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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.
| 201. |
Without using distance formula, show that points `(2, 1)`, `(4, 0)`, `(3, 3)`and `(3, 2)`are the vertices of a parallelogram. |
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Answer» Please refer to video for the figure.Here, slope of `AB = (0-(-1))/(4-(-2)) = 1/6` Slope of `BC = (3-0)/(3-4) = -3` Slope of `CD = (2-3)/(-3-3) = 1/6` Slope of `DA = (2-(-1))/(-3-(-2)) = -3` As, slope of `AB` = slope of `CD` and slope of `BC` = slope of `AD` It means, `AB||CD` and `BC||AD` So, ABCD is a parallelogram. |
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| 202. |
The triangle formed by the straight lines `x=-y`, `x+y=4` and `x+3y=4` is :A. isoscelesB. equilateralC. right-angledD. None of these |
| Answer» Correct Answer - A | |
| 203. |
Find the value of k so that the lines 3x-y-2=0, 5x+ky-3=0 and 2x+y-3=0 are concurrent. |
| Answer» Correct Answer - k=-2 | |
| 204. |
Find the area of triangle formed by the lines :` x+y-6=0, x-3y-2=0 a n d 5x-3y+2=0` |
| Answer» Correct Answer - 12 sq units | |
| 205. |
Show thatthe points `A(1, -2), B(3, 6), C(5, 10)`and `D(3, 2)`are thevertices of a parallelogram. |
| Answer» Show that AB=CD and BC=AD. | |
| 206. |
Consider two intersecting (non-perpendicular lines `L_(1)=0 and L_(2)=0` and a point `P_(1)`. Image of `P_(1)` and `L_(1)=0` is `P_(2)`, image of `P_(2)` in `L_(2)=0` is `P_(3)` , image of `P_(3)` in `L_(1)` =0 is `P_(4)` and so on. Which of the following statements are incorrect ?A. `vec(P_(3)P_(5))=vec(P_(4)P_(6))`B. `vec(P_(1)P_(4))=vec(P_(2)P_(3))`C. `P_(i)` are concylicD. `P_(5)` is `P_(14)` |
| Answer» Correct Answer - D | |
| 207. |
Find the angle between the lines whose slopes are `sqrt3 and 1/sqrt3.` |
| Answer» Correct Answer - `30^(@)` | |
| 208. |
If the sum of the slopes of the lines given by `4x^2+2lambdaxy-7y^2=4` is equal to the product of the slopes,then `lambda` is equal to? |
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Answer» Let we have below two lines that represent the given equation. `y-m_1x = 0` and `y-m_2x=0` Combined equation of these two lines will be, `(y-m_1x)(y-m_2x) = 0` `y^2-(m_1+m_2)xy+m_1m_2x^2=0->(1)` Now, the given equation is, `4x^2+2lambdaxy -7y^2= 4` `y^2-4/7x^2- 2/7lambdaxy = 0` Comparing this with the given equation (1), `m_1+m_2 = 2/7lambda` `m_1m_2 = -4/7` As we are given sum and product of roots are equal. So, `2/7lambda = -4/7=> lambda = -2` |
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| 209. |
Find the angle between the lines whose slopes are `(2- sqrt3) and (2+ sqrt3)`. |
| Answer» Correct Answer - `60^(@)` | |
| 210. |
If a point `P(x, y)` is equidistant from the points `A(6. - 1)` and `B(2,3)` relation between `x` and `y`. |
| Answer» Correct Answer - x-y=3 | |
| 211. |
Find the distance between `P(x_1, y_1)a n d Q(x_2, y_2)`when i. `P Q`is parallelto the y-axis ii. PQ is parallel to the x-axis. |
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Answer» Correct Answer - (i) `|x_(2)-x_(1)|` (ii) `|y_(2)-y_(1)|` (i) When AB is a parallel to the x-axis, then A and B have the same ordinate. Then, `A(x_(1),y) and B(x_(2),y)` are the given points. So, AB`=|x_(2)-x_(1)|` (ii) When AB is parallel to the y-axis. Then A and B have the same abscissa. Then, `A(x,y_(1)) and B(x,y_(2))` are the given points. So, AB=`|y_(2)-y_(1)|` |
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| 212. |
The equation of straight line passing through point (1,2) and having intercept of length 3 between straight line 3x+4y=12 and 3x+4y=24 isA. 7x + 24y - 55 = 0B. 24x + 7y - 38 = 0C. 24x - 7y - 10 = 0D. 7x - 24y + 41 = 0 |
| Answer» Correct Answer - D | |
| 213. |
Find the image of the point P(1, 2) in the line x-3y+4=0 |
| Answer» `((6)/(5),(7)/(5))` | |
| 214. |
The points (-a, -b), (0, 0), (a, b) and (`a^2`,ab) are (a)Collinear (b) vertices of a reotangle (c) Vertices of an parallelogram (d) None of these |
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Answer» `y=mx,m=b/a` `y=b/a*x` `(-a,-b)=-b=b/a(-a)` `(a,b)=b=b/a(a)` `(a^2,ab)=ab=b/a(a^2)` its a Collinear. |
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| 215. |
The locus of a point P which divides the line joining `(1,0) and (2 costheta, 2 sintheta)` intermally in the ratio `1:2` for all `theta`. isA. straight lineB. circleC. pair of straight linesD. parabola |
| Answer» Correct Answer - B | |
| 216. |
The equation to the straight line passing through the point `(a "cos"^(3) theta, a "sin"^(3) theta)` and perpendicular to the line `x "sec" theta + y"cosec" theta = a` isA. `x "cos " theta-y "sin" theta = a "cos" 2theta`B. `x "cos " theta+y "sin" theta = a "cos" 2theta`C. `x "sin" theta+y "cos" theta = a "cos" 2theta`D. none of these |
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Answer» Correct Answer - A The line perpendicular to `x "sec" theta + y "cosec" theta = a` is `x "cosec" theta-y "sec"theta = lambda` This line passes through the point `(a "cos"^(3) theta, a "sin"^(3) theta)`. Then, `(a "cos"^(3) theta)"cosec" theta -(a "sin"^(3)theta) "sec" theta=lambda` `"or " lambda = a(( "cos"^(3) theta)/("sin"theta) - ("sin"^(3) theta)/("cos"theta))` `= a(( "cos"2 theta)/("cos"theta "sin"theta))` |
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| 217. |
Difference of slopes of the lines represented by the equation `x^2(sec^2 theta - sin ^2 theta) -2xytan theta + y^2 sin^2 theta=0` is(A) `4`(B) `3`(C) `2`(D) None of these |
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Answer» Let we have below two lines that represent the given equation. `y-m_1x = 0` and `y-m_2x=0` Combined equation of these two lines will be, `(y-m_1x)(y-m_2x) = 0` `y^2-(m_1+m_2)xy+m_1m_2x^2=0->(1)` Now, the given equation is, `x^2(sec^2theta -sin^2theta)-2xytantheta+y^2sin^2theta = 0` `=> x^2((sec^2theta -sin^2theta)/sin^2theta)-2tantheta/sin^2thetaxy+y^2=0` `=>y^2-2secthetacosecthetaxy+(sec^2thetacosec^2theta-1)x^2` Comparing this with the given equation (1), `m_1+m_2 = 2secthetacosectheta` `m_1m_2 = sec^2thetacosec^2theta-1` `(m_1-m_2)^2 = (m_1+m_2)^2-4m_1m_2` `=>(m_1-m_2)^2 = 4sec^2thetacosec^2theta-4sec^2thetacosec^2theta+4` `=>(m_1-m_2)^2 = 4` `|m_1-m_2| = 2` |
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| 218. |
If `p_(1) and p_(2)` are the lengths of the perpendicular form the orgin to the line `x sec theta+y cosec theta=a and xcostheta-y sin theta=a cos 2 theta` respectively then prove that `4p_(1)^(2)+p_(2)^(2)=a^(2)` |
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Answer» We have `x sectheta+y "cosec" theta =a Rightarrow (x)/(cos theta)+(y)/(sin theta)-a=0.......(i)` Now, `p_(1)` is the length of perpendicular from the origin to line (i), so we have `p_(1)=(|(1)/(cos theta)xx0+(1)/(sin theta)xx0-a|)/(sqrt((1)/(cos^(2)theta)+(1)/(sin^(2)theta)))=|a|sin theta cos theta=(|a|)/(2)sin 2theta` `Rightarrow 2p_(1)=|a|sin 2theta` `Rightarrow 4p_(1)^(2)=a^(2)sin^(2) 2theta........(ii)` The other line is `x cos theta-y sin theta-a cos 2theta=0`.......(ii) Now, `p_(2)` is the length of perpendicular from the origin to line (ii), so we have `P_(2)=(|0xx cos theata-0xxsin theta-acos2 theta|)/(sqrt(cos^(2)theta+sin^(2)theta))=|a cos 2theta|` `Rightarrow p_(2)^(2)=a^(2) cos^(2)2 theta...(iv)` Adding (ii) and (iv), we get `4p_(1)^(2)+p_(2)^(2)=a^(2)(sin^(2)2theta+cos^(2)2theta)` `Rightarrow 4p_(1)^(2)+p_(2)^(2)=a^(2)` |
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| 219. |
Three vertices of a quadrilateral in order are `(6,1)(7,2)" and "(-1,0)`. If the area of the quadrilateral is 4 sq. unit. Then the locus of the fourth vertex has the equation.A. x-7 y=1B. x-7 y+15=0C. x=7y+15=0D. `(x-7y)^(2)+14(x-7y)-15=0` |
| Answer» Correct Answer - C | |
| 220. |
Find apoint on the x-axis which is equidistant from the points `(7, 6)`and `(-3, 4)`. |
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Answer» Correct Answer - P(3,0) Let the required point be P(x,0) Then, `AP=BP Leftrightarrow AP^(2)=BP^(2)` |
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| 221. |
Find the equation of the line passing through the point (2, 3) and perpendicular to the line 4x+3y=10. |
| Answer» Correct Answer - `3x-4y+6=0` | |
| 222. |
The diagonals of a parallelogram PQRS are along the lines x+3y =4 and 6x-2y = 7, Then PQRS must be : |
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Answer» `L_1:x+3y=4` `m_(L1)=-1/3` `L_2:6x-2y=7` `m_(L2)=3` `m_(L1)*m_(L2)` `-1/3*3=-1`. |
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| 223. |
Family of lines `x sec^(2) theta + y tan^(2)theta -2=0` for different real `theta`, isA. not concurrentB. concurrent at (1,1)C. concurrent at (2,-2)D. concurrent at (-2,2) |
| Answer» Correct Answer - C | |
| 224. |
· 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 | |
| 225. |
Find the equation of the line through the intersection of the lines x-7y+5=0 and 3x+y-7=0 and which is parallel to x-axis. |
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Answer» Correct Answer - y=1 When a line is parallel to x-axis then coefficient of x should be 0. |
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| 226. |
Find the equationof the line through the intersection of the lines `2x+" "3y " "4" "=" "0`and `x " "5y=" "7`that has its x-intercept equal to `" "4`. |
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Answer» The given line are 2x+3y-4=0 and x-5y+7=0. The equation of any line through the intersection of given line is of the form `(2x+3y-4)+k(x-5y+7)=0` `Rightarrow (2+k)x+(3-5k)y+(7k-4)=0.....(i)` If this line has x-intercept-4, then the point (-4,0) lies on (i), `therefore (2+k)(-4)+(7k+4)=0 Rightarrow -8-4k+7k-4=0` `Rightarrow 3k=12 Rightarrow k=4.` Substituting k=4 in (i), we get 6x-17y+24=0, which is the required equation. |
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| 227. |
Find the equation of the line parallel to the y-axis and drawn through the point of intersection of the lines `x-7y+15=0 and 2x+y=0`. |
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Answer» The given intersecting line are: `x-7y+15=0.....(i)` `2x+y=0.....(ii)` On solving (i) and (ii), we get x=-1 and y=2. Thus, the given lines intersect at the point P(-1,2) The line parallel to the y-axis and drawn through P(-1,2) is given by x=-1, i.e., x+1=0 |
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| 228. |
The line `y=(3x)/4`meets the lines `x-y=0`and `2x-y=0`at points `Aa n dB`, respectively. If `P`on the line `y=(3x)/4`satisfies the condition `P AdotP B=25 ,`then the number of possible coordinates of `P`is____ |
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Answer» Correct Answer - 3 Point P which lies on the line y=3x/4 can be chosen as P(h, 3h/4). If `theta` is the angle that the line y=3x/4 makes with the positive direction of the x-axis, then `"tan " theta = (3)/(4) " or cos " theta =(4)/(5) " and sin" theta = (3)/(5)` Now, the coordinates of points A and B which lie on the line y=3x/4 can be chosen as `A-=(h+(4r_(1))/(5), (3h)/(4) + (3r_(1))/(5)) " and "B-=(h+(4r_(2))/(5), (3h)/(4) + (3r_(2))/(5))` Since A lies on the line x-y+1 =0, we have `(h+(4r_(1))/(5)) -((3h)/(4) + (3r_(1))/(5)) +1 =0` `"or " r_(1) = (-5)/(4)(h+4)` Since B lies on the line 2x-y-5=0, we have `2(h+(4r_(2))/(5))- ((3h)/(4) + (3r_(2))/(5))-5=0` `"or " r_(2) = (-5)/(4)(h-4)` According to the given condition, we have `PA * PB =25` `i.e., |r_(1)| * |r_(2)| = 25` `i.e., (25)/(16)(h^(2) -16) = +- 25` `i.e., h^(2) = 16+-16 = 32, 0` `i.e., h=+-4sqrt(2), 0` Hence, the required points are (0, 0), `(4sqrt(2), 3sqrt(2)), " and " (-4sqrt(2), -3sqrt(2)).` |
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| 229. |
Find the equation of the line drawn through the point interseciton of the line 4x-3y+7=0 and 2x+3y+5=0 and passing through the point (-4,5). |
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Answer» The equation of any line through the point of intersection of the given line is of the form. `(4x-3y+7)+k(2x+4y+5)=0` `Rightarrow (4+2k)x+(3k-3)y+(5k+7)=0......(i)` If it passes through the point (-4,5), we have `(4+2k)(-4)+(3k-3).5+(5k+7)=0` `Rightarrow -16-8k+15k-15+5k+7=0 Rightarrow 12k=24 Rightarrow k=2` Substituting k=2 in (i), we get 8x+3y+17=0, which is the required equation |
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| 230. |
Find the equation of the line through the intersection of the lines `2x+ 3y 4 = 0` and `x - 5y= 7`that has its x-intercept equal to ` 4`. |
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Answer» Equation of the required line can be given as `2x+3y-4+k(x-5y-7) = 0` `(2+k)x+(3-5k)y - (4+7k) = 0->(1)` As, x- intercept is given, we can put `y = 0` `:.(2+k)x = 4+7k` `=> (2+k)(-4) = 4+7k` `=>-8-4k = 4+7k=>k = -12/11` So, putting value of `k` in (1), we get the required equation. `(2-12/11)x+(3+60/11)-(4-84/11) = 0` `=>10x+93y+40 = 0`, which is the required equation. |
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| 231. |
Find the equation of line parallel to the y-axis and drawn through the point of intersection of `x 7y+ 5 = 0`and `3x+y 7 = 0`. |
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Answer» The equation of any line through the point of intersection of the given line is of the form `(x-7y+5)+k(3x+y-7)=0` `Rightarrow (1+3k)x+(k-7)y+(5-7k)=0........(i)` If the line is parallel to y-axis then coefficient of y should be 0,i.e. k-7=0, which gives k=7. Substituting k=7 in (i), we get 22x-44=0 `therefore` x-2=0, which is the required equation. |
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| 232. |
Find the equation of the line through the intersection of lines `3x +4y=7` and `x-y+2= 0` and whose slope is 5. |
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Answer» The given lines are 3x+4y-7=0 and x-y+2=0 The equation of any line through the point of intersection of the given line is of the form, `(3x+4y-7)+k(x-y+2)=0........(i)` `Rightarrow (3+k)x+4(4-k)y+(2k-7)=0` `Rightarrow (4-k)y-(3+k)y+(9-2k)` `Rightarrow y=(-(3+k))/(4-k)x+((7-2k))/((4-k))` `Rightarrow y=((k+3))/((k-4))x+((7-2k))/((4-2k)` `"Slope of this line is"((k+3))/((k-4))` `therefore (k+3)/(k-4)=5 Rightarrow k+3=5k-20 Rightarrow 4=23 Rightarrow (23)/(4)` `"Substituting k"=(23)/(4)"in (i), we get "` `(3x+4y-7)+(23)/(4)(x-y+2)=0` `Rightarrow 4(3x+4y-7)+23(x-y+2)=0` `Rightarrow 25x-7y+18=0,` which is the required equation. |
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| 233. |
Find the equation of the line through the intersection of `5x-3y=1`and `2x-3y -23= 0`and perpendicular to the line `5x - 3y -1 =0`. |
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Answer» Let `m1` be the slope of required line. `m2=5/3`(slope of given line) As, `m1*(m2)=-1` so, `m1=-3/5`.On Solving given two equations, `5x-3y-1=0, 2x+3y-23=0` We get,`(x,y)=(24/7, 113/7)`. Equation of required line,`(y-113/7)=(-3/5)(x-24/7)` `63x+105y-781=0` |
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| 234. |
Points A (1, 3) and C (5, 1) are opposite vertices of a rectangle ABCD. If the slope of BD is 2, then its equation isA. `2x-y=4`B. `2x+y=4`C. `2x+y-7=0`D. `2x+y+7=0` |
| Answer» Correct Answer - A | |
| 235. |
In a plane there are two families of lines `y=x+r, y=-x+r`, where `r in {0, 1, 2, 3, 4 }`. The number of squares of diagonals of length 2 formed by the lines is: |
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Answer» Correct Answer - 9 Each family has parallel lines having the distance between them as `1//sqrt(2)` unit. Both the families are perpendicular to each other. So, to form a square of diagonal 2 units, lines of alternate pair are to be chosen. Both the families have three such pairs. So, the number of squares possible is `3 xx 3 =9.` |
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| 236. |
If the equation `x^2+(lambda+mu)x y+lambdau y^2+x+muy=0`represents two parallel straight lines, then prove that `lambda=mudot`A. (3,-1)B. `-3,1`C. (1,1)D. none of these |
| Answer» Correct Answer - A | |
| 237. |
A variable line L is drawn through O(0,0) to meet the line `L_(1) " and " L_(2)` given by y-x-10 =0 and y-x-20=0 at Points A and B, respectively. Locus of P, if `OP^(2) = OA xx OB`, isA. `(y-x)^(2) = 100`B. `(y+x)^(2) = 50`C. `(y-x)^(2) = 200`D. none of these |
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Answer» Correct Answer - C `r^(2) = (10 xx 20)/("sin" theta - "cos"theta)^(2)` `"or " (r "sin" theta - r" cos" theta)^(2) = 200` Hence, the locus is `(y-x)^(2) = 200`. |
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| 238. |
If the foot of the perpendicular from the origin to a straight line isat `(3,-4)`, then find the equation of the line. |
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Answer» Correct Answer - 3x-4y=25 Let P(3,-4) be the foot of the perpendicular from the origin O on the required line. Then the slope of OP is `(-4-0)/(3-0) = -(4)/(3)` Therefore, the slope of the required line is 3/4. Hence, its equation is `y+4 = (3)/(4)(x-3)` or 3x-4y = 25 |
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| 239. |
Find the equation of the line through the intersection of the lines 3x+y-9=0 and 4x+3y-7=0 and which is perpendicular to the line 5x-4y+1=0. |
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Answer» `5x-4y+1=0 Rightarrow y=(5)/(4)x+(1)/(4)` Slope of this given line is `(5)/(4)` Now, the equatoin of any line through the intersection of the given line is of the form `(3x+y-9)+k(4x+3y-7)=0......(i)` `Rightarrow (3+4k)x+(1+3k)y-(9+7k)=0` `Rightarrow (1+3k)y=-(3+4k)x+(9+7k)` `Rightarrow y=-((3+4k))/((1+3k))x+((9+7k))/((1+3k))..............(ii)` Let m be the slope of the line perpendicular to the required line. `"Then,"mxx(5)/(4)-1 Rightarrow m=(-4)/(5)` `"Then,"mxx(5)/(4)-1 Rightarrow m=(-4)/(5)` `therefore "we must have"(-(3+4k))/((1+3k))=(-4)/(5)` `Rightarrow 15+20k=4+12k Rightarrow 8k=-11 Rightarrow k=(-11)/(8)` Substituting `k=(-11)/(8)(4+3y-7)=0` `(3x+y-9)-(11)/(8)(4+3y-7)=0` `Rightarrow (24x+8y-72)-44x-33y+77=0` `Rightarrow 20x+25y-5=0 Rightarrow 4x+5y-1=0` |
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| 240. |
Find the equation of the line through the intersection of lines `x+ 2y 3 = 0`and `4x y+ 7 =0` and which is parallel to `5x+ 4y 20 = 0` |
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Answer» `5x+4y-20=0 Rightarrow y=(-5)/(4)x+5` `therefore "slope of the given lne"=(-5)/(4)` and slope of the required line =`(-5)/(4)` Now, the equation of any line through the intersection of hte given line of the form `(x+2y-3)+k(4x-y+7)=0` `Rightarrow (1+4k)x+(2-k)y+(7k-3)=0` `Rightarrow (2-k)y=-(1+4k)x+(3-7k)` `Rightarrow y=(-(1+4k))/((2-k))x+((3-7k))/((2-k))` `Rightarrow y=((1+4k))/((k-2))x+((3-7k))/((2-k))` Slope of this line =`((1+4k))/((k-2))` `therefore ((1+4k))/((k-2))=(-5)/(4)Rightarrow 4+16k=-5k+10` `Rightarrow 21k=6Rightarrow =(6)/(21)=(2)/(7)` Substituting `k=(2)/(7)` in (i), we get `(x+2y-3)+(2)/(7)(4x-y+7)=0` `Rightarrow (7x+14y-21)+(8x-2y+14)=0` `Rightarrow 15x+12y-7=0,` which is the required equation. |
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| 241. |
Find the equation of the line drawn through the point of intersection of the lines x+y=9 and 2x-3y+7=0 and whose slope is `(-2)/(3)` |
| Answer» Correct Answer - 2x+3y-23=0 | |
| 242. |
Find the equation of the line through the intersection of the lines 5x-3y=1 and 2x+3y=23 and which is perpendicular to the line 5x-3y=1 |
| Answer» Correct Answer - 63x+105y-781=0 | |
| 243. |
Find the equation of the line through the intersection of the lines 2x-3y+1=0 and x+y-2=0 and drawn parallel to y-axis. |
| Answer» Correct Answer - x=1 | |
| 244. |
. The points (1,3), (5, 1) are the opposite vertices of a rectangle. The other two vertices lie on the line `y=2x+c`. Find c and remaining two vertices.A. 4B. `-4`C. 2D. none of these |
| Answer» Correct Answer - B | |
| 245. |
Consider the family of lines `(x-y-6) + lambda(2x + y + 3) = 0` and `(x + 2y-4) + mu(3x-2y-4)=0` . If the lines of these 2 families are at right angle to each other then the locus of their point of intersection, is |
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Answer» `l_1:(x-y-6)+lamda(2x+3+y)` , intersection point of these two lines is`(1,-5)`(found out by elimination) `l_2:(x+2y-4)+mu(3x-2y-4)`,intersection point of these two lines is`(2,1)`(found out by elimination) Let point of intersection be `(h,k)` given the two lines are perpendicular to each other `(k+5)/(h-1) xx (k-1)/(h-2)=-1` =>`h^2+k^2-3h+4k-3=0` .(locus of point of intersection) |
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| 246. |
A line which makes an acute angle `theta`with the positive direction of the x-axis is drawn through the point `P(3,4)`to meet the line `x=6`at `R`and `y=8`at `Sdot`Then,`P R=3sectheta``P S=4cos e ctheta``P R=+P S=(2(3sintheta+4costheta_)/(sin2theta)``9/((P R)^2)+(16)/((P S)^2)=1`A. `PR = 3 "sec"theta`B. `PS = 4 " cosec"theta`C. `PR+PS = (2(3"sin" theta + 4 "cos" theta))/("sin" 2theta)`D. `(9)/(PR)^(2) + (16)/(PS)^(2) = 1` |
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Answer» Correct Answer - A::B::C::D The equation of any line making an acute angle `theta` with the positive direction of the x-axis and passing through P(3, 4) is `(x-3)/("cos"theta) = (y-4)/("sin" theta) " " (i)` Where |r| is the distance of any point (x,y) from P. Therefore, `A(r " cos "theta +3, r " sin " theta +4)` is a general point on line (i). If A is R, then `r " cos "theta +3 = 6 " or " r =(3)/("cos "theta)= 3 "sec"theta` `"Since "theta "is acute, " cos" theta gt 0`. Therefore, `PR = r = 3 " sec " theta " "(ii)` `"If " A-=S, r " sin "theta +4 =8. "Therefore",` `r = 4 " cosec " theta` `therefore PS = 4 " cosec " theta` `"Also, " PR+PS = (3)/("cos"theta) + (4)/("sin" theta)` ` = (2(3"sin" theta + 4"cos" theta))/("sin" 2theta)` `"and " (9)/((PR)^(2)) + (16)/((PS)^(2)) = "cos"^(2)theta + "sin"^(2)theta = 1` Therefore, (1),(2), (3), and (4) all are correct. |
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| 247. |
Find the equation of the straight line passing through the intersectionof the lines `x-2y=1`and `x+3y=2`and parallel to `3x+4y=0.` |
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Answer» Correct Answer - 3x+4y-5=0 Point of intersection of x-2y=1 and x+3y=2 is (7/5, 1/5) Any line parallel to 3x+4y=0 is 3x+4y+K=0. As this line passes through (7/5, 1/5), we get `(21)/(5) + (4)/(5)+K =0` or K=-5 Therefore, the required line is 3x+4y-5=0 |
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| 248. |
Find the equation of the line drawn through the point of intersection of the lines x-y=1 and 2x-3y+1=0 and which is parallel to the line 3x+4y=12 |
| Answer» Correct Answer - 3x+4y-24=0 | |
| 249. |
The distance of the point of intersection of the lines `2x-3y+5=0` and `3x+4y=0` from the line `5x-2y=0` isA. `(130)/(17sqrt129)`B. `(13)/(7sqrt29)`C. `(130)/(7)`D. None of these |
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Answer» Correct Answer - A Given equation of lines `2x-3y+5=0` and `3x+4y=0` From Eq.(ii),put the volume of `x=(-4y)/(3)` in Eq.(i), we get `2((-4y)/(3))-3y+5=0` `rArr -8y-9y+15=0` `rArry=(15)/(17)` From Eq.(ii), `3x+4.(15)/(17)=0` `rArr x=(-60)/(17.3)=(-20)/(17)` So, the point of intersection is `((-20)/(17),(15)/(17))` `therefore` Required distances form the line `5x-2y=0` is, `d=|-5xx(20)/(17)-2((15)/(17))|/(sqrt(25+4))=|(-100)/(17)-(30)/(17)|/(sqrt29)=(130)/(17sqrt29)` `[because distance of a point `p(x_1y_1)` form the line `ax+by+c=0` is `d =|(ax_1+by_1+c)|/(sqrt(a^2+b^2`]` |
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| 250. |
The number of integral values of `m`for which the x-coordinate of the point of intersection of the lines `3x+4y=9`and `y=m x+1`is also an integer is2 (b) 0(c) 4 (d) 1A. 2B. 0C. 4D. 1 |
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Answer» Correct Answer - A `"Solving "3x+4y=9, y=mx+1, "we get" ` `x=(5)/(3+4m)` Here, x is an integer if 3+4m=1,-1,5,-5. Hence m=-2/4, -4/4, 2/4,-8/4. So, m has two integral values. |
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