Test: Straight Lines- 1 - JEE MCQ

x₁ = (1*3 - 1*1)/(1-1)
y₁ = (1*4 - 1*2)/(1-1)
Hence the denominator becomes zero, here we can say that any point cannot be divided in the ratio of 1:1 externally.

Step-by-step explanation:  We are given to find the distance of the point (x, y) from the origin.

We know that the co-ordinates of the origin are (0, 0).

And, the distance between any two points (a, b) and (c, d) is given by the following distance formula:

Therefore, the distance between the point (x, y) and the origin will be

Distance is a metric, a bifunction d, which is always non-negative, along with being symmetric, satisfies the triangle inequality, and the identity of indiscernibles (i.e., d(x,y)=0⟺x=y)
d(x,y)=0⟺x=y). The “nearest” distance from a point (x1,y1) to the y-axis in 2-space is along the line y=y1. (orthogonal to the y-axis). This line interests the y-axis exactly at the point (0,y1). Using the Euclidean distance metric on R², one obtains:
d((x,y),(0,y))= √(x−0)²+(y−y)²
= |x|

Slope of line passing through (a,b) and (−a,−b) is given by (b+b)/(a+a) = b/a
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a²,ab) lie on the above line

Slope of a line is not defined if the line is parallel to x axis as all points on the line have same x coordinate.

5x − 12y + 65 = 0, 5x − 12y − 39 = 0 
Since, these lines are parallel .
a = 5,  b = -12,  c₁ = 65    c₂ = -39
So, d = |c1 − c2|/[a² + b²]^1/2
⇒ d = 104/13
= 8

Given parallel lines3x +4y+13=0
perp length from origin P1= 13/√(3)² +(4)²
=13/5
and another line 3x+4y-13=0 P2= -13/√(3)²+(4)²
=-13/5
Now distance b/w parallel lines
D=P1-P2
=13/5+13/5
=26/5 units

Explanation:- x+y=0 .....  (i) 
3x+y−4=0 ..... (ii) 
x+3y−4=0 ....... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4
⟹ x=2, y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4 = (4−x)/3
⟹ −9x+12 = 4−x
⟹x=1, y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
-x = (4−x)/3
⟹ −3x = 4−x
⟹ x=−2, y=2
So, AB,BC,AC form a triangle ABC
Now, AB = [(1−2)² + (1+2)²]^1/2 = (10)^1/2
BC = [(−2−1)² + (2−1)2]^1/2 = (10)^1/2
AC = [(−2−2)² + (2+2)²]^1/2 = (32)^1/2
​∴ AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle.

Correct option (A) 2x - 3y + 1 = 0

Explanation:

Let the required line is 2x - 3y + k = 0

it passes through (1, 1)

:. 2 - 3 + k = 0

k = 1

:. required line is 2x - 3y + 1 = 0

Given vertex of ΔABC A(−2,3)
Equation of side AB perpendiuclar to x−y−4=0 is 
AB:x+y−λ=0
Side AB passes through point A(−2,3) 
−2+3−λ=0
λ=1
Hence AB:x+y−1=0−−−−(1)
Equation of side AC perpendiuclar to 2x−y−5=0 is 
AC:x+2y−λ=0
Side AC passes through point A(−2,3) 
−2+6−λ=0
λ=4
Hence AC:x+2y−4=0−−−−(2)
The right bisectors of all sides are meeting at point P(3/2 , 5/2)
Right bisector from point B on side AC is perpendicular
Hence Equation of right bisector perpendicular to x+2y−4=0 is 
2x−y−λ=0
Here right bisector is passing through point P (2 × 3/2 − 5/2​)λ = 0
=> 6/2 − 5/2 = λ
λ=1/2 
Equation of right bisector BD be 
2x−y−1/2 =0
4x−2y−1=0−−−−(3)
Here On solving equation of AB and BD we get point of intersection i.e. point B
4(1−y)−2y−1=0
4−4y−2y−1=0
−6y=−3
y = 1/2
​x = 1 − y = 1− 1/2 = 1/2
Point B(1/2,1/2)
Right bisector from point A on side BC is passing through point P
Hence equation of AD from point A and P
y−3 = {[5/2−3](x+2)/(3/2+2)}
y−3 = −1/7(x+2)
Slope of AD is mAD = −1/7
Hence slope of side BC perpendicular to AD is −1/7mBC = −1
mBC = 7
Equation of side BC from point B(1/2,1/2) with slope mBC=7
y − 1/2 = 7(x−1/2)
2y−1=14x−7
14x−2y−6=0

Given lines are 4x + 3y = 11 and 4x + 3y = 15/2
Distance between two parallel lines = |c₁ - c₂ / √a² + b²|
= | 11 - 15/2/ √16 + 9 |
= | 7 / 2 × 5 |
= 7/10

Line y = 0  is X-axis slope of this line = 0 = m1 
line x-y = 0 is y = x which is passes through (0,0)  slope of this line = 45 = m2
tanФ = (m2-m1)/(1-m1m2) = (1-0)/(1-1*0)
        = 1/1 = 1
tanФ =1 
Ф = 45