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51.

What is the equation of the line which passes through (4, -5) and is perpendicular to `3x+4y+5=0` ?A. `4x-3y-31=0`B. `3x-4y-41=0`C. `4x+3y-1=0`D. `3x+4y+8=0`

Answer» Correct Answer - A
`3x+4y+5=0` or `y=(-3)/4 x+ (-5)/4`
Slope `=(-3)/4`
Slope of required line, `m=(-1)/(- 3/4)=4/3`
Also line passes through `(4, -5)`
Equation of line, `y+5=4/3 (x-4)`
`implies 3y+15=4x-16`
`implies 4x-3y-31=0`
Discussion
52.

A points `P` moves such that its distances from `(1,2)` and `(-2,3)` are equal. Then, the locus of `P` isA. straight lineB. ParabolaC. ellipseD. hyperbola

Answer» Correct Answer - A
Let moving point be `P(x, y)`
`sqrt((y-2)^(2)+(x-1)^(2))=sqrt((y-3)^(2)+(x+2)^(2))`
`implies (y+2)^(2)+(x-1)^(2)=(y-3)^(2)+(x+2)^(2)`
`implies y^(2)+4+4y+x^(2)+1-2x=y^(2)+9-6y+x^(2)+4x+4`
`implies 10y-6x-8=0`
`:.` Locus of P is a straight line.
Discussion
53.

For what value of k are the two straight lines `3x+4y=1` and `4x+3y+2k=0` equidistant from the point (1, 1) ?A. `1/2`B. `2`C. `-2`D. `- 1/2`

Answer» Correct Answer - D
`d_(1)=|(3xx1+4xx1-1)/sqrt(3^(2)+4^(2))|=6/5`
`d_(2)=|(4xx1+3xx1+2K)/sqrt(3^(2)+4^(2))|=(7+2K)/5`
`d_(1)=d_(2)`
`7+2K=6`
`k=-1/2`
Discussion
54.

What is the equation of the line joining the origin with the point of intersection of the lines `4x+3y=12` and `3x+4y=12` ?A. `x+y=1`B. `x-y=1`C. `3y=4x`D. `x=y`

Answer» Correct Answer - D
The equation of given lines are
`4x+3y=12` ...(i)
and `3x+4y=12` ...(ii)
On simplifying (i) and (ii), we get
`x=12/7 and y=12/7`
`:.` Point of intersection of given line is `(12/7, 12/7)`. Hence, the equation of line passing through (0, 0) and `(12/7, 12/7)` is
`(y-0)/(x-0)=(12/7-0)/(12/7-0) implies y=x`
Discussion
55.

An equilateral triangle has one vertex at (0, 0) and another at `(3, sqrt(3))`. What are the coordinates of the third vertex ?A. `(0, 2sqrt(3))` onlyB. `(3, sqrt(3))` onlyC. `(0, 2sqrt(3)) or (3, -sqrt(3))`D. Neither `(0, 2sqrt(3)) nor (3, -sqrt(3))`

Answer» Correct Answer - C
Let ABC is equilateral triangle with A(0, 0) and `B(3, sqrt(3))` and C to be known.
`:. AB=sqrt((3-0)^(2)+(sqrt(3)-0)^(2))=sqrt(9+3)=sqrt(12)`
Take option (a) i.e. `C (0, 2 sqrt(3))`
`CA=sqrt(0^(2)+(2sqrt(3))^(2))=sqrt(12)`
`CB=sqrt((3)^(2)+(sqrt(3))^(2))=sqrt(12)`
Take option (b) i.e. `C (3, -sqrt(3))`
`CA=sqrt(3^(2)+(sqrt(3))^(2))=sqrt(12)`
`CB=sqrt((0)^(2)+(2sqrt(3))^(2))=sqrt(12)`
`:.` Both option (a) and (b) are correct.
Discussion
56.

Consider the two lines `x+y+1=0` and `3x+2y+1=0` What is the equation of the line passing through the point of intersection of the given lines and parallel to y-axis ?A. `x+1=0`B. `x-1=0`C. `x-2=0`D. `x+2=0`

Answer» Correct Answer - B
Equations of lines
`x+y+1=0`
`3x+2y+1=0`
`{:(3x+3y+3=0),(3x+2y+1=0),("- - - -"),(bar(" y = -2 ")),(" "x=1):}`
Points of intersection (1, -2)
Equation of y-axis
`x=0`
Equation of line parallel to y-axis is
`x=k`
If this line passes through (1, -2 then)
`x=1`
Hence equation of line which passes through point of intersection of given line (1, -2) and parallel to y-axis `x=1`
`implies x-1=0`
Discussion
57.

Consider the two lines `x+y+1=0` and `3x+2y+1=0` What is the equation of the line passing through the point of intersection of the given lines and parallel to x-axis ?A. `y+1=0`B. `y-1=0`C. `y-2=0`D. `y+2=0`

Answer» Correct Answer - D
Equations of lines
`x+y+1=0`
`3x+2y+1=0`
`{:(3x+3y+3=0),(3x+2y+1=0),("- - - -"),(bar(" y = -2 ")),(" "x=1):}`
Points of intersection (1, -2)
Equation of x-axis
`y=0`
Line parallel to x axis is
`y=k`
If this line passes through (1, -2) then
`k=-2`
`implies y=-2`
`implies y+2=0`
Equation of line passing through (1, -2) and parallel to x-axis is
`y+2=0`
Discussion
58.

If (a, b) is at unit distance from the line `8x+6y+1=0`, then which of the following conditions are correct ? 1. `3a-4b-4=0` 2. `8a+6b+11=0` 3. `8a+6b-9=0` Select the correct answer using the code given below :A. 1 and 2 onlyB. 2 and 3 onlyC. 1 and 3 onlyD. 1, 2 and 3

Answer» Correct Answer - B
Here `(|8a+6b+1|)/sqrt(8^(2)+6^(2))=1 implies |8a+6b+1|=10`
`implies 8a+6b+1= pm 10`
`implies 8a+6b+1=10` & `8a+6b+1=-10`
`implies 8a+6b-9=0` & `8a+6b+11=0`
Discussion
59.

What is the equation of the straight line which passes through the point of intersection of the straight lines `x+2y=5` and `3x+7y=17` and is perpendicular to the straight line `3x+4y=10` ?A. `4x+3y+2=0`B. `4x-y+2=0`C. `4x-3y-2=0`D. `4x-3y+2=0`

Answer» Correct Answer - D
Intersecting lines are : `x+2y=5` & `3x+7y=17`
On solving these we get : `x=1` & `y=2`
Equation of perpendicular line is
`3x+4y=10 or y=(-3)/4x+10`
So, slope `=(-3)/4`
`implies` Slope of required line `=4/3`
`:.` Equation of given line is
`(y-2)=4/3 (x-1) or 4x-3y+2 =0`
Discussion
60.

What is the equation of the straight line parallel to `2x+3y+1=0` and passes through the point (-1, 2) ?A. `2x+3y-4=0`B. `2x+3y-5=0`C. `x+y-1=0`D. `3x-2y+7=0`

Answer» Correct Answer - A
The equation of line parallel to
`2x+3y+1=0` is `2x+3y+K=0`
It is passing through point (-1, 2)
`:. 2(-1)+3(2)+K=0`
`implies -2+6+K= 0 implies K=-4`
`:.` Eqn. is `2x+3y-4=0`
Discussion
61.

The angle between the lines `y=(2-sqrt3)x+5 and y=(2+sqrt3)x-7` isA) `30^@` B) `60^@` C) `45^@`A. `60^(@)`B. `45^(@)`C. `30^(@)`D. `15^(@)`

Answer» Correct Answer - A
The given lines are
`y=(2-sqrt(3))x+5` and `y=(2+sqrt(3))x-7`
Therefore, slope of first line `m_(1)=2-sqrt(3)` and
slope of second line `m_(2)=2+sqrt(3)`
`:. tan theta=|(m_(2)-m_(1))/(1+m_(1)m_(2))|=|(2+sqrt(3)-2+sqrt(3))/(1+(4-3))|`
`=|(2sqrt(3))/2|=sqrt(3)="tan" pi/3 implies theta=pi/3=60^(@)`
Discussion
62.

What is the minimum value of `a^(2)x+b^(2)y` where `xy=c^(2)` ?A. abcB. 2abcC. 3abcD. 4abc

Answer» Correct Answer - B
Let `p=a^(2)x+b^(2)y` and `xy=c^(2)`
`implies y=c^(2)/x` ...(1)
`implies P=a^(2) x+b^(2) (c^(2)/x)`
Now, `(dp)/(dx)=0implies a^(2)- (b^(2) c^(2))/x^(2)=0`
`:. Y=c^(2)/((bc)/a)=(ac^(2))/(bc)=(ac)/b`
`implies a^(2)=(b^(2)c^(2))/x^(2)`
`implies x=(bc)/a :. P_("min")=a^(2)((bc)/a)+b^(2) ((ac)/b)`
`=abc+abc=2abc`.
Discussion
63.

the lines `(p+2q)x+(p-3q)y=p-q` for different values of `p&q` passes trough the fixed point is:A. `(3/2, 5/2)`B. `(2/5, 2/5)`C. `(3/5, 3/5)`D. `(2/5, 3/5)`

Answer» Correct Answer - D
As given As given, `(p+2q)x+(p-3q)y=p-q`
`implies px+2qx+py-3qy=p-q`
`implies p(x+y)-q(3y-2x)=p-q`
Equation co-efficient of p and q
`implies x+y=1 and 3y-2x=1`
Solving these, we get
`x=2/5, y=3/5`.
So, line passes through `(2/5, 3/5)`.
Discussion