JEE Advanced Level Test: Straight Lines- 1 - EmSAT Achieve MCQ

 let the two vectors be a and b,
According to the question,
|a-b| = |a+b|,
from parallelogram law of vectors, we know that,
|p±q| = √(p² + q² ± 2pq cos ∆)
where, ∆ = angle between the vectors p and q,
=> √(a²+b²+2abcos∆) = √(a²+b²-2abcos∆)
squaring on both sides and cancelling a and b both sides,
=> 2abcos∆ = -2abcos∆
=> 4abcos∆ = 0,
|a| ≠ |b| ≠ 0,
=> cos∆ = 0,
=> ∆ = odd integral multiples of 90°,
=> ∆ = 90°

We know that
If a line joining point A(a,b) and B (c,d) is divided by point O(x,y)
In ratio m:n
Then X = (m×c + n a) / (m+n)
And Y = (m×d + n× b) / (m+n)
Here we have point lies in between A(3,-4)and B(-5,6) is O (x,0)
Let the line is divided by point O in ratio of k:1
Then
⇒ 0 = (k × 6 + 1×-4 ) / ( k+1)
⇒ 6k - 4 = 0
⇒ 6k = 4
⇒ K = 4/6
 = 2/3

Step-by-step explanation:

The given points when plotted show a right angled triangle

Since we know that the circumcentre of a right angled triangle lies on the midpoint of the hypotenuse, using section formula:

C=(3/2,4/2) => (3/2,2)

The slope-intercept form of a straight line  is y=mx+c
Here, m = tanθ = 3/5, c=−3
Writing an equation with given values we get
y=3/5​
x−3⇒5y=3x−15
⟹5y−3x+15=0 is the required equation of a line.

Let the equation of straight line be x/A+ y/B=1
Where A is x intercept and B is y intercept 
And coordinate of point which divide portion between axes in 5:3 is 
(3A/8 , 5B/8)
As this line passes through (-3,5), then 
A=B=8 
Hence, equation of line is x-y+8=0

A(1,2) and B(−2,0) are given point 
∴ midpoint p=((1+(−2))/2, (2+0)/2)
∴ midpoint p=(−1/2,1)
Also slope of AB = [0−(2)]/(-2-1)   = −2/-3​
∴ slope of AB=2/3
slope of line perpendicular to AB = (−1/(2/3)) = -3/2
∴ equation of perpendicular bisector
y−1=−3/2(x+1/2)
∴ y−1=−3/2(x+1/2)
∴ 2y−2=−3(x+1/2)
∴ 4y−4=−3(2x+1)
∴ 4y−4=−6x−3
∴ 4y+6x−4+3=0
6x+4y−1=0
∴ equation of line 6x+4y−1 = 0

Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.

Equation of line parallel to ax+by+c=0 is ax+by+k=0 where k is constant.
As this line passes through (c,d) then
ac+bd+k=0⇒k=−ac−bd
Therefore equation of line is a(x−c)+b(y−d)=0

We haveP(8,−9)
L1: 2x+3y−4=0
L2: 6x+9y+8=0
L1(P)=16−27−4=−15
L2(P)=48−81+8=−25
L1(P)×L2(P)=(−15)×(−25)>0 [Product will be positive]
As L1(P)×L2(P)>0, hence
Point lies on same side of line.

L₁ : x + y = 5, L₂ : y – 2x = 8
L₃ : 3y + 2x = 0, L₄ : 4y – x = 0
L₅ : (3x + 2y) = 6
vertices of quadrilateral
0(0, 0), A (4, 1), B (–1, 6), C(–3, 2)
L5 (0) = – 6 < 0 L5 (A) = 12 + 2 – 6 = 8 > 0
L5 (B) = – 3 + 12 – 6 = 3 > 0
L5 (C) = –9 + 4 – 6 = – 11 < 0
O & C points are same side
& A & B points are other same side w.r.t to L5
So L5 divides the quadrilateral in two quadrialteral
Aliter:
If abscissa of A is less then abscissa of B
⇒ Alies left of B
otherwise A lies right of B

Let S(x, y), then
(x+1)² + y² + (x-2)² + y² = 2[(x-1)² + y²
2x + 1 + 4 -4x
= -4x + 2
x = -3/2
Hence it is a straight line parallel to y-axis.

The given equation of line is 

(p+2q)x+(p−3q)y=p−q

px+2qx+py−3qy−p+q=0

p(x+y−1)+q(2x−3y+1)=0

For different value of p the given equation can be 0 only if 

x+y−1=0 and 2x−3y+1=0

i.e x+y=1.......(1)

2x−3y=−1.......(2)

Solving (1) and (2), we get

2(1−y)−3y=−1

2−2y−3y=−1

−5y=−3

y=3/5

Putting in (1)

x=1−3/5​

x=2/5

Hence the given line will pass through (x,y)=(2/5 , 3/5)

Given line a(3x+4y+6)+b(x+y+2)=0
Here L1 : 3x+4y+6=0
L2 : x+y+2=0
On solving L1,L2  we get 
3x+4y=−6
4x+4y=−8
x=−2,y=0
Point of intersection of L1,L2 is (−2,0)
Now given point is P(2,3) 
Equation of line from (−2,0) and (2,3)
y=3/4(x+2)
4y=3x+6
So the slope of above line is 3/4​
Line perpendicular to line 4y=3x+6 passing through the common point will be at greatest distance 
Hence Slope of required line be −4/3
y=−4/3(x+2)
3y=−4x−8
4x+3y+8=0

⇒ (sinA–sinB)(sinB–sinC)(sinC–sinC)=0
⇒ A = B or B = C or C = A
any two angles are equal
Δ is isosceles.

Let y=m₁x and y=m₂x be the lines represented by ax²+2hxy+by²=0
Then their images in y=0 are y=−m₁x and y=−m₂x and So their combined equation is y²+m₁​m₂x²+xy(m₁+m₂)=0
⇒y²+a/bx²+xy(−2h/b)=0
⇒ax²−2hxy+by² = 0