` is the line : It is equidistant from (1, -2) and (3,4) `therefore |(1+2 + lambda)/(sqrt2)| = |(3 - 4 + lambda)/(sqrt2)|` `implies } lambda + 3| = |lambda - 1| implies lambda + 3 = - (lambda -1) implies lambda = -1` Thus , the required line is x - y - 1 = 0 |
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| 326. |
Find the value of `lambda`, if the line `3x-4y-13=0,8x-11 y-33=0a n d2x-3y+lambda=0`are concurrent. |
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Answer» The given lines are concurrent if `|{:(3,-4,13),(8,-11,-33),(2,-3,lambda):}| = 0` `" or " 3(-11lambda-99) +4(8lambda+66)-13(-24+22)=0` ` " or " -lambda-7=0` `" or " lambda=-7` Alternative method: The given equations are `3x-4y-13=0 " " (1)` `8x-11y-33=0 " " (2)` `" and " 2x-3y+lambda=0 " " (3)` Solving (1) and (2), we get x =11 and y=5 Thus, (11,5) is the point of intersection of (1) and (2). The given lines will be concurrent if they pass through a common point, i.e., the point of intersection of any two lines lies on the third. Therefore, (11,5) lies on (3), i.e., `2 xx 11-3 xx 5+lambda=0` `" or " lambda=-7` |
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| 327. |
Find the value of `lambda`, if the line `3x-4y-13=0,8x-11 y-33=0a n d2x-3y+lambda=0`are concurrent.A. 7B. 6C. `-7`D. 4 |
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Answer» Correct Answer - C The given lines are concurrent , if `|{:(3,-14,-13),(8,-11,-33),(2,-3,lambda):}|=0` `implies 3(-11lambda-99)+4(8lambda+66)-13(-24+22)=0` `implies -lambda-7=0 implies lambda=-7` |
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| 328. |
Line joining the points (3,-4) and (-2,6) is perpendicular to the line joining the points (-3,6) and (9,-18). |
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Answer» Correct Answer - False Given points are A(3,-4), B(-2,6), P(-3,6) and Q(9,-18) Now, slope of AB `=(6+4)/(-2-3)=-2` and slope of PQ `=(-18-6)/(9+3)=-2` So, line AB is parallel to line PQ. |
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| 329. |
The point `A (2, 1)` is shifted by `3sqrt2` unit distance parallel to the line `x + y = 1` in the direction of increasing ordinate to reach a point B. Find the image of B by the line `x+y= 1`.A. (5,-2)B. (-3,2)C. (5,4)D. (-1,4) |
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Answer» The equation of a line passing through (2,1) and parallel to x + y = 1 is `(x - 2)/(cos 3pi //4) = (y-1)/(sin pi//4) or , (x -1)/(-1 //sqrt2) = (y-1)/(1//sqrt2)` The coordinates of points on this line which are at a distance `3sqrt2` from (2,1) are given by `(x-2)/(-1//sqrt2) = (y-1)/(1//sqrt2) = pm 3 //sqrt2 implies x = 2 pm 3 , y = 1 pm 3` But the ordinate of Q is more than that of P (2,1) . So , the coordinates of Q are (-1,4). The image of Q(-1,4) in the line x + y = 1 is given by `(x +1)/(1) = (y-4)/(1) = -2((-1 + 4 -1)/(1^(2) + 1^(2)))` `implies (x +1)/(1) = (y-4)/(1) = -2 implies x = -3 , y =2` Thus , the coordinates of the required point are (-3,2) |
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| 330. |
Point `P(2, 4)` is translated through a distance `3sqrt2` units measured parallel to the line `y-x-1 = 0` in the directionof decreasing ordinates to reach at Q. If R is the imageof Q with respect to the line `y-x-1 = 0`, then thecoordinates of R are given by(1) `(5,7)`(2) `(-1,1)`(3) `(6,6)`(4) `(0,0)`A. (-1,1)B. (5,7)C. (6,6)D. (0,0) |
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Answer» The equation of a line through P( 2, 4) and parallel to y - x - 1 = 0 is `(x -2)/(cos pi //4) = (y-4)/(sin pi//4)` The coordinates of Q are given by gt `(x -2)/(cos pi //4) = (y-4)/(sin pi//4) = -3sqrt2 implies x = -1 , y = 1` Thus , the coordinates of Q are (-1,1). R is the image of Q with respect to the line y - x - 1 = 0 . Therefore , coordinate of R is given by `( x + 1)/(-1) = (y-1)/(1) = -2 ((1 + 1+ -1))/((-1)^(2) + (1)^(2))` `implies (x + 1)/(-1) = (y-1)/(1) = -1 implies x = 0 , y = 0 ` Hence , the coordinates of R are (0,0) . |
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| 331. |
If the lines `a x+2y+1=0,b x+3y+1=0a n dc x+4y+1=0`are concurrent, then `a ,b ,c`are ina. A.P. b. G.P. c. H.P. d. none of these |
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Answer» Correct Answer - False Given lines are `ax+2y+1=0` `bx+3y+1=0` From Eq. (i), on putting `y=(-ax-1)/(2)` in Eq. (ii), we get `bx-(3)/(2)(ax+1)+1=0` `rArr 2bx-3ax-3+2=0` `rArr x(2b-3a)=1rArrx=(1)/(2b-3a)` Now, using `x=(1)/(2b-3a)` in Eq. (i), we get `(a)/(2b-3a)+2y+1=0` `rArr 2y=-[(a+2b-3a)/(2b-3a)]` `rArr 2y=(-(-2b-2a))/(2b-3a)` `rArr y=((a-b))/(2b-3a)` So, the point of intersection is `((1)/(2b-3a),(a-b)/(2b-3a))` Since, this point lies on `cx+4y+1=0`, then `(c)/(2b-3a)+(4(a-b))/(2b-3a)+1=0` `rArr c+4a-4b+2b-3a=0` `rArr -2b+a+c=0rArr2b=a+c` Hence, the given statement is false. |
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| 332. |
If one vertex of an equilateral triangle is at `(2.-1)` 1base is `x + y -2 = 0`, then the length of each side, isA. `sqrt(3//2)`B. `sqrt(2//3)`C. `2//3`D. `3//2` |
| Answer» Correct Answer - B | |
| 333. |
Statement - 1 : Equations `(2 pm sqrt3) x - y = 1 pm 2sqrt3` represent two sides of an equilateral triangle having one vertex (2 , 3) and x + y - 2 = 0 as the opposite side . Statement - 2 : The equation of the lines passing through `(x_(1) , y_(1))` and making constant angle `alpha` with the line y = mx + c are given by `y - y_(1) = (m pm tan alpha)/( 1 pm tan alpha) ( x - x_(1))`A. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 10B. Statement-1 is True , Statement-2 is True , Statement -2 is not a correct explanation for Statement - 1 .C. Statement-1 is True , Statement - 2 is False .D. Statement - 1 is False , Statement -2 is True . |
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Answer» Correct Answer - A Statement- 2 is true . Using statement - 2 , the equations of the sides are `y - 3 = (-1 pm tan 60^(@))/(1 pm tan 60^(@)) (x -2 ) or , (2 pm sqrt3 ) x - y = 1 pm 2 sqrt3` So , statement -1 is also true . Also , statement - 2 is a correct explanation for statement - 1. |
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| 334. |
The locus of a point that is equidistant from the lines `x+y - 2sqrt2 = 0` and `x + y - sqrt2 = 0` is(a) `x+y-5sqrt2=0`(b) `x+y-3sqrt2=0`(c) `2x+2y-3 sqrt2=0`(d) `2x+2y-5sqrt5=0`A. `x+y-5sqrt(2)=0`B. `x+y-3sqrt(2)=0`C. `2x+2y-3sqrt(2) = 0`D. `2x+2y-5sqrt(2) = 0` |
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Answer» Correct Answer - C For any point P (x,y) that is equidistant from the given line, we have `x+y-sqrt(2) = -(x+y-2sqrt(2))` `"or " 2x+2y-3sqrt(2) =0` |
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| 335. |
If the points `(a,a^(2)) and (1,2)` lie in the same angular region between the lines `3x+4y-1=0` and `2x+y-3=0` , thenA. `a lt -3` or , `a gt 1`B. `a in [-3,1]`C. `a lt (1)/(4)` or , ` a gt -1`D. none of these |
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Answer» Correct Answer - A Given points will lie in the same angular region between the lines `3x+4y-1=0 and 2x+y-3=0` , if they lie on the same side of the two lines. `therefore (3a +4a^(2)-1)(3+8-1) gt 0 and (2a + a^(2)-3)(2+2-3) gt 0` `implies 4a^(2)+ 3a -1 gt 0 and a^(2)+ 2a-3 gt 0` `implies (4a-1)(a+1) gt 0 and (a+3)(a-1) gt 0` `implies (a lt -1 ` or , `a gt (1)/(4)) and (a lt -3 ` or , `a gt 1) implies a lt -3 ` or, `a gt 1` |
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| 336. |
Two sides of a square lie on the lines `x+y=1a n dx+y+2=0.`What is its area? |
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Answer» Clearly, the length of the side of the square is equal to the distance between the parallel lines. `x+y-1=0 " " (1)` `" and " x+y+2=0 " " (2)` Hence, side length `=(|2-(-1)|)/(sqrt((1+1))) = (3)/(sqrt(2))` Therefore, area of square `=(9)/(2)` sq. units. |
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| 337. |
The number of lines that are parallel to `2x + 6y -7= 0` and have an intercept `10` between the coordinate axes isA. 1B. 2C. 4D. infinitely many |
| Answer» Correct Answer - B | |
| 338. |
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope. |
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Answer» Given equation is 6x + 3y + 8 = 0, which can be written as 3y = – 6x – 8 y = -6x/3 - 8/3 y = -2x - 8/3 This is of the form y = mx + c with m = -2 y = -2x – 8/3 is in slope-intercept form with slope = -2 |
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| 339. |
If lines `p x+q y+r=0,q x+r y+p=0a n dr x+p y+q=0`areconcurrent, then prove that `p+q+r=0(w h e r ep ,q ,r`are distinct`)dot`A. `p+q+r=0`B. `p^(2)+q^(2)+r^(2)=pq+qr+rp`C. `p^(3)+q^(3)+r^(3)=3pqr`D. none of these |
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Answer» Correct Answer - C The lines `px+qy+r=0, qx+ry+p=0` and `rx+py+q=0` are concurrent, if `|{:(p,q,r),(q,r,p),(r,p,q):}|=0` `implies (p+q+r)^(2)(p^(2)+q^(2)+r^(2)-pq-qr-rp)=0` `implies p^(3)+q^(3)+r^(3)-3pqr=0implies p^(3)+q^(3)=3pqr` |
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| 340. |
Condition when variable is involved :(ii) Find the locus of point of intersection of the lines `xcosalpha+ysinalpha=a` and `xsinalpha-ycos alpha=b` where b is the variable.A. `x^(2) + y^(2) = a^(2) - b^(2)`B. `x^(2) - y^(2) =a^(20 - b^(2)`C. `x^(2) + y^(2) = a^(2) +b^(2)`D. none of these |
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Answer» Correct Answer - C Let P(h,k) be the point of intersection of the given lines. Then, `h cos alpha + k sin alpha = a" "(i)` and `h sin alpha - k cos alpha =b" "(ii)` Here `alpha` is a variable . So we have to eliminate `alpha`. Squaring and adding (i) and (ii) we, get `(h cos alpha + k sin alpha)^(2) + (h sin alpha -k cos alpha )^(2) = a^(2) + b^(2) ` `rArr h ^(2) + k^(2) = a^(2) + b^(2)` Hence , locus of (h,k) is `x ^(2) + y^(2) = a^(2) + b^(2)`. |
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| 341. |
Let `a x+b y+c=0`be a variable straight line, whre `a , ba n dc`are the 1st, 3rd, and 7th terms of an increasing AP, respectively.Then prove that the variable straight line always passes through a fixedpoint. Find that point. |
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Answer» Let the common difference of A.P.be d. Then b=a+2d and c=a+6d. So, given variable straight line will be ax+(a+2d)y+a+6d=0 or a(x+y+1) + d(2y+6)=0 This is the equation of family of straight lines concurrent at point of intersection of lines x+y+1=0 and 2y+6=0 which (2,-3). |
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| 342. |
Let `u-=a x+b y+a b3=0,v-=b x-a y+b a3=0,a ,b in R ,`be two straight lines. The equations of the bisectors of the angleformed by `k_1u-k_2v=0`and `k_1u+k_2v=0`, for nonzero and real `k_1`and `k_2`are`u=0`(b) `k_2u+k_1v=0``k_2u-k_1v=0`(d) `v=0`A. u=0B. `k_(2)u + k_(1)v = 0`C. `k_(2)u - k_(1)v = 0`D. v=0 |
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Answer» Correct Answer - A::D Note that the lines are perpendicular, Assume the coordinates axes to be directed along u=0 and v = 0, Now, the lines `k_(1) u-k_(2)v=0 " and " k_(1) u + k_(2)v =0` are equally inclined with the u-v axes. Hence, the bisectors are u=0 and v=0. |
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| 343. |
Find the equation of the bisectors of the angles between the coordinate axes. |
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Answer» Correct Answer - `x-y=0 or x+y=0` `m=tan45^(@)=1 or m=tan135^(@)=-1."And" c=0` |
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| 344. |
Line `a x+b y+p=0`makes angle `pi/4`with `cosalpha+ycosalpha+ysinalpha=p ,p in R^+`. If these lines and the line `xsinalpha-ycosalpha=0`are concurrent, then`a^2+b^2=1`(b) `a^2+b^2=2``2(a^2+b^2)=1`(d) none of theseA. `a^(2) +b^(2) = 1`B. `a^(2) +b^(2) = 2`C. `2(a^(2) +b^(2)) = 1`D. none of these |
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Answer» Correct Answer - B `"Lines "x"cos" alpha + y"sin"alpha =p " and "x"sin"alpha-y"cos"alpha=0` are mutually perpendicular. Thus, ax+by+p=0 will be equally inclined to these lines and would be the angle bisesctor of these lines. Now, the equations of angle bisectors are `x"sin" alpha - y"cos"alpha =+-(x"cos"alpha+y"sin"alpha-p` `"or " x("cos" alpha - "sin"alpha)+ y("sin"alpha+"cos"alpha)=p` `"or " x("sin" alpha + "cos"alpha)- y("cos"alpha-"sin"alpha)=p` Comparing these lines with ax+by+p=0, we get `(a)/("cos" alpha-"sin"alpha) = (b)/("sin" alpha+"cos"alpha)= 1` `rArr a^(2) + b^(2) = 2` `(a)/("sin" alpha+"cos"alpha) = (b)/("sin" alpha-"cos"alpha)= 1` `rArr a^(2) + b^(2) = 2` |
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| 345. |
Let `O`be the origin. If `A(1,0)a n dB(0,1)a n dP(x , y)`are points such that `x y >0a n dx+y |
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Answer» Correct Answer - A Since `xy gt 0,` P lies either in the first quadrant or in the third quadrant. The inequality `x+y lt 1` represents all the points below the line x+y=1 so that `xy gt 0` and `x+y lt 1 imply that P lies either inside the triangle OAB or in the third quadrant. |
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| 346. |
Let `O`be the origin. If `A(1,0)a n dB(0,1)a n dP(x , y)`are points such that `x y >0a n dx+y |
| Answer» Correct Answer - A | |
| 347. |
In a triangle `A B C`, the bisectors of angles `Ba n dC`lies along the lines `x=ya n dy=0.`If `A`is `(1,2)`, then the equation of line `B C`is`2x+y=1`(b) `3x-y=5``x-2y=3`(d) `x+3y=1`A. 2x+y=1B. 3x-y=5C. x-2y=3D. x+3y=1 |
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Answer» Correct Answer - B The reflections of A in the two angle bisectors will lie on the line BC, So, (2,1) and (1,-2) will lie on BC. The equation of BC will be `y+2 = ((1+2)/(2-1))(x-1)` i.e., 3x-y=5 |
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| 348. |
In ` A B C`, the coordinates of the vertex `A`are `(4,-1)`, and lines `x-y-1=0`and `2x-y=3`are the internal bisectors of angles `Ba n dC`. Then, the radius of the encircle of triangle `A B C`is`4/(sqrt(5))`(b) `3/(sqrt(5))`(c) `6/(sqrt(5))`(d) `7/(sqrt(5))`A. `4//sqrt(5)`B. `3//sqrt(5)`C. `6//sqrt(5)`D. `7//sqrt(5)` |
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Answer» Correct Answer - C Incentre I is point of intersection of two angle bisectors which is (2,1). `IA = sqrt(2^(2) + 2^(2)) = 2 sqrt(2)` `"Also, AI" = r " cosec" (A)/(2)` `angleBIC = (pi)/(2) + (A)/(2)` `"or tan"((pi)/(2)+(A)/(2)) = |(1-2)/(1+2)|` `"or tan "(A)/(2) = 3` `"So, r" = (AI)/("cosec"(A)/(2)) = (2sqrt(2))/(sqrt(1+(1)/(9))) (2sqrt(2) xx 3)/(sqrt(10)) = (6)/(sqrt(5))` |
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| 349. |
In a triangle ABC , right angled at the vertex A , if the position vectors of A , B and C are respectively `3 hati + hatj-hatk , -hati + 3hatj + phatk` and `5 hati + q hatj - 4 hatk` , then the point (p,q) lies on a lineA. parallel to y-axisB. making an acute angle with the positive direction of x-axisC. parallel to x-axisD. making an obtuse angle with the positive direction of x axis |
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Answer» Correct Answer - B We find that `vecAB = - 4hati + 2 hatj + (p+1) hatk` and `vecAC = 2 hati + (q -1) hatj - 3hatk` . It is given that the triangle is right angled at A . `therefore vec(AB) * vec(AC) = 0 implies -8 + 2q -2- 3p - 3 = 0 implies 3p-2p + 13 = 0` Hence , (p,q) lies on the line 3x - 2y + 13 = 0 . This line makes an acute angle with the positive direction of x - axis . |
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| 350. |
A straight line `L`is perpendicular to the line `5x-y=1`. The area of the triangle formed by line `L ,`and the coordinate axes is 5. Find the equation of line `Ldot`A. `x + 5 y + 5 = 0`B. `x + 5y pm sqrt2 = 0 `C. `x + 5y pm sqrt5 = 0`D. `x + 5y pm 5sqrt2 = 0` |
| Answer» Correct Answer - D | |