Test: Straight Lines- 2 - Commerce MCQ

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:
a₁x + b₁y + c₁ = 0,
a₂x + b₂y + c₂ = 0
a₃x + b₃y + c₃ = 0.
These lines are said to be concurrent if the following condition holds:
Determinant of
a₁     b₁     c₁
a₂     b₂     c₂   =  0
a₃     b₃     c₃
Now
l       m     n  
m     n      l     =   0
n      l      m
l(nm - l²) - m(m² - nl) + n(ml - n²) = 0
lmn - l³ - m³ + lmn + lmn - n³ = 0
l³ + m³ + n³ = 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.