1 Crore+ students have signed up on EduRev. Have you? 
The lines x + (k – 1) y + 1 = 0 and 2x+k^{2}y−1 = 0 are at right angles if
A line is equally inclined to the axis and the length of perpendicular from the origin upon the line is √2. A possible equation of the line is
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/(√1^{2}+1^{2})
This implies c = 2
Hence the equation of the line is x+y=2
The lines y = mx , y + 2x = 0 , y = 2x + λ and y =  mx + λ form a rhombus if m =
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
Two points (a , 0) and (0 , b) are joined by a straight line. Another point on this line is
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 coordinates 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.
The coordinates of the foot of perpendicular from (0 , 0) upon the line x + y = 2 are
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.
The area of the triangle whose sides are along the lines x = 0 , y = 0 and 4x + 5y = 20 is
A line is drawn through the points (3 , 4) and (5 , 6) . If the is extended to a point whose ordinate is – 1, then the abscissa of that point is
Equation of line is (y4)/(x3) = (64)/(53)
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
The line x + y – 6 = 0 is the right bisector of the segment [PQ]. If P is the point (4, 3) , then the point Q is
Given, the line equation : x+y−6=0 .....(i)
Coordinates of P are (4,3)
Let the coordinates of Q be (x,y)
Now, the slope of the given line is
y = 6x
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) * (b3)/(a4) = 1
= (b3)/(a4) = 1
= b3 = a4
=> ab1 = 0...............(2)
Solving (1) and (2), we get
a = 3, b = 2
The acute angle between the lines ax + by + c = 0 and (a + b)x = (a – b)y, a ≠ b, is
ax + by + c = 0 and (a + b)x = (a – b)y
m_{1} = a/b, m_{2} = (a+b)/(ab)
tanx = [(m_{1}m_{2})/(1+m_{1}×m_{2})]
=> {(a/b) (a+b)/(ab)}/{1+(a/b)[(a+b)/(ab)]}
=> {a^{2}+ababb^{2}}/{b(ab)} * {bab^{2}a^{2}ab}/{b(ab)}
=> (a^{2}b^{2})/{1/(a^{2}b^{2})
tanx = 1
x = tan1(1)
Angle = 45^{o}
The lines x + 2y – 3 = 0, 2x + y – 3 = 0 and the line l are concurrent. If the line I passes through the origin, then its equation is
The triangle formed by the lines x + y = 1, 2x + 3y – 6 = 0 and 4x – y + 4 = 0 lies in
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.
The area of triangle formed by the lines y = x, y = 2x and y = 3x + 4 is
If the points representing the complex numbers  4 +3i , 2 – 3i and 0 + π are collinear , then the value of p is
The lines ix + my + n = 0 , mx + ny + l = 0 and nx + ly +m = 0 are concurrent if
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
The equations of the lines through ( 1 ,  1) and making angles of 45^{0} with the line x + y = 0 are
Given the 4 lines with equations x + 2y – 3 = 0, 2x + 3y – 4 = 0, 3x + 4y – 5 = 0, 4x + 5y – 6 = 0 , then these lines are
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.
The orthogonal projection of the point (2 ,  3) on the line x + y = 0 is
Three points A , B and C are collinear if the area of triangle ABC is
The lines 8x + 4y = 1, 8x + 4y = 5, 4x + 8y = 3, 4x + 8y = 7 form a
The number of points on X axis which are at a distance of c units (c < 3) from (2 , 3) is
If (x , y) are the coordinates of point in the plane , then represents
156 videos176 docs132 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
156 videos176 docs132 tests









