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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.
| 1. |
Find the ratio in which the x-axis cuts the join of the points A(4,5) and B(-10,-2). Also, find the point of intersection. |
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Answer» Correct Answer - (5:2, P(-6,0) Let the required ratio be k:1 and let P(x,0) be the point of divison. Then, `(kxx(-2)+1xx)/(k+1)=0 Rightarrow k=(5)/(2)` So, the required ratio=5:2 |
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| 2. |
If the origin is shifted to the point (-3,-2) by a translation of the axes, find the new coordinates of the point (3,-5). |
| Answer» Correct Answer - (6,-3) | |
| 3. |
The distance between the lines `5x-12y+65=0` and `5x-12y-39 = 0` is :A. 4B. 16C. 2D. 8 |
| Answer» Correct Answer - D | |
| 4. |
Prove that the locus of the centroid of the triangle whose vertices are`(acost ,asint),(bsint ,-bcost),`and `(1,0)`, where `t`is a parameter, is circle.A. `(3x -1)^(2) + (3y)^(2) = a^(2) - b^(2)`B. `(3x -1)^(2) + (3y)^(2) = a^(2) + b^(2)`C. `(3x + 1)^(2) + (3y)^(2) = a^(2) + b^(2)`D. ` (3x + 1)^(2) + (3y)^(2) = a^(2) - b^(2)` |
| Answer» Correct Answer - B | |
| 5. |
Prove that the locus of the centroid of the triangle whose vertices are`(acost ,asint),(bsint ,-bcost),`and `(1,0)`, where `t`is a parameter, is circle.A. `(3x+1)^(2) + (3y)^(2) = a^(2) - b^(2)`B. `(3x - 1)^(2) = a^(2) - b^(2)`C. `(3x -1)^(2) + (3y)^(2) = a^(2) + b^(2)`D. `(3x + 1)^(2) + (3y)^(2) = a^(2) + b^(2)` |
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Answer» Let (h , k) be the coordinates of the centroid . Then , h = `( a cos t + b sin t + 1)/(3) ` and `k = (a sin t - b cos t + 0)/(3)` `implies 3h - 1 = a cos t + b sin t ` and 3k = a sin t - b cos t `implies (3h-1)^(2) + (3k)^(2) = (a cos t + b sint)^(2) + (a sin t- b cos t)^(2)` `implies (3h-1)^(2) + (3k)^(2) = a^(2) + b^(2)` Hence , the locus of (h , k) is `(3x-1)^(2) + (3y)^(2) = a^(2) + b^(2)` |
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| 6. |
For `a gt b gt c gt 0`, if the distance between `(1,1)` and the point of intersection of the line `ax+by-c=0` and `bx+ay+c=0` is less than `2sqrt2` then,(A) `a+b-cgt0` (B) `a-b+clt0` (C) `a-b+cgt0` (D) `a+b-clt0`A. `a+b-c gt 0`B. `a-b+c lt 0`C. `a-b+c gt0`D. `a+b-c lt 0` |
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Answer» Correct Answer - A Solving given lines for their point of intersection, we get the point of intersection as (-c/(a+b), -c/(a+b)). Its distance from (1, 1) is `sqrt((1+(c)/(a+b))^(2) + (1+(c)/(a+b))^(2)) lt 2sqrt(2) " " ("given")` `"or " (a+b+c)^(2) lt 4(a+b)^(2) " or " (a+b+c)^(2)-(2a+2b)^(2) lt 0` `"or " (c-a-b)(c+3a+3b) lt 0` `"Since "a gt b gt c gt 0, (c-a-b) lt 0 " or "a+b-c gt 0` |
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| 7. |
ABC is a triangle formed by the lines xy = 0 and x + y = 1 . Statement - 1 : Orthocentre of the triangle ABC is at the origin . Statement - 2 : Circumcentre of `Delta`ABC is at the point (1/2 , 1/2) .A. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 12B. 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 - B The equation of the sides of `DeltaABC` are x = 0 , y = 0 and x + y = 1 . Clearly , `DeltaABC` is right angled triangle with right angle at the origin and coordinates the end-points of hypotenuse as (1,0) and (0,1) . So , orthocentre of the triangle is at the origin and circumcentre at the mid-point of its hypotenuse i.e. at (1/2 , 1/2) . |
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| 8. |
Find the equations of the altitudes of a `triangle ABC`, whose vertices are A(2,-2), B(1,1) and C(-1,0). |
| Answer» Correct Answer - 2x+y-2=0, 3x-2y-1=0, x-3y+1=0 | |
| 9. |
The orthocentre of the triangle formed by the lines `x=2,y=3` and `3x+2y=6` is at the pointA. (2,0)B. (0,3)C. (2,3)D. none of these |
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Answer» Correct Answer - C We know that the orthocentre of a right triangle is at the vertex forming right angle. Here, x=2 and y=3 are perpendicular sides of the given triangle. So, their point of intersection i.e., (2,3) is the orthocentre. |
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| 10. |
If A(0, 0), B(2, 4) and C(6, 4) are the vertices of a `triangle ABC`, find the equations of its sides. |
| Answer» Correct Answer - y=4,2x-3y=0,2x-y=0 | |
| 11. |
Find the equation of a line which is equidistant from the lines `x=4a n dx=8.`A. x=2B. x=6C. y=2D. y=6 |
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Answer» Correct Answer - A The given lines are both parallel ot y-axis and the required line is equidistant from these lines. So, it is also parallel ot y-axis at a distance of `(1)/(2)(-4+8)=2` units from it Hence , its equation is x=2. |
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| 12. |
Find the value of `x`for which the points `(x-1), (2,1)a n d (4,5) `are collinear. |
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Answer» Let A(x,-1), B(2,1) and C(4,5) be the given collinear points. Then, by collinearity of A,B,C we have: slope of AB=slope of BC `(1-(-1))/(2-x)=((5-1))/((4-2)) Rightarrow (2)/(2-x)=2 Rightarrow 2-x=1 Rightarrow x=1` Hence, x=1 |
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| 13. |
The medians BE and AD of a triangle with vertices `A(0, b), B(0, 0) and C(a, 0) ` are perpendicular to each other, if .A. `b=sqrt(2)a`B. `a=pmsqrt(b)b`C. `b=-sqrt(2)a`D. `a=-2b` |
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Answer» Correct Answer - B The coordinate of D and E are (a/2, 0) and (a/2,b/2) respectively. Now, `m_(1)`=Slope of AD `=(b-0)/(0-a//2)=-(2b)/(a)` `m_(2)` =Slope of BE `=(b//2-0)/(a//2-0)=(b)/(a)` Since AD and BE are perpendicular . Therefore, `m_(1)m_(2)=-1-(2b)/(a)xx(b)/(a)=-1 implies a=pmsqrt(2)b` |
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| 14. |
If A(-1, 6), B(-3, -9) and C(5, -8) are the vertices of a `triangle ABC`, find the equations of its medians. |
| Answer» Correct Answer - 29+4y+5=0, 8x-55y-21=0, 13x+14y+47=0 | |
| 15. |
Prove that the points `(5,1), (1,-1)a n d (11 ,4)`are collinear. Also find the equation of the straight line on whichthese points lie. |
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Answer» Let A(5,1), B(1,-1) and C(11,4) be the given points. Then, `"Slope of AB"=((-1-1))/((1-5))=(-2)/(-4)=(1)/(2)` `and "Slope of BC"=(4-(-1))/(11-1)=(5)/(10)=(1)/(2)` `therefore` slope of AB=slope of BC `Rightarrow AB|| BC` and have a point B in common `Rightarrow A,B,C "are collinear"` |
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| 16. |
Find the equation of a line passing through the points (-1,1) and (2,-4) |
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Answer» Let the given points be A(-1,1) and (2,-4) . We know that the equation of a line passing through the points. `(x_(1),y_(1)) and (x_(2),y_(2))"is given by"((y-y_(1)))/((x-x_(1)))=((y_(2)-y_(1)))/((x_(2)-x_(1)))` Here, `x_(1)=-1, y_(1)=1, x_(2)=2 and y_(2)=-4`. So, the equation of a line passing through the given points is `((y-1))/(x-(-1))=((-4-1))/(2-(-1)) Leftrightarrow ((y-1))/((x+1))=(-5)/(3)` `Leftrightarrow 3(y-1)=-5(x+1)` `Leftrightarrow 5x+3+2=0`. Hence, the required equation is 5x+3y+2=0. |
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| 17. |
Show that the three points (3,0),(-2,-2) and (8,2) are collinear. Also, find the equation of the straight line on which these points lie. |
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Answer» Let A(3,0),B(-2,-2) and C(8,2) be the given points. Then, the equation of line AB is given by `(y-0)/(x-3)=(-2-0)/(-2-3)" "["using"(y-y_(1))/(x-x_(1))=(y_(2)-y_(1))/(x_(2)-x_(1))]` `Rightarrow (y)/(x-3)=(2)/(5)........(i)` `Rightarrow 2x-5y-6=0,` which is the required equation. Putting x=8 and y=2 in (i), we get `LHS=(2xx8)-(5xx2)-6=0=RHS` Thus, the point C(8,2) also lies on (i). Hence, the given points lie on the same striaght line, whose equation is 2x-5y-6=0. |
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| 18. |
Two points `(a,0)` and `(0,b)` are joined by a straight line. Another point on this line , is(A)`(3a,-2b)` (B) `(a^2,ab)` (C) `(-3a,2b)` (D) `(a,b)`A. `(3a,-2b)`B. `(a^(2),ab)`C. `(-3a,2b)`D. (a,b) |
| Answer» Correct Answer - A | |
| 19. |
Find the value of k for which the points A(-2, 3), B(1, 2) and C(k, 0) are collinear. |
| Answer» Correct Answer - k=7 | |
| 20. |
Show that the points (a,0),(0,b) and (3a,-2b) are collinear. Also, find the equation of line containing them. |
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Answer» Let `A(a,0),B(0,b) and C(3a,-2b)` be the given points. Then, the equation of the line AB is given by `(y-0)/(x-a)=(b-0)/(0-a) Rightarrow -ay=bx-ab` `Rightarrow bx+ay-ab=0` Thus, the equation of the line AB is bx+ay-ab=0…………(i) Putting x=3a and y-02b (i), we get `LHS=b.3a+a(-2b)-ab=3ab-2ab=0=RHS` Thus, the point C(3a,-2b) also lies on AB. Hence, the given points are collinear and the equation of hte line containing them is bx+ay-ab=0 |
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| 21. |
If the points A(-2,-5), B(2,-2) and C(8,a) are collinear , then a =A. `-5//2`B. `5//2`C. `3//2`D. `1//2` |
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Answer» Correct Answer - B Since A,B,C are collinear points . `therefore ` Slope of AB=Slope of AC `implies (3)/(4)=(a=5)/(10)implies a=(5)/(2)` |
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| 22. |
The vertices of a diagonal of a square are `(-2,4)` and `(-2,-2)` Find the other verticesA. (1,-1),(5,1)B. (1,1),(5,-1)C. (1,1),(-5,1)D. none of these |
| Answer» Correct Answer - C | |
| 23. |
The points A(-2,1), B(0,5) and C(-1,2) are collinear. |
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Answer» Correct Answer - False Given points are A(-2,1)B(0,5)and C(1,2). Now, slope of `AB=(5-1)/(0+2)=2` Slope of `BC=(2-5)/(-1-0)=3` Slope of `AC=(2-1)/(-1+2)=1` Since, the slopes are different. Hence, A,Band C are not collinear. So, statements is false. |
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| 24. |
If `A(1,1), B(sqrt3+1,2) and C(sqrt3,sqrt3+2)` are three vertices of a square, then the diagonal through B isA. `y = (sqrt3 - 2) x + (3 - sqrt3) `B. `y = 0`C. y = xD. none of these |
| Answer» Correct Answer - D | |
| 25. |
If the vertices of a triangle have rationalcoordinates, then prove that the triangle cannot be equilateral. |
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Answer» Correct Answer - True We know that, if the verticles of a triangle have integral coordinates, then the triangle cannot be equilateral. Hence, the given statements is true. Since, in equilatteral triangle, we get `tan 60^@=sqrt3` =slope of the line, so with integral coordinates as vertices, the triangle cannot be equilateral. |
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| 26. |
If `)-4,0)`and `(1,-1)`are two vertices of a triangle of area `4s qdotu n i t s ,`then its third vertex lies on`y=x`(b) `5x+y+12=0``x+5y-4=0`(d) `x+5y+12=0`A. y = xB. 5x + y + 12 = 0C. x + 5y - 4 = 0D. none of these |
| Answer» Correct Answer - C | |
| 27. |
If two vertices of an equilateral triangle have rational co-ordinates, then for the third vertex which one is most applicable?A. integral coordinatesB. coordinates which are rationalC. at lest one coordinate irrationalD. coordinates which are irrational |
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Answer» Correct Answer - C Let `A(x_(1),y_(1)),B (x_(2),y_(2))` and `Cx_(3),y_(3))` be the verticales of an equilateral triangle ABC such that `x_(1),x_(2)` and `y_(1),y_(2)` are integers. If we assume that none of the coordinates of the vertex C are irrational, then we find that `Delta = "Area of " Delta ABC` `rArr Delta = (1)/(2)|{:(,x_(1),y_(1),1),(,x_(2),y_(2),1),(,x_(3),y_(3),1):}|="A rationa number"` But, `Delta = (sqrt(3))/(4)("Side")^(2)=(sqrt(3))/(4) xx" A rational number"` `rArr " Delta = "An irrational number"` Thus , we are arrive at contradiction. Therefore, our supposition is worng . Hence, at lest one coordinate of C is irrational. |
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| 28. |
Find the locus of image of the veriable point `(lambda^(2), 2 lambda)` in the line mirror x-y+1=0, where `lambda` is a peremeter. |
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Answer» Let the image of `(lambda^(2), 2 lambda)` in the line mirror x-y+1=0 be (h,k). `therefore (h-lambda^(2))/(1) = (k-2lambda)/(-1) = (-2(lambda^(2)-2lambda +1))/(2)` `therefore h+1 = 2lambda " " (1)` `" and " k= lambda^(2) + 1 " " (2)` Putting the value of `lambda` from (1) in (2), we get `k-1 = ((h+1)/(2))^(2)` or 4(y-1) = `(x+1)^(2)` This is the required locus. |
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| 29. |
The range of values of `a` such that the angle `theta` between the pair of tangents drawn from `(a,0)` to the circle `x^2+y^2=1` satisfies `pi/2 |
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Answer» a>1 whe a=1 `theta=2pi/2=pi` `1ltaltsqrt2` `-sqrt2ltalt-1` `a in{(-sqrt2-1)uu(1,sqrt2)}` option b is correct. |
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| 30. |
The angle between the lines `xcos alpha_1+ysin alpha_1=p_1` and `xcos alpha_2+ysin alpha_2=p_2` is(A) `|alpha_1+alpha_2|` (B) `|alpha_1-alpha_2|` (C) `|2alpha_1|` (D) `|2alpha_2|` |
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Answer» Equation of line 1, `L_1 =>xcos alpha_1+ysin alpha_1=p_1` `=>y = -x(cos alpha_1)/(sin alpha_1) +p_1/sin alpha_1` `=> y = -cot alpha_1 x +p_1/sin alpha_1` Comparing it with,`y = mx+c` Slope of `L_1(m_1) = -cotalpha_1` Equation of line 1, `L_2 =>xcos alpha_2+ysin alpha_2=p_2` `=>y = -x(cos alpha_2)/(sin alpha_2) +p_2/sin alpha_2` `=> y = -cot alpha_2 x +p_1/sin alpha_2` Comparing it with,`y = mx+c` Slope of `L_2(m_2) = -cotalpha_2` So, `tan alpha = |(m_1-m_2)/(1+m_1m_2)|` `tan alpha = |(cot alpha_2 - cot alpha_1)/(1+cot alpha_1cot alpha_2)|` `=>tan alpha = |(sin alpha_1cosalpha _2 - sin alpha_2cosalpha_1)/(cos alpha_1cosalpha _2 - sin alpha_2sinalpha_1)|` `=>tan alpha = |(sin(alpha_1-alpha_2))/(cos(alpha_1-alpha_2))|` `=>tan alpha = |tan(alpha_1 - alpha_2)|` `=> alpha = |alpha_1 - alpha_2|` So, angle between these two lines will be `|alpha_1 - alpha_2|`. |
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| 31. |
The straight lines `x+2y-9=0,3x+5y-5=0`, and `a x+b y-1=0`are concurrent, if the straight line `35 x-22 y+1=0`passes through the point`(a , b)`(b) `(b ,a)``(-a ,-b)`(d) none of theseA. (a,b)B. `(b,c)`C. `(a,b)`D. `(-a,b)` |
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Answer» Correct Answer - A The three lines of are concurrent if `|{:(,1,2,-9),(,3,5,-5),(,a,b,-1):}|` `rArr 35a - 22b + 1 = 0 rArr 35x - 22y + 1=0` passes through (a,b) |
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| 32. |
The centres of two circles C1 and `C2` each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1` and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is |
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Answer» in `/_ PDO` `/_PRO = 90^@`angle between tangent let `/_ POR = alpha` `/_APD = alpha = /_EPB` (V.O.A) In `/_ PEB` `BE=1` (radius) `BP= 3` `PE^2 = BD^2 - BE^2` `PE^2 = 9 - 1` `PE = 2 sqrt(2)` `tan alpha = (BE)/(PE) = 1/(2sqrt(2))` in ` /_PDO ` `tan alpha = (PD)/(DO) = (PD)/R` `R= (PD)/tan alpha = (2 sqrt(2))/(1/(2sqrt2))` `R=8`units |
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| 33. |
Find the distance between the points `(0,-3)` and `(3,0)`A. A(2,-3) and B(-6,3)B. C(-1,1) and D(8,11)C. P(-8,-3) and Q(-2,-5)D. R(a+,a-b) and S(a-b,a+b) |
| Answer» Correct Answer - A::B | |
| 34. |
The lines `x + y=|a| and ax-y = 1` intersect each other in the first quadrant. Then the set of all possible values of a is the interval:A. `[1,oo)`B. `(-1,oo)`C. `(-1,1)`D. `(0,oo)` |
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Answer» Correct Answer - A Given lines intersect at the point `P((1+|a|)/(1+a),(a|a|-1)/(1+a))` This point will lie in first quadrant iff `1+a gt0 and a|a| -1 gt 0` `implies a gt -1 and a|a| -1 ge 0` `implies a^(2)-1 ge 0 " if " a ge 0 " or " , -a^(2)-1 gt 0 " if " 0 lt a lt -1` `implies a in [1,oo)` |
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| 35. |
The point P(a,b), Q(c,d), R(a,d) and S(c,b) , where a,b,c ,d are distinct real numbers, areA. collinearB. vertices of a squareC. vertices of a rhombusD. concyclic |
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Answer» Correct Answer - B We have, Slope of PS=0=Slope of QR `implies` PS and QR are parallel to x-axis PR and QS are parallel to y-axis. `implies ` PQRS is a rectangle. Also, PQ=RS. Hence , PQ=RS. Hence, PQRS is a square. |
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| 36. |
If the x intercept of the line `y = mx + 2` is greater than `1/2` then the gradient of the line lies in the intervalA. (-1,0)B. `((-1)/(4),0)`C. `(-oo, -4)`D. (-4,0) |
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Answer» Correct Answer - D Given equation of line y=mx+2 `rArr (x)/(-(2)/(m))+(y)/(2) =1` It is given that -x-intercept of the above line is greater than `(1)/(2).` `therefore -(2)/(m) gt(1)/(2)` `rArr -(2)/(m) -(1)/(2) gt 0` `rArr -((4+m)/(2m)) gt 0` `rArr (4+m)/(2m) lt 0` `rArr -4 lt m lt 0` |
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| 37. |
(i) Find the gradient of a straight line which is passes through the point (-3. 6) and the mid point of (4,-5) and (-2, 9)A. `(pi)/(4)`B. `(pi)/(6)`C. `(pi)/(3)`D. `(3pi)/(4)` |
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Answer» Correct Answer - D The coordinates of the mid-point D of the segment BC are (1,2). `therefore ` Slope of AD`=(6-2)/(-3-1)=-1` `implies theta =-1` , where `theta` is the angle made by AD with x-axis `implies theta=(3pi)/(4)` So, AD makes angles `(3pi)/(4)` with x -axis |
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| 38. |
Angle made with the x-axis by a straight line drawn through (1, 2) sothat it intersects `x+y=4`at a distance `(sqrt(6))/3`from (1, 2) is`105^0`(b) `75^0`(c) `60^0`(d) `15^0`A. `pi//6 and pi//3`B. `pi//8 and 3pi//8`C. `pi//12 and 5pi//12`D. none of these |
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Answer» Correct Answer - C Suppose a line passing through (1,2) makes an angle `theta` with x-axis . Then, its equation is `(x-1)/(cos theta)=(y-2)/(sin theta)` The coordinates of a point on this line at a distance `(sqrt(16))/(3)=sqrt((2)/(3))` from (1,2) are given by `(x-1)/(costheta)=(y-2)/(sin theta)=sqrt((2)/(3))` or, `(1+sqrt((2)/(3)) costheta, 2+sqrt((2)/(3)) sin theta)` This point lies on `x+y=4` `therefore 1+sqrt((2)/(3)) costheta + 2+ sqrt((2)/(3)) sin theta=4` `implies cos theta + sin theta=sqrt((3)/(2))` `implies sin (theta+(pi)/(4))=(sqrt(3))/(2)implies theta=(pi)/(12) and (5pi)/(12)` |
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| 39. |
If the sum of the distances of a point from two perpendicular lines ina plane is 1, then its locus isa square(b) a circlea straight line (d) two intersecting linesA. a circleB. an ellipseC. a hyperbolaD. none of these |
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Answer» Let the mutually perpendicular lines be the co-ordinate axes and P(h,k) be a variable point . Then , `|h| + |k| = 1 " "` [Given] `therefore` Locus of P (h,k) is |x| + |y| = 1 Clearly , it represents a square . |
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| 40. |
`A B C`is an isosceles triangle. If the coordinates of the base are `B(1,3)`and `C(-2,7)`, the coordinates of vertex `A`can be`(1,6)`(b) `(-1/2,5)``(5/6,6)`(d) none of theseA. (1,6)B. (1/2 , 5)C. (5/6 , 6)D. none of these |
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Answer» Clearly , vertex P lies on the perpendicular bisector of QR whose equation is `y - 5 = (3)/(4) (x + (1)/(2)) or , 6x - 8y + 43 = 0` Clearly , (5/6, 6) satisfies this equation . ltbRgt Hence , the vertex P is (5/6 , 6). |
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| 41. |
A line through the point A(2,4) intersects the line `x+y=9` at the point. The minimum distance of AP , isA. `(5)/(sqrt(2))`B. `(7)/(sqrt(2))`C. `(2)/(sqrt(2))`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - C The equation of a line passing through A(2,4) is `(x-2)/(cos theta)=(y-4)/(sin theta)` Suppose it cuts the line `x+y=9` at point P whose coordinates are given by `(x-2)/(cos theta)=(y-4)/(sin theta)=r` I.e. `x=2+r cos theta, y=4+r sin theta` `therefore 2+rcos theta+4+rsin theta=9` `implies r(cos theta+sin theta)=3` `implies r=(3)/(cos theta+sin theta)` `implies r ge (3)/(sqrt(2)) [because cos theta + sin theta le sqrt(2) therefore (1)/(cos theta + sin theta ge (1)/(sqrt(2)))]` Hence , the minimum value of AP is `(3)/(sqrt(2))` |
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| 42. |
distance of the lines `2x-3y-4=0` from the point `(1, 1)` measured paralel to the line `x+y=1` isA. `sqrt2`B. `(5)/(sqrt2)`C. `(1)/(sqrt2)`D. 6 |
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Answer» The equation of the line through the point P(1,1) parallel to the line x + y = 1 is `(x -1)/(cos (3pi)/(4)) = (y-1)/(sin (3pi)/(4))` Suppose it meets the line 2 x - 3y = 4 at `Q (1 + r cos (3pi)/(4) , 1 + r sin (3pi)/(4))` , where PQ = |r| As Q lies on the line 2x - 3y = 4 . `therefore 2 (1 - (r)/(sqrt2)) - 3 (1 + (r)/(sqrt2) ) = 4` `implies -1- (5r)/(sqrt2) =4 implies (5r)/(sqrt2) = -5 implies r = - sqrt2` Hence , PQ = `sqrt2` |
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| 43. |
The distance of a point (2,3) from the line 2x -3y + 9 = 0 measured along a line x-y +1 =0 is :A. 4B. `2sqrt(2)`C. `8sqrt(2)`D. `4sqrt(2)` |
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Answer» Correct Answer - D The slope of the line `x-y+1=0`. So, it makes and angle of `45^(@)` with x-axis. The equation of a line passing through (2,3) and making an angle of `45^(@)` is given by `(x-2)/(cos 45^(@))=(y-3)/(sin 45^(@))=r` Co-ordinate of any point on his line are ltbr? `(2+r cos 45^(@), 3+r sin 45^(@))` or, `(2+(r)/(sqrt(2)),3+(r)/(sqrt(3)))` If this point lies on the line `3=2x-3y+9=0` `4+rsqrt(2)-9-(3r)/(sqrt(2))+9=0 implies r=4sqrt(2)` units. |
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| 44. |
Find the distance of the point `(3, 5)` from the line `2x + 3y = 14` measured parallel to the line `x-2y = 1.`A. `7//sqrt5`B. `7//sqrt(13)`C. `sqrt5`D. `sqrt(13)` |
| Answer» Correct Answer - C | |
| 45. |
Distance of a point (2,5) from the line 2x-y-4=0 measured parallel to the line 3x-4y + 8 = 0 is : |
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Answer» (y-5)=3/4(x-2) 4y-20=3x-6 4y-3x-14=0-(1) Given 2x-y-4=0-(2) From equation 1 and 2 4(2x-4)-3-14=0 8x-3x-30=0 x=6 y=8 A(6,8) AB=`sqrt((6-2)^2+(8-5)^2)` `=sqrt(4^2+3^2)` =5 Units. |
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| 46. |
The ratio in which the line 3x+4y+2=0 divides the distance between 3x+4y+5=0 and 3x+4y-5=0 is? |
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Answer» We know, Distance=`|(C_2-C_2)/sqrt(A^2+B^2)|` From 1 and 2 m=`|(5-2)/sqrt(3^2+4^2)|=3/5` n=`|(2+5)/sqrt(3^2+4^2)|=7/5` Ration of m and n m:n=`(3/5)/(7/5)=3/7`. |
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| 47. |
The ratio in which the line 3x+4y+2=0 divides the distance between 3x+4y+5=0 and 3x+4y-5=0 is?A. `7 : 3`B. `3 : 7`C. `2 : 3`D. none of these |
| Answer» Correct Answer - B | |
| 48. |
The area of the parallelogram formed by the lines `xcosalpha + y sin alpha =p, x cos alpha + y sin alpha = q,x cos beta + y sin beta=r` and `x cosbeta+y sinbeta = s` for given values of p, q, r and s is least, if `(alpha - beta)=` (A) `+-pi/2` (B) `pi/4` (C) `pi/6` (D) `pi/3`A. `pm (pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(3)` |
| Answer» Correct Answer - B | |
| 49. |
If `P`is a point `(x ,y)`on the line `y=-3x`such that `P`and the point (3, 4) are on the opposite sides of the line `3x-4y=8,`then`x >8/(15)`(b) `x >8/5``y |
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Answer» Correct Answer - A::C Let L=3x-4y-8. Then the value of L at (3,4) is `3 xx 3-4 xx 4-8 = -15 lt 0.` Hence, for the point P(x,y), we should have `L lt 0` `"or " 3x-4y-8 gt 0` `"or " 3x-4(-3x)-8 gt 0 " " [because P(x,y) " lies on "y = -3x]` `"or " x gt (8)/(15)` `"and " -y-4y-8 gt 0` `"or " y lt -(8)/(5)` |
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| 50. |
If `"sin"(alpha + beta) "sin" (alpha-beta) = "sin" gamma(2"sin" beta + "sin"gamma) " where " 0 lt alpha, beta, lt pi,` then the straight line whose equation is `x "sin" alpha+y "sin" beta-"sin" gamma = 0` passes through the pointA. (1,1)B. (-1,1)C. (1,-1)D. none of these |
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Answer» Correct Answer - C `"sin"(alpha+beta)"sin"(alpha-beta)="sin" gamma(2"sin"beta + "sin"gamma)` `"or ""sin"^(2)alpha-("sin"beta + "sin" gamma)^(2)=0` `"or "("sin"alpha+"sin" beta+"sin"gamma)("sin"alpha-"sin"beta-"sin"gamma)=0` `"Since " 0 lt alpha, beta, gamma lt pi," we have"` `"sin"alpha+"sin"beta+"sin"gamma ne 0` `therefore " sin"alpha-"sin"beta-"sin"gamma = 0` `"So, "x"sin" alpha+y"sin" beta-"sin"gamma=0` passes through the fixed point (1,-1). |
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