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QUESTION: 1

Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)

Solution:

AT FIRST FIND THE COMMON PONT OF THIS FAMILY WHICH IS A (-1/3 , 1/3)

THEN FOR THE LINE TO BE AT THE FARTHEST DISTANCE FROM B(1,-3) THE LINE SHOULD BE PREPENDICULAR TO AB THEREFORE THE SLOPE OF LINE m= 2/5

EQUATION OF LINE IS 2x + 2/3 = 5y - 5/3 WHICH IS HENCE 6x-15y-7=0

QUESTION: 2

One side of an equilateral triangle is 3x+4y=7 and its vertex is (1,2). Then the length of the side of the triangle is

Solution:

QUESTION: 3

Find the equation of the bisector of the angle between the lines x+2y–11 = 0, 3x–6y–5 = 0 which contains the point (1,–3)

Solution:

**x+2y−11=0 ..(1)**

**3x−6y−5=0 ..(2)**

**Point of intersection of (1) and (2)**

**x=19/3**

**y=7/3**

**Equation passing through (19/3,7/3) and (1,-3)**

**y-7/3=1616(x-19/3)**

**x-y-4=0**

QUESTION: 4

A straight line which make equal intercepts on +ve x and y axes and which is at a distance '1' unit from the origin intersects the straight line y = 2x + 3 + √2 at (x_{0}, y_{0}) then 2x_{0} + y_{0}

Solution:

Equation of the straight line having equal intercepts is x + y = k and proceed.

QUESTION: 5

The locus of the image of the point (2, 3) in the line (x -2y +3) + λ (2x -3y + 4) = 0 (λ∈R) is

Solution:

A(2,3),B(1,2),P(x,y)

AB = BP

QUESTION: 6

A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q. As L varies, the absolute minimum value of OP+OQ is (O is origin)

Solution:

The equation of the line L be x coordinates of P and Q are 3x + 4y = 9 and y = mx + 1 and P (-3, 4) .

So Q (0, 1)

So, absolute minimum value of x

QUESTION: 7

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 = mx + 1 is also an integer is

Solution:

Solving given equations we get is an integer, if

(-2, 0) y = mx So m has two integral values.

QUESTION: 8

Area of the parallelogram formed by the lines y = mx, y = mx+ 1, y = nx and y = nx+ 1 equals (m ≠ n)

Solution:

If be the distance between parallel sides and θ be the angle between adjacent sides, then

Requried area Whre y = mx + 1, x (distance between | | lines) and π/4 (0,3)

QUESTION: 9

If the midpoint of sides of a triangle are (a, 0), (0, b) and (-a, -b), then the centroid of the triangle is

Solution:

QUESTION: 10

The triangle ABC has medians AD, BE, CF. AD lies along the line y = x + 3, BE lies along the line y = 2x + 4, AB has length 60 and angle C = 90°, then the area of ΔABC is

Solution:

QUESTION: 11

Let A(1, 0), B(0, –2) and P(a, b) be a third point that lies on the straight line 'x - y= 0' such that, |PA + PB| is minimum then the values of |a + b| =

Solution:

QUESTION: 12

Let A(0, 0), B(3, 0), C(2, 1) then area of the square inscribed in the triangle such that, one side of the square is taken along

Solution:

QUESTION: 13

Let O(0, 0), A(–1, 1) the point ‘A’ is rotated 60^{o} in anticlockwise sense about origin then the slope of line joining original and the new positions of ‘A’ is

Solution:

QUESTION: 14

Let y = 2x + c_{1}, y = 2x + c_{2} meet X-axis at A_{1}, A_{2} respectively and Y-axis at B_{1}, B_{2} respectively then the locus of the point of intersection of lines

Solution:

QUESTION: 15

Let A(2, 3) and B is the image of A with respect to the lines (2x - 3y + 4) + k (x - 2y + 3) = 0 then the locus of B is

Solution:

The given line is

(2x - 3y + 4) + k (x - 2y + 3) = 0

as x−2y=−3

2x−3y=−4

multiplying eqn(2) by 2 and subtracting we get, y=2

so, x=−3+4=1

so point of intersection P is (1,2)

A′ is the reflection of A

as, AP=BP=A′P

so, √(2-1)^{2}+(3-2)^{2}] =[(h-1)^{2}+(k-2)^{2}]^{1/2}

1+1=h^{2}+k^{2}-2h-4k+1+4

h^{2}+k^{2}−2h−4k+3=0

x^{2}+y^{2}-2x-4y+3=0

QUESTION: 16

Let 'P' be a point inside the equilateral triangle ABC, PA = 3, PB = 4, PC = 5 then the area of the triangle ABC (in square units)

Solution:

Rotate the triangle by 60^{o}

AC^{2} = AP^{2} + PC^{2} - 2AP.PCcos(θ + 60°)

QUESTION: 17

Let a straight line passing through P(1, 4) with negative slope cuts the coordinate axes at A, B then the area of the triangle OAB when OA + OB is minimum is

Solution:

OA + OB = 1 + 4cot θ + 4 + tan θ

= 5 + tan θ + 4 cot θ

Attains minima only if tan θ = 4 cot θ

QUESTION: 18

The image of the curve with respect to the straight line x + y= 1 is

Solution:

Replace x by 1 – Y

y by 1 – X

QUESTION: 19

Let (a, b) lies on obtuse angle bisector of x + 2y + 1 = 0, 2x + y + 1 = 0 then the least value of (a - 1)^{2} + (b - 2)^{2} is

Solution:

Find short dist. From (1, 2)

QUESTION: 20

Let A(0, 0) and x + y - 1 = 0, 2x + 1 = 0 are the altitude, median from the vertices B, C respectively of ΔABC then the area of the triangle

Solution:

QUESTION: 21

Let (x - 1)^{2} + (y - 2)^{2} = 1/4 then the minimum value of " x + 3y+ 4"

Solution:

QUESTION: 22

Two vertical poles 20 mts, and 80 mts stand apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of others (in metres)

Solution:

The height "a" of the point E of intersection of the lines AD & BC drawn from the tops B & D of two poles AB & CD , to the feet of the other pole is given by LADDER THEOREM .

1/a = 1/AB + 1/CD

1/a = 1/20 + 1/80

= 5/80 = 1/16

So. a = 16 m

QUESTION: 23

A(3, 1), B(13, 6), C(13, 21), D(3, 16) forms a parallelogram, then the y-intercept of the line passing through (1, 0) and dividing the parallelogram into two equal areas

Solution:

QUESTION: 24

If the straight lines ax + by + p = 0 and x cosa + y sina = p are inclined at an angle p/4 and concurrent with the straight line x sina – y cosa = 0, then the value of a^{2} +b^{2} is

Solution:

ax+by+p=0 ...(1)

⇒by=−ax+p

⇒y=−a/bx+p/b

So, Slope,m1 = −a/b

and xcosθ+ysinθ−p=0 ..(2)

⇒ysinθ=−xcosθ+p

⇒y=−cosθ/sinθx+p/sinθ

So, Slope,m_{2}=−cosθ/sinθ

Now, straight lines (1) and (2) include an angle π/4

i,e tanπ/4 = ∣∣(m_{1}−m_{2})/(1+m_{1}m_{2})∣∣

⇒ 1 = ∣∣(m_{1}−m_{2})/(1+m_{1}m_{2})∣∣

⇒1+m_{1}m_{2}|=|m_{1}−m_{2}|

⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣

=∣∣−a/b−(−cosθ/sinθ)∣∣

⇒∣∣(bsinθ+acosθ)/bsinθ∣∣

=∣∣(−asinθ+bcosθ)/bsinθ∣∣

⇒|bsinθ+acosθ|=|−asinθ+bcosθ|

Squaring both sides, we get

b^{2}sin^{2}θ+a^{2}cos^{2}θ+2absinθ cosθ=a2sin2θ+b2cos2θ−2absinθ cosθ ...(3)

Given three lines are concurrent

Now, we will find the point of intersection of

xcosθ+ysinθ=p and xsinθ−ycosθ=0

which is x=pcosθ and y=psinθNow, point of intersection will satisfy (1)

i.e apcosθ+bpsinθ+p=0

⇒apcosθ+bpsinθ=−p

⇒acosθ+bsinθ=−1

Squaring both sides, we get

⇒a^{2}cos^{2}θ+b2sin^{2}θ+2absinθ cosθ=1 ...(4)

Put the value of (4) in (3), we get

a^{2}sin^{2}θ+b^{2}cos^{2}θ−2absinθ cosθ =1 ....(5)

Adding (4) and (5), we get

a^{2}cos^{2}θ+b^{2}sin^{2}θ+2absinθ cosθ+a^{2}sin2θ+b^{2}cos^{2}θ−2absinθ cosθ=1+1

⇒a^{2}(sin^{2}θ+cos^{2}θ)+b2(sin^{2}θ+cos^{2}θ)=2

⇒a^{2}+b^{2}=2

QUESTION: 25

Let (1, 2), (3, 8), (4, 1) are three vertices of a parallelogram the sum of all possible x coordinates of the fourth vertex of the parallelogram is

Solution:

Let ABCD be a parallelogram.

then A (1,2), B (3,8) and C (4,1). Let the coordinates of D be (x,y)

Since the diagonals of a parallelogram bisect each other

Coordinates of the mid-point of AC = Coordinates of the mid-point of BD

(1+4)/2 , (2+1)/2 = (3+x)/2 , (8+y)/2

5/2 , 3/2 (3+x)/2 , (8+y)/2

2.5, 1.5 , (3+x)/2 , (8+y)/2

(3+x)/2 = 2.5 , 3/2 = (8+y)/2

(3+x) = 5 , -5 = y

2 = x , -5 = y

∴ Coordinates of the fourth vertex are (2, -5)

QUESTION: 26

The orthocenter of the triangle formed by the lines x - 7y + 6 = 0, 2x - 5y - 6 = 0 and 7x + y - 8 = 0 is

Solution:

x-7y+6 = 0

m1 = -1/-7 = 1/7

7x+y-8 = 0

m2 = -7/1

Slope = 1/7 * (-7)

= 1

B is orthocentre of triangle ABC

x-7y = -6 ---------------------------(1)

7x+y = 8 ---------------------------(2)

y =8-7x ----------------------------(3)

Put eq (3) in eq(1)

x-7(8-7x) = -6

x-56+49x = -6

50x = 50

x = 1

Similarly y = 1

Points are (1,1)

QUESTION: 27

The equations x = t^{3} + 9 and represents a straight line where t is a parameter. Then y-intercept of the line is :

Solution:

x=0

Put into x= t^{3} + 9

0 = t^{3} + 9

t^{3} = -9

y = 3(t^{3})/4 + 6

y = 3(-9)/4 + 6

y= -¾

QUESTION: 28

The vertices of triangle ABC are A(–1,–7), B(5, 1) and C(1, 4). The equation of the bisector of the angle ABC of ΔABC is :

Solution:

QUESTION: 29

y - 1 = m_{1} x - 3 and y - 3 = m_{2} (x - 1) are two family straight lines (m_{1}, m_{2}R), at right angles to each other. The locus of their point of intersection is :

Solution:

Locus of P (h, k) is circle with AB as diameter

(x -3)(x -1) + (y -1)(y -3) = 0

QUESTION: 30

The equation of the line passing through the intersection of the lines 3x + 4y = -5, 4x + 6y = 6 and perpendicular to 7x - 5y + 3 = 0 is :

Solution:

Equation of altitude on BC

x + 4y= 13

Equation of altitude on AB

7x - 7y + 19 = 0

⇒

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