Straight Lines Test - 2

Since the line is equally inclined the slope of the line should be -1, because it makes 135^o in the positive direction of the X axis
This implies the equation of the line is y= -x +c  
i.e; x+y - c =0
distance of the line from the origin is given as √2
Therefore √2 = |c|/(√1² +1² )
This implies c = 2
Hence the equation of the line is x+y=2

Given Lines:-
L1 : y=mx
L2: y+2x=0
L3 : y=2x+k  
L4 : y+mx=k
To form a rhombus, opposite sides must be parallel to each other.
∵Slopes of parallel lines are always equal
L1 ∥L3
L2 ∥L4
​
Therefore,m = 2

Equation of line x/a + y/b = 1
Given that (a,0) and (0,b) lie on a straight line.
We know that a straight line is represented by y = mx + c
Substituting co-ordinates in equation, we get
(1) 0 = am+c
(2) b=a(0)+c = c
⇒m = −b/a, c = b
∴Equation of line is ay=ab −bx
Substituting options we see that (3a,−2b) lies on this line.

Let the perpendicular line of x+y=2 is y −x=λ
It passes through (0,0), then λ=0
∴y −x=0
The point of intersection of y −x=0 and x+y=2 is (1,1), which is the required coordinates.

Equation of line is (y-4)/(x-3) = (6-4)/(5-3)
y - 4 = x - 3
=>x - y + 1 = 0
Let abscissa of the point is a  
:. (a, -1) should satisfy the equation  
:. a- (-1) + 1 = 0  
a = - 2

Given, the line equation : x+y −6=0 .....(i) 
Co-ordinates of P are (4,3)
Let the co-ordinates of Q be (x,y) 
Now, the slope of the given line is  
y = 6-x
slope m = -1
So, the slope of PQ will be −1/m
​[As the product of slopes of perpendicular lines is −1]
Slope of PQ = -1/(-1) = 1
R lies on x + y - 6 = 0
(a+4)/2 + (b+3)/2 - 6 = 0
=>a + 4 + b + 3 - 12 = 0
=>a + b - 5 = 0.......(1)
Slope of LM = slope of PQ = -1
= -(1) * (b-3)/(a-4) = -1
= (b-3)/(a-4) = 1
= b-3 = a-4
=>a-b-1 = 0...............(2)
Solving (1) and (2), we get
a = 3, b = 2

ax + by + c = 0 and (a + b)x = (a –b)y  
m₁ = -a/b,   m₂ = (a+b)/(a-b)
tanx = [(m₁ -m₂ )/(1+m₁ ×m₂ )]
=>{(-a/b)- (a+b)/(a-b)}/{1+(-a/b)[(a+b)/(a-b)]}
=>{-a² +ab-ab-b² }/{b(a-b)} * {ba-b² -a² -ab}/{b(a-b)}
=>(-a² -b² )/{1/(-a² -b² )
tanx = 1
x = tan-1(1)
Angle = 45^o

We have,

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

2x+3y=6......(2)

4x −y=−4......(3)

From equation (1) and (2) to and we get,

(x+y=1)×2

2x+3y=6

2x+2y=2

2x+3y=6

On subtracting and we get,

y=4 put in (1) and we get,

2x+2y=2

2x+2(4)=2

2x+8=2

2x=−6

x=−3




Hence, this is the answer.

Suppose we have three straight lines whose equations are:
a ₁x + b ₁y + c ₁= 0,
a ₂x + b ₂y + c ₂= 0
a ₃x + b ₃y + c ₃= 0.
These lines are said to be concurrent if the following condition holds:
Determinant of
a ₁  b ₁  c ₁
a ₂  b ₂  c ₂ =  0
a ₃  b ₃  c ₃
Now
l     m   n  
m   n    l   =  0
n    l    m
l(nm - l ²) - m(m ²- nl) + n(ml - n ²) = 0
lmn - l ³- m ³+ lmn + lmn - n ³= 0
l ³+ m ³+ n ³= 3lmn
this condition true if an only if
l + m + n = 0

 x + 2y –3 = 0,(1)
 2x + 3y –4 = 0,(2)
 3x + 4y –5 = 0,(3) 
4x + 5y –6 = 0,(4)
 Solving equation (1) and (2), we get
 3x + 6y −9 = 0 [Multiplying (1) by 3] 
3x + 4y −7 = 0  
 ⇒2y −2 = 0  
 y = 1  
Putting value of y in (1), we get  
x + 2 −3 = 0  
x = 1  
The point (1, 1) lies on 3x + 4y −7 = 0 but not on 2x + 3y −4 = 0 and 4x + 5y −6 = 0.
Hence, they are neither concurrent, nor they can form a quadrilateral nor parallel.

xy = 0
either x = 0,y=some constant or y = 0,x = some constant
So it will be a pair of perpendicular lines.