JEE Advanced Level Test: Straight Line- 3 - EmSAT Achieve MCQ

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

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

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

A(2,3),B(1,2),P(x,y)
AB = BP

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

Solving given equations we get  is an integer, if
(-2, 0) y = mx So m has two integral values.

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) 

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=h²+k²-2h-4k+1+4
h²+k²−2h−4k+3=0
x²+y²-2x-4y+3=0

Rotate the triangle by 60^o


AC² = AP² + PC² - 2AP.PCcos(θ + 60°)

OA + OB = 1 + 4cot θ + 4 + tan θ
= 5 + tan θ + 4 cot θ
Attains minima only if tan θ = 4 cot θ

Replace x by 1 – Y
y by 1 – X

Find short dist. From (1, 2)

 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

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,m₂=−cosθ/sinθ
Now, straight lines (1) and (2) include an angle π/4
i,e tanπ/4 = ∣∣(m₁−m₂)/(1+m₁m₂)∣∣
⇒ 1 = ∣∣(m₁−m₂)/(1+m₁m₂)∣∣
⇒1+m₁m₂|=|m₁−m₂|
⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣
=∣∣−a/b−(−cosθ/sinθ)∣∣
⇒∣∣(bsinθ+acosθ)/bsinθ∣∣
=∣∣(−asinθ+bcosθ)/bsinθ∣∣
⇒|bsinθ+acosθ|=|−asinθ+bcosθ|
Squaring both sides, we get
b²sin²θ+a²cos²θ+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
⇒a²cos²θ+b2sin²θ+2absinθ cosθ=1   ...(4)
Put the value of (4) in (3), we get
a²sin²θ+b²cos²θ−2absinθ cosθ =1   ....(5)
Adding (4) and (5), we get
a²cos²θ+b²sin²θ+2absinθ cosθ+a²sin2θ+b²cos²θ−2absinθ cosθ=1+1
⇒a²(sin²θ+cos²θ)+b2(sin²θ+cos²θ)=2
⇒a²+b²=2

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)

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)

x=0
Put into x= t³ + 9
0 = t³ + 9
t³ = -9
y = 3(t³)/4 + 6
y = 3(-9)/4 + 6
y= -¾

Locus of P (h, k) is circle with AB as diameter 
(x -3)(x -1) + (y -1)(y -3) = 0

Equation of altitude on BC
x + 4y= 13
Equation of altitude on AB
7x - 7y + 19 = 0
⇒