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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.
| 351. |
Consider a triangle PQR with coordinates of its vertices as P(-8,5), Q(-15, -19), and R (1, -7). The bisector of the interior angle of P has the equation which can be written in the form ax+2y+c=0. The radius of the in circle of triangle PQR is The radius of the circle of triangle PQR isA. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - B The coordinates of the incenter are given as follows: `bar(x) = (20(-8)+15(-15)+25(1))/(20+15+25) = -6` `bar(y) = (20(5)+15(-19)+25(-7))/(60) = -6` Hence, incenter I is (-6, -6). Now, the equation of side PR is `y+7 = (12)/(-9)(x-1)` or 4x-4 = -3y-21 or 4x+3y+17 = 0 Inradius is given as the distance of I from side PR, i.e., `(|-24-18+17|)/(5) = 5` |
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| 352. |
Let ABCD be a parallelogram the equation of whose diagonals are `AC : x+2y =3`; BD: 2x + y = 3. If length of diagonal` AC =4` units and area of `ABCD = 8` sq. units.(i) The length of the other diagonal is (ii) the length of side AB is equal toA. `2sqrt(10)//3`B. `4sqrt(10)//3`C. `8sqrt(10)//3`D. none of these |
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Answer» Correct Answer - A In `DeltaCPB,` `"cos " theta = (PC^(2) + PB^(2)-BC^(2))/(2PC xx PB)` `"or "BC = (2sqrt(10))/(3)` |
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| 353. |
In a rhombus ABCD the diagonals AC and BD intersect at the point (3,4) . If the point A is (1,2) the diagonal BD has the equationA. x - y - 1 = 0B. x + y - 1 = 0C. x - y + 1 = 0D. x + y - 7 =0 |
| Answer» Correct Answer - D | |
| 354. |
The straight lines `x+y=0, 3x+y-4=0 and x+3y-4=0` form a triangle which is (A) isosceles (B) right angled (C) equilateral (D) scaleneA. isoscelesB. equilateralC. right angledD. none of these |
| Answer» Correct Answer - A | |
| 355. |
Find the equation of the line which cuts off intercepts 4 and -6 on the x-axis and y-axis respectively. |
| Answer» Correct Answer - `3x-2y-12=0` | |
| 356. |
Find the equation of the line which passes through the point (3, -5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign. |
| Answer» Correct Answer - x-y-8=0 | |
| 357. |
In what ratio, the line joining `+(-1,1)a n d(5,7)`is divided by the line `x+y=4?`A. `3 : 2`B. `2 : 3`C. `1 : 2`D. `2 :1` |
| Answer» Correct Answer - C | |
| 358. |
If the line segment joining (2,3) and (-1,2) is divided internally in the ratio 3:4 by the line `x+2y=lambda` , then `lambda`=A. `(41)/(7)`B. `(5)/(7)`C. `(36)/(7)`D. `(31)/(7)` |
| Answer» Correct Answer - A | |
| 359. |
If each of the points `(x-1,4),(-2,y_1)`lies on the line joining the points `(2,-1)a n d(5,-3)`, then the point `P(x_1,y_1)`lies on the line.`6(x+y)-25=0``2x+6y+1=0``2x+3y-6=0`(d) `6(x+y)+25=0`A. x = 3yB. x = - 3yC. y = 2x + 1D. 2x + 6y + 1 = 0 |
| Answer» Correct Answer - D | |
| 360. |
Find the length of perpendicular from the origin to each of the following (i)7x+24y=50 (ii)4x+3y=9 (iii)x=4 |
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Answer» Correct Answer - (i)2 units (ii)`(9)/(2)`units (iii)4 units The given line is x+0. y-4=0 |
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| 361. |
Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (4,7). |
| Answer» Correct Answer - x+y-11=0 | |
| 362. |
A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is :A. `x+y=7`B. `3x-4y+7=0`C. `4x+3y=24`D. `3x+4y=25` |
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Answer» Correct Answer - C Let the equation of the line be `(x)/(a)+(y)/(b)=1` . It cuts the coordinates axes at P(a,0) and Q(0,b) . It is given that PQ is bisector at A(3,4). `therefore (a+0)/(2)=3 and (0+b)/(2)=4 implies a=6,b=8`. Hence, the equation of the line is `(x)/(6)+(y)/(8)=1` or , `4x+3y=24` . |
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| 363. |
Find the perpendicular distance between the lines `3x+4y+9=0` and to `6x+8y+15=0` isA. `3//2`B. `3//10`C. 6D. none of these |
| Answer» Correct Answer - B | |
| 364. |
Find the distance of the point (4, 2) from the line joining the points (4, 1) and (2,3). |
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Answer» Correct Answer - `(sqrt2)/(2)` units Equation of the line joining the points A(4,1) and B(2,3) is `(y-1)/(x-4)=(3-1)/(2-4),i.e. x+y-5=0` |
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| 365. |
A straight line l with negative slope passes through (8,2) and cuts the coordinate axes at P and Q. Find absolute minimum value of 'OP+OQ`` where O is the origin- |
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Answer» Correct Answer - 18 Let the equation of line be (y-2) = m(x-8) where `m lt 0.` `rArr P-= (8-(2)/(m), 0)" and " Q -=(0 ,2-8m)` `"Now, "OP+OQ =|8-(2)/(m)| + |2 -8m|` `= 10+(2)/(-m) + (-8m)` `ge 10+2sqrt((2)/(-m) xx(-8m)) ge 18` |
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| 366. |
Reduce the equation 5x-12y=60 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes. |
| Answer» Correct Answer - `(x)/(12)+(y)/(-5)=1,13"units"` | |
| 367. |
What is the equation of the straight line which is perpendicular to `y=x` and passes through (3, 2) ?A. `x-y=5`B. `x+y=5`C. `x+y=1`D. `x-y=1` |
| Answer» Correct Answer - B | |
| 368. |
Find the distance of the point (3,-5) from the line 3x-4y=27. |
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Answer» Correct Answer - `(2)/(5)"units"` The givenline is 0. x+1. y-4=0 |
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| 369. |
Reduce the equation 3x-4y+12=0 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes. |
| Answer» Correct Answer - `(x)/(-4)+(y)/(3)=1,5"units"` | |
| 370. |
One of the diagonals of a square is the portion of the line x/2+y/3=2 intercepted between the axes. Then the extremitites of the other diagonal areA. (5,5), (-1,1)B. (0,0), (4,6)C. (0,0),(-1,1)D. (5,5),(4,6) |
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Answer» Correct Answer - A The extremities of the given diagonal are (4,0) and (0,6) Hence, the slope of this diagonal is -3/2 and the slope of other diagonal is 2/3. The equation of the other diagonal is `(x-2)/(3//sqrt(13)) = (y-3)/(2//sqrt(13)) =r` For the extremities of the diagonal, `r =+-sqrt(13).` Hence, `x-2 = +-3, y-3=+-2` x=5,-1 and y5,1 Therefore, the extremities of the diagonal are (5,5) and (-1,1). |
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| 371. |
If the straight line `(x)/(a)+(y)/(b)=1` passes through the points (8,9) and (12,-15) find the values of a and b. |
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Answer» Correct Answer - a=2,b=3 Since `(x)/(a)+(y)/(b)=1` passes through the points A(8,-9) and B(12,-15) we have `(8)/(a)-(9)/(b)=1 and (12)/(a)-(15)/(b)=1` On solving we get `(3)/(b)=1 and (8)/(a)=4`. This gives a=2,b=3. |
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| 372. |
A straight line passes through the point (-5,2) and the portion of the line intercepted between the axes is divided at this point in the ratio 2:3. Find the equation of the line. |
| Answer» Correct Answer - 3x-5y+25=0 | |
| 373. |
Find the equation of the line which passes through the point (22,-6) and whose intercept on the x-axis exceeds the intercept on the y-axis by 5. |
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Answer» Correct Answer - 6x+11y-66=0 or x+2y-10=0 Let the required equation be `(x)/(a+5)+(y)/(a)=1` Since it passes through (22,-6), we have `(22)/(a+b)-(6)/(a)=1` This gives `a^(2)-11a+30=0 Rightarrow (a-6)(a-5)=0 Leftrightarrow a=6 or a=5` So, the required equations are `(x)/(11)+(y)/(6)=1 or (x)/(10)+(x)/(5)=1` |
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| 374. |
Find the equation of the line whose portion intercepted between the coordinate axes is divided at the point (5, 6) in the ratio 3:1. |
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Answer» Correct Answer - 2x+5y-40=0 Let the required equations be `(x)/(a)+(y)/(b)=1`. Then, it cuts the axes at A(a,0) and B(0,b) Let P(5,6) divides AB in the ratio 3:1. `"Then",(3xx0+1xxa)/(3+1)=5 and (3xxb+1xx0)/(3+1)=6 Leftrightarrow a=20 and b=8` |
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| 375. |
Find the equation of the line whose portion intercepted between the axes is bisected at the point (3,-2) |
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Answer» Correct Answer - 2x-3y-12=0 Let the required be `(x)/(a)+(y)/(b)=1` Then, it cuts the axes at A(a,0) and B(0,b). `therefore ((a+0)/(2)=3 and (0+b)/(2)=-2) Leftrightarrow a=6,b=-4` |
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| 376. |
If PM is the perpendicular from P(2,3) on the line `x+y=3`, then the coordinate of M areA. (2,1)B. (-1,4)C. (1,2)D. (4,-1) |
| Answer» Correct Answer - C | |
| 377. |
A ray of light coming fromthe point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5,3). The coordinates of the point A is :A. (13/5,0)B. (5/13,0)C. (-7,0)D. none of these |
| Answer» Correct Answer - A | |