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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. |
What is the area of the triangle whose vertices are (3, 0), (0, 4) and (3, 4) ?A. 6 sq. unitB. 7.5 sq. unitC. 9 sq. unitD. 12 sq. unit |
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Answer» Correct Answer - A Required Area `=1/2|(3,0,1),(0,4,1),(3,4,1)|=1/2 [3(4-4)+1(0-12)]=6` |
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| 2. |
The numerical value of the perimeter of a square exceeds that of its area by 4. what is the side of the square ?A. 1 unitB. 2 unitC. 3 unitD. 4 unit |
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Answer» Correct Answer - B Let the side of the square `= x` units Area of square `=x^(2)` units and perimeter of square `=4x` unit According to question, `x^(2)+4=4x` `implies x^(2)-4x+4=0` `implies (x-2)^(2)=0` `implies x=2` `:.` Side of square `=2` unit |
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| 3. |
What is the maximum number of straight lines that can be drawn with any four points in a plane such that each line contains at least two of these points ?A. 2B. 4C. 6D. 12 |
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Answer» Correct Answer - C Required number of lines `=.^(4) C_(2)= (4!)/(2!2!)=6` |
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| 4. |
If the straight lines x - 2y=0 and kx + y=1 intersectsat the point (1,1/2) then what is the value of k?A. 1B. 2C. `1//2`D. `-1//2` |
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Answer» Correct Answer - C Since. The straight lines `x-2y=0` and `kx+y=1` intersect at the point `(1, 1/2)` `:.` The point `(1, 1/2)` satisfies the equation `kx+y=1` `:.` Put `x=1`, and `y=1/2` in `eq^(n) kx+y=1`, we get `k.1+1/2 = 1 implies k=1/2` |
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| 5. |
What is the equation of line passing through `(0,1)` and making an angle with the Y-axis equal to the inclination of the line `x - y = 4` with X-axis?A. `y=x+1`B. `x=y+1`C. `2x=y+2`D. None of the above |
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Answer» Correct Answer - A Given line is `x-y=4` slope `=1` i.e. `m=1` Since required line passes through (0, 1) `:. y-1=m (x-0)` `implies y-1=1 (x) implies y=x+1` |
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| 6. |
Two straight line paths are represented by the equation `2x-y=2` and `-4x+2y=6`. Then the paths willA. cross each other at one pointB. not cross each otherC. cross each other at two pointsD. cross each other at infinitely many points |
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Answer» Correct Answer - B Given lines are `2x-y=2` and `2x-y=-3` since, Both lines are parallel so they will never meet. |
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| 7. |
Prove that the line `x+2y-9=0 and 2x+4y+5=0` are parallel.A. 2B. `-1`C. 1D. 0 |
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Answer» Correct Answer - A Given equation of lines are `x+2y-9=0 implies 2y=-x+9` `implies y=-1/2x+9/2` ...(1) and `kx+4y+5=0 implies 4y=-kx-5` `implies y=(-k)/4 x- 5/4` ...(2) Since line (1) and line (2) are parallel therefore their slopes are equal. `(-1)/2=(-k)/4 implies k=2` |
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| 8. |
What is the inclination of the line `sqrt(3)x-y-1=0` ?A. `30^(@)`B. `60^(@)`C. `135^(@)`D. `150^(@)` |
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Answer» Correct Answer - B Given equation can be written as `y=sqrt(3)x-1` on comparing with `y=mx+c` We get `tan theta=sqrt(3) implies theta =60^(@)` |
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| 9. |
Consider the following points : 1. (0, 5) 2. (2, -1) 3. (3, -4) Which of the above lie on the line `3x+y=5` and at a distance `sqrt(10)` from (1, 2) ?A. 1 onlyB. 2 onlyC. 1 and 2 onlyD. 1, 2 and 3 |
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Answer» Correct Answer - C All three points (0, 5, (2, -1) and (3, -4) lie on `3x+y=5` `sqrt((0-1)^(2)+(5-2)^(2))=sqrt(10)` `sqrt((2-1)^(2)+(-1-2)^(2))=sqrt(10)` `sqrt((3-1)^(2)+(-4-2)^(2))=sqrt(40)=2sqrt(10)` |
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| 10. |
The line through the points (4, 3) and (2, 5) cuts off intercepts of length `lambda` and `mu` on the axes. Which one of the following is correct ?A. `lambda gt mu`B. `lambda lt mu`C. `lambda gt - mu`D. `lambda = mu` |
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Answer» Correct Answer - D Let the equation of line be `x/lambda+y/mu=1` The line passes through (4, 2) and (2, 5). `:. 4/lambda +3/mu=1`, is possible when `lambda=mu=7` and `2/lambda+3/mu=1`, is possible when `lambda=mu=5` So, `mu=lambda` |
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| 11. |
The equation of the line, the reciprocals of whose intercepts on the axes are m and n, is given byA. `nx+my=mn`B. `mx+ny=1`C. `mx+ny=mn`D. `mx-ny=1` |
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Answer» Correct Answer - B Let line be `x/a+y/b=1` given that `1/a=m` and `1/b=n` `a=1/m, b=1/n` equation of line, `mx+ny=1` |
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| 12. |
What is the equation of straight line passing through the point (4, 3) and making equal intercepts on the coordinate axes ?A. `x+y=7`B. `3x+4y=7`C. `x-y=1`D. None of these |
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Answer» Correct Answer - A Let equation of line be `x/a+y/a=1` or `x+y=a` line passing through (4, 3), then `a=0` Required equation, `x+y=7` |
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| 13. |
The line `y=0` divides the line joining the points (3, -5) and (-4, 7) in the ratio :A. `3 : 4`B. `4 : 5`C. `5 : 7`D. `7 : 9` |
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Answer» Correct Answer - C Let P( x, y) be the point of division that divides the line joining (3, -5) and (-4, 7) in the ratio of `k:1` Now, `y=(7k-5)/(k+1)` ...(i) Since, P lies on `y=0` or x-axis then, from eq. (i) `0=(7k-5)/(k+1) implies 7k=5 implies k=5/7` |
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| 14. |
What angle does the line joining the points (5,2) and (6,-15) subtend on (0,0)?A. `pi/6`B. `pi/4`C. `pi/2`D. `(3pi)/4` |
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Answer» Correct Answer - C Slope of line joining (5, 2) and (0, 0) `tan A=m_(1) =(2-0)/(5-0)=2/5` Slope of line joining (6, -15) and (0, 0) `tan B=m_(2)=(-15)/6=(-5)/2` Now, `m_(1) m_(2)=2/5(- 5/2)=-1` Hence, both lines are perpendicular and than angle between them `=pi/2` |
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| 15. |
The length of latus rectum of the ellipse `4x^(2)+9y^(2)=36` isA. `4/3`B. `8/3`C. 6D. 12 |
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Answer» Correct Answer - B `4x^(2)+9y^(2)=36` `x^(2)/3^(2)+y^(2)/2^(2)=1` Length of latus rectum `=2xx2^(2)/3=8/3` |
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| 16. |
What is the equation to the straight line passing through (5, -2) and (-4, 7) ?A. `5x-2y=4`B. `-4x+7y=9`C. `x+y=3`D. `x-y=-1` |
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Answer» Correct Answer - C Equation of line `y+2=(7+2)/(-4-5) (x-5)` `implies y+2=-x+5` `implies x+y=3` |
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| 17. |
The points (a, b), (0, 0), (-a, -b) and `(ab, b^(2))` areA. the vertices of a parallelogramB. the vertices of a rectangleC. the vertices of a squareD. collinear |
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Answer» Correct Answer - B Given points, `A(a, b), B(0, 0), C(-a, -b), D(ab, b^(2))` Slope of `AB=(b-0)/(a-0)=b/a` Slope of `BC=b/a` Slope of `AC=b/a` Slope of `BD=b/a` So, the points are collinear. |
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| 18. |
The distance of the point (1, 3) from the line `2x+3y=6`, measured parallel to the line `4x+y=4`, isA. `5/sqrt(13)` unitsB. `3/sqrt(17)` unitC. `sqrt(17)` unitsD. `sqrt(17)/2` units |
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Answer» Correct Answer - D The line `4x+y=4` can be written as `y=-4x+4`. So, slop is -4. The line parallel to `4x+y=4` will have slope `-4` only. Given point `=(1, 3)` Equation of line passing through (1, 3) and slope -4 is `y-3=-4(x-1)` `implies y-3=-4x+4 implies 4x+y=7`. Solving the two equations, we get `2x+3y=6 implies 4x+6y=12` `{:(4x+y=7),(cancel((-))(-) (-)),(bar(5y=5implies y=1)):}` `2x+3y=6 implies 2x+3(1)=6` `implies 2x=3 implies x=3/2`. Distance between the points (1, 3) and `(3/2, 1)` is `sqrt((3/2-1)^(2)+(1-3)^(2))=sqrt((1/2)^(2)+(-2)^(2))=sqrt(1/4+4)=sqrt(17/4)=sqrt(17)/2`. |
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| 19. |
Which of the following is correct in respect of the equations `(x-1)/2=(y-2)/3` and `2x+3y=5` ?A. They represent two lines which are parallel.B. They represent two lines which are perpendicular.C. They represent two lines which are neither parallel nor perpendicular.D. The first equation does not represent a line. |
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Answer» Correct Answer - B Given equation of line is `(x-1)/2=(y-2)/3` `implies 3x-3=2y-4 implies 3x-2y+1 =0` `implies y=(3x)/2+1/2` and equation of second line is `2x+3y=5` `implies y=(-2)/3 x+5/3` `:.` Slope of first line, `m_(1)=3/2` and slope of second line, `m_(2)=-2/3` `:. m_(1)m_(2)=-1` Hence, two lines are perpendicular to each other. |
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| 20. |
What is the angle between the lines `x+y=1` and `x-y=1` ?A. `pi/6`B. `pi/4`C. `pi/3`D. `pi/2` |
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Answer» Correct Answer - D Slope of `x+y=1` is -1 Slope of `x-y=1` is 1 Let `tan A=-1, tan B=1` `A=(3pi)/4, B=pi/4` `A-B=pi/2` |
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| 21. |
If p be the length of the perpendicular from the origin on the straight line `ax+by=p` and `b=sqrt(3)/2`, then what is the angle between the perpendicular and the positive direction of x-axis ?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - C Equation of line is `ax+by-p=0`, then length of perpendicular, from the origin. `p=|(axx0+bxx0-p)/sqrt(a^(2)+b^(2))| or p=|(-p)/sqrt(a^(2)+b^(2))|` or `|1/sqrt(a^(2)+b^(2))|=1 or a^(2)+b^(2)=1` `b=sqrt(3)/2 or b^(2)=3/4` `a^(2)+3/4=1` `a^(2)=1/4 implies a=1/2 " "[a=-1/2" not taken since angle is with +ve direction to x-axis."]` Equation is `1/2 x+sqrt(3)/2 y=p` or `x cos 60^(@)+y sin 60^(@)=p` Angle `=60^(@)` |
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| 22. |
The centroid of the triangle with vertices (2, 3), (-2, -5) and (3, 5) is atA. (1, 1)B. (2, -1)C. (1, -1)D. (1, 2) |
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Answer» Correct Answer - A Centroid `=((2-2+3)/3, (3-5+5)/3)` `=(1, 1)` |
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| 23. |
Write the number of diagonals of an n-sided polygon.A. `(n(n-1))/2`B. `(n(n-3))/2`C. `n^(2)-n`D. `(n(n+1))/2` |
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Answer» Correct Answer - B A n-sided regular polygon have `(n(n-3))/2` diagonals. |
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| 24. |
The equation of the curve whose slope is given by `(dy)/(dx)=(2y)/x ; x >0, y >0`and which passes through the point (1,1) is`x^2=y`b. `y^2=x`c. `x^2=2y`d. `y^2=2x`A. CircleB. ParabolaC. EllipseD. hyperbola |
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Answer» Correct Answer - B `(dy)/(dx)=(2y)/x` `implies (dy)/y=2 (dx)/x` On integration `int (dy)/y=2 int (dx)/x` `implies log y=2 log x+log a` `implies log y=log x^(2)+log a` `implies log y=log (a^(2).a)` `implies y=x^(2)a` at `(1, 1), a=1` `implies x^(2)=y=4 (1/4) y` `implies` the curve is parabola. |
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| 25. |
If (p,q) is a point on x-axis, which is equidistant from (1,2) and (2,3). Find p and q :A. `p=0, q=4`B. `p=4, q=0`C. `p=3//2, q=0`D. `p=1, q=0` |
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Answer» Correct Answer - B Since (p, q) is the point on the x-axis `:. q=0` Let `P=(p, 0)` `A=(1, 2)` and `B=(2, 3)` Given : `PA=PB` `implies PA^(2)=PB^(2)` `implies (1-p)^(2)+4=(2-p)^(2)+9` `implies 1+p^(2)-2p-4-p^(2) +4p=5` `implies 2p=8` `implies p=4` Hence, `p=4, q=0` |
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| 26. |
If the area of a triangle with vertices `(-3,0),(3,0)` and `(0,0` is 9 sq. units. Then the value of `k` will beA. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - A Let the vertices of the `Delta ABC` be `A(-3, 0), B(3, 0)` and `C (0, k)`. `:.` Area of `Delta ABC=1/2 |(,,),(,,),(,,)|` Given, area is 9 `implies 9=1/2 {-3(-k)+1(3k)}` `implies 18=3k+3k` `implies k=18/6=3` |
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| 27. |
What is the slope of the line perpendicular to the line `x/4+y/3=1` ?A. `3/4`B. `- 3/4`C. `-4/3`D. `4/3` |
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Answer» Correct Answer - D Given equation of the line `x/4+y/3=1` can be written as `3x+4y=12 implies 4y=-3x+12` `implies y=(-3)/4 x+12/4` The slope of the line `x/y+y/3=1` is `(-3)/4` `:.` Slope of the line perpendicular to this line `=-((-1)/(3//4))=4/3` |
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| 28. |
If the mid point between the points `(a+b, a-b)` and `(-a, b)` lies on the line `ax+by=k`, what is k equal to ?A. `a//b`B. `a+b`C. `ab`D. `a-b` |
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Answer» Correct Answer - C Given points are `(a+b, a-b)` and `(-a, b)` Mid point is `((a+b-a)/2, (a-b+b)/2)=(d/2, a/2)` Since, it lies on `ax+by=k` `:. a(b/2)+b(a/2)=k implies k=ab` |
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| 29. |
If A(2, 3), B(1, 4), C(0 - 2) and D (x, y) are the vertices of a parallelogram, then what is the value of (x, y) ?A. (1, -3)B. (2, 4)C. (1, 1)D. (0, 0) |
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Answer» Correct Answer - A As given : A(2, 3), B(1, 4), C(0, -2) and D (x, y) are the vertices of a parallelogram. Diagonals of a parallelogram bisect each other So, mid-point are same for both diagonals AC and BD. `(2+0)/2=(1+x)/2` and `(3-2)/2=(4+y)/2` `implies x=1 and y=-3` `implies D(x, y)=(1, -3)` |
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| 30. |
If O be the origin and `A(x_(1), y_(1)), B(x_(2), y_(2))` are two points, then what is `(OA) (OB) cos angle AOB` ?A. `x_(1)^(2)+x_(2)^(2)`B. `y_(1)^(2)+y_(2)^(2)`C. `x_(1)x_(2)+y_(1)y_(2)`D. `x_(1)y_(1)+x_(2)y_(2)` |
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Answer» Correct Answer - C Let `O(0, 0), A(x_(1), y_(1))` and `B(x_(2), y_(2))` be three points `OA=sqrt(x_(1)^(2)+y_(1)^(2)), OB=sqrt(x_(2)^(2)+y_(2)^(2))` `AB=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))` In `Delta AOB`, `s` `cos angle AOB=(OA^(2)+OB^(2)-AB^(2))/(2.OA.OB)` `implies OA.OB cos angle AOB=(OA^(2)+OB^(2)-AB^(2))/2` `=(x_(1)^(2)+y_(1)^(2)+x_(2)^(2)+y_(2)^(2)-{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)})/2` `implies OA.OB. cos angle AOB` `=(x_(1)^(2)+y_(1)^(2)+x_(2)^(2)+y_(2)^(2)-{x_(2)^(2)+x_(1)^(2)-2x_(1)x_(2)+y_(2)^(2)+y_(1)^(2)-2y_(1)y_(2)})/2` `=(2(x_(1)x_(2)+y_(1)y_(2)))/2=x_(1)x_(2)+y_(1)y_(2)` |
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| 31. |
If (a, b), (c, d) and (a-c, b-d) are collinear, then which one of the following is correct ?A. `bc-ad=0`B. `ab-cd=0`C. `bc+ad=0`D. `ab+cd=0` |
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Answer» Correct Answer - A Let A, B and C having co-ordinates (a, b), (c, d) and `{(a-c), (b-d)}` respectively be the points If these points are collinear then `|(a,b,1),(c,d,1),(a-c,b-d,1)|=0` `R_(2) rarr R_(2)-R_(1)` gives `|(a,b,1),(c-a,d-b,0),(a-c,b-d,1)|=0` `R_(3) rarr R_(2)+R_(3)` gives `|(a,b,1),(c-a,d-b,0),(0,0,1)|=0` `implies 1. {a(d-b)-b(c-a)}=0` `implies ad-ab-bc+ab=0` `implies bc-ad=0` |
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| 32. |
Consider the lines `y=3x, y=6x` and `y=9` The centroid of the triangle is at which one of the following points ?A. (3, 6)B. `(3/2, 6)`C. (3, 3)D. `(3/2, 9)` |
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Answer» Correct Answer - B Coordinates of O, A, B are (0, 0) `(3/2, 9), (3, 9)` respectively. `:.` Centroid `C=[((0+3/2+3)/3)"," ((0+9+9)/3)]=(3/2, 6)` |
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| 33. |
What is the foot of the perpendicular from the point (2, 3) on the line `x+y-11=0` ?A. (1, 10)B. (5, 6)C. (6, 5)D. (7, 4) |
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Answer» Correct Answer - B We jnow a line perpendicular to a given line `ax-by+c=0` is `bx+ay+k=0` `:.` The equation of line perpendicular to given line `x+y-11=0` ...(i) is `-x+y+lambda=0` ...(ii) Since, this equation passes through (2, 3). therefore (2, 3) satisfies the equation of line `:. -2+3+lambda=0` `implies lambda=-1` `:.` From Eq. (ii), `-x+y-1=0` `implies y=x+1` And from eq. (i), `x+x+1-11=0` `implies 2x=10` `implies x=5` Hence, coordinates of foot of perpendicular from (2, 3) to given line is (5, 6) |
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| 34. |
The product of y the perpendiculars from the two points `(pm 4, 0)` to the line `3x cos phi +5y sin phi =15` isA. 25B. 16C. 9D. 8 |
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Answer» Correct Answer - C If the length of perpendicular be `p_(1)` from the point (4, 0) `p_(1)=|(12 cos phi -15)/sqrt((3 cos phi)^(2)+(5 cos phi)^(2))|` `=(15-12 cos phi)/sqrt((3 cos phi)^(2)+(5 cos phi)^(2))` If length of perpendicular be `p_(2)` from the point (-4, 0) `P_(2)=|(-12 cos phi -15)/sqrt((3 cos phi)^(2)+(5 cos phi)^(2))|` `=((12 cos phi +15))/sqrt((3 cos phi)^(2)+(5 sin phi)^(2))` `p_(1).p_(2)=((15-12 cos phi)(12 cos phi +15))/((3 cos phi)^(2)+(5 sin phi)^(2))` `=((225-144 cos^(2) phi))/(9 cos^(2) phi + 25 sin^(2) phi)=(9(25-16 cos^(2) phi))/(25-16 cos^(2) phi)` `=9` |
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| 35. |
Consider the following statements : 1. The equation to a straight line parallel to the axis of x is `y=d`, where d is a constant. 2. The equation to the axis of x is `x=0`. Which of the statement (s) given above is/are correct ?A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2 |
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Answer» Correct Answer - A We know that, the equation of x-axis is `y=0` Statement 1 says that the equation to a straight line parallel to the axis of x is `y=d`. Since, d is constant therefore it can be zero. Thus, only statement 1 is correct. |
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| 36. |
The middle point of the segment of the straight line joining the points `(p, q) and (q,-p)` is `(r/2, s/2).` What is the length of the segment?A. `[(s^(2)+r^(2))^(1//2)]//2`B. `[(s^(2)+r^(2))^(1//2)]//4`C. `(s^(2)+r^(2))^(1//2)`D. `s+r` |
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Answer» Correct Answer - C Two joining points are (p, q) and (q, -p) Mid point of (p, q) and (q, -p) is `((p+q)/2, (q-p)/2)` But it is given that the mid-point is `(r/s, s/2)` `:. (p+r)/2=r/2 and (q-p)/2=s/2` `implies p+q=r` and `q-p=s` Now, length of segment `=sqrt((p-q)^(2)+(q+p)^(2))` (by distance formula) `=sqrt(s^(2)+r^(2))=(s^(2)+r^(2))^(1//2)` |
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| 37. |
If `(a, 0), (0, b)` and `(1, 1)` are collinear, what is `(a+b-ab)` equal to ?A. 2B. 1C. 0D. `-1` |
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Answer» Correct Answer - C Since, `(a, 0), (0, b)` and `(1, 1)` are collinear. `:. |(a,0,1),(0,b,1),(1,1,1)|=0` `implies a(b-1)+1(0-b)=0` `implies ab-a-b=0` `implies ab-a-b=0` `implies a+b-ab=0` |
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| 38. |
If the points A(1, 2), B(2, 4) and C(3, a) are collinear, what is the length BC ?A. `sqrt(2)` unitB. `sqrt(3)` unitC. `sqrt(5)` unitD. 5 unit |
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Answer» Correct Answer - C Since the points are collinear. `|(1,2,1),(2,4,1),(3,a,1)|=0` Expanding the determinant `implies 1|(4,1),(a,1)|-2|(2,1),(3,1)|+1|(2,4),(3,a)|=0` `implies (4-a)-2(2-3)+1(2a-12)=0` `implies 4-a+2+2a-12=0` `implies a-6=0` `implies a=6` Thus, Coordinates of C are (3, 6). Thus, `BC=sqrt((3-2)^(2)+(6-4)^(2))` `=sqrt(1+4)=sqrt(5)` unit |
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| 39. |
What is the product of the perpendiculars from the two points `(pm sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi =ab` ?A. `a^(2)`B. `b^(2)`C. `ab`D. `a//b` |
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Answer» Correct Answer - A Given equation of line is `ax cos phi +by sin phi -ab =0` Let `d_(1)` be the perpendicular distance from `(sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi-ab=0` and `d_(2)` from `(-sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi -ab =0` At point `(sqrt(b^(2)-a^(2)), 0)` `d_(1)=(asqrt(b^(2)-a^(2)) cos phi -ab)/sqrt(a^(2) cos^(2) phi +b^(2) sin^(2) phi)` At point `(-sqrt(b^(2)-a^(2)), 0)` `d_(2)=(-asqrt(b^(2)-a^(2))cos phi-ab)/sqrt(a^(2) cos^(2) phi +b^(2) sin^(2) phi)` `:. d_(1)d_(2)=-([a^(2)(b^(2)-a^(2)) cos^(2) phi-a^(2)b^(2)])/(a^(2) cos^(2) phi+b^(2) sin^(2) phi)` `= - (a^(2) (-b^(2) sin^(2) phi -a^(2) cos^(2) phi))/(a^(2) cos^(2) phi+b^(2) sin^(2) phi)=a^(2)` |
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| 40. |
What is the acute angle between the lines `Ax+By=A+B` and `A( x-y)+B(x+y)=2B` ?A. `45^(@)`B. `tan^(-1) (A/sqrt(A^(2)+B^(2)))`C.D. `60^(@)` |
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Answer» Correct Answer - A Lines are `L_(1) equiv Ax+By=A+B` and `L_(2) equiv A(x-y)+B(x+y)=2B` Slope of `L^(1)` is `- A/B` `m_(1)=-A/B" "[m_(1)" is the side of line "L_(1)]` For line `L_(2)` : `Ax-Ay+Bx+By=2B` `(A+B)x-(A-B)y=2B`. Slope of line `L_(2)` in `((A+B))/(A-B)` `m_(2)=((A+B))/((A-B))" "[m_(2)" is the slope of line "L_(2)]` If angle between line `L_(1)` and `L_(2)` is `theta` then `tan theta=(m_(1)-m_(2))/(1+m_(1)m_(2))` `=(-A/B-(A+B)/(A-B))/(1+(-A/B)xx((A+B)/(A-B)))=((-A(A-B)-B(A+B))/(B(A-B)))/((B(A-B)-A(A+B))/(B(A-B)))` `=(-A^(2)+AB-AB-B^(2))/(AB-B^(2)-A^(2)-AB)=(-B^(2)-A^(2))/(-B^(2)-A^(2))=1` so, `theta=pi/4` |
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| 41. |
If p be the length of the perpendicular from the origin on the straight line `x+2by =2p`.then what is the value of b?A. `1/p`B. `p`C. `1/2`D. `sqrt(3)/2` |
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Answer» Correct Answer - D Length of perpendicular from the origin on the straight line `x+2by-2p=0` is `|(0+2bxx0-2p)/sqrt(1^(2)+(2b)^(2))|=p` or `p=|(-2p)/sqrt(1^(2)+4b^(2))|` or `p^(2)=(4p^(2))/(1+4b^(2))` `4/(1+4b^(2))=1` `implies 1+4b^(2)=4 or 4b^(2)=3` `implies b^(2)=3/4 implies b = pm sqrt(3)/2 implies b = sqrt(3)/2` matches with the given option. |
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| 42. |
In what ratio does the line `y-x+2=0` cut the line joining (3, -1) and (8, 9) ?A. `2 : 3`B. `3 : 2`C. `3 : -2`D. `1 : 2` |
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Answer» Correct Answer - A Let the point of intersection divide the line segment joining points, (3, -1) and (8, 9) in K : 1 ratio then : Then point is `((8k+3)/(k+1), (9k-1)/(k+1))` Since this point lies on the line `y-x+2=0` We have, `(9k-1)/(k+1)-(8k+3)/(k+1)+2=0` `=(9k-1-8k-3)/(k+1)+2=0=(k-4)/(k+1)+2=0` `=k-4+2k+2=0=3k-2=0` `k=2/3 : 1` i.e. `2 : 3` |
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| 43. |
The points (2, -2), (8, 4), (4, 6) and (-1, 1) in order are the vertices of which one of the following quadrilaterals ?A. SquareB. RhombusC. Rectangle (but not square)D. Trapezium |
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Answer» Correct Answer - D Let points be A(2, -2), B(8, 4), C(4, 6) and D(-1, 1) in order and are vertices of a quadrilateral. `AB^(2)=(8-2)^(2)+(4+2)^(2)=36+36=72` `BC^(2)=(4-8)^(2)+(6-4)^(2)=16+4=20` `CD^(2)=(4+1)^(2)+(6-1)^(2)=25+25=50` `AD^(2)=(2+1)^(2)+(-2-1)^(2)=9+9=18` Thus `AB ne BC ne CD ne AD` Hence, quadrilateral is a trapezium. |
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| 44. |
Consider a parallelogram whose vertices are A (1, 2), B (4, y), C (x, 6) and D (3, 5) taken in order. What is the point of intersection of the diagonals ?A. `(7/2, 4)`B. (3, 4)C. `(7/2, 5)`D. (3, 5) |
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Answer» Correct Answer - A Point of intersection (a, b) is `(7/2, 4)`. |
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| 45. |
Consider a parallelogram whose vertices are A (1, 2), B (4, y), C (x, 6) and D (3, 5) taken in order. What is the area of the parallelogram ?A. `7/2` square unitsB. 4 square unitsC. `11/2` square unitsD. 7 square units |
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Answer» Correct Answer - D Area of parallelogram `=2` area of `Delta ADB` `vec(a)=vec(AB)=(4-1) hat(i)+(3-2)hat(j)` `vec(b)=vec(AB)=(3-1) hat(i)+(5-2) hat(j)` `:.` Area of parallelogram `=2[1/2| vec(a)xxvec(b)|]=|vec(a)xxvec(b)|` Now, `vec(a)xxvec(b)=|(hat(i),hat(j),hat(k)),(3,1,0),(2,3,0)|=7 hat(k)` `:.` Area `=|vec(a)xxvec(b)|=|7hat(k)|=sqrt(49)=7` square unit |
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| 46. |
A line passes through the point `(2,2)` and is perpendicular to the line `3x + y =3,` then its `y`-intercept isA. `3/4`B. `4/3`C. `1/3`D. 3 |
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Answer» Correct Answer - B A line passes through (2, 2) and is perpendicular to the line `3x+y=3` Slope of line `3x+y=3` is `-3` Slope of line which passes through (2, 2) is `1/3` `:.` Equation of line passes through (2, 2) and having slope `(1/3)` is `(y-2)=1/3 (x-2)` `:. 3y-6=x-2` `:. x-3y+4=0` In oder to find y-intercept of the line Put `x=0` in `x-3y+4=0` `:. -3y=-4` `:. y=4/3` `:.` Option (b) is correct. |
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| 47. |
Let p, q, r, s be the distances from origin of the points (2, 6), (3, 4), (4, 5) and (-2, 5) respectively. Which one of the following is a whole number ?A. pB. qC. rD. s |
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Answer» Correct Answer - B Let A(2, 6), B(3, 4), C(4, 5) and D(-2, 5) are the given points. Let O be the origin, i.e. O (0, 0) `OA=sqrt((2-0)^(2)+(6-0)^(2))=sqrt(40)=2sqrt(10)` units `OB=sqrt((3-0)^(2)+(4-0)^(2))=sqrt(9+16)=5` units `OC=sqrt((4-0)^(2)+(5-0)^(2))=sqrt(16+25)=sqrt(41)` units `OD=sqrt((-2-0)^(2)+(5-0)^(2))=sqrt(4+25)=sqrt(29)` units So, `q=OB=5` units is the correct answer. |
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| 48. |
The equation to the locus of a point which is always equidistant from the points (1, 0) and (0, -2) is :A. `2x+4y+3=0`B. `4x+2y+3=0`C. `2x+4y-3=0`D. `4x+2y-3=0` |
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Answer» Correct Answer - A Let P(x, y) be the point. Let `A=(1, 0)` and `B=(0, -2)` then `PA=PB` `implies (PA)^(2)=(PB)^(2)`. `implies (x-1)^(2)+y^(2)=x^(2)+(y+2)^(2)`. `implies 1-2x=4y+4` `implies 2x+4y+3=0` |
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| 49. |
The locus of a point equidistant from three collinear points isA. a straight lineB. a pair of pointsC. a pointD. the null set |
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Answer» Correct Answer - D There can not be any point which is equidistant from three collinear points. `:.` Locus = null set. |
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| 50. |
What is the perpendicular distance between the parallel lines `3x+4y=9` and `9x+12 y+28=0` ?A. `7/3` unitsB. `8/3` unitsC. `10/3` unitsD. `11/3` units |
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Answer» Correct Answer - D The given lines are :- `3x+4y=9` `implies y=9/4-3/4 x` ...(i) and `9x+12y+28=0 implies y= (-7)/3-(3x)/4` ...(ii) We have, `m=(-3)/4, C_(1)=9/4, C_(2)=(-7)/3` Now, Distance `=(|C_(1)-C_(2)|)/sqrt(1+m^(2))=(|9/4-(-7/3)|)/sqrt(1+9/16)=11/3` units |
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