If the lines are at right angles to each other, then the product of their slopes = -1. Slope of any line = -(coefficient of x/coeffiecient of y)
Therefore slope of line 1 = -(1/(k-1)
Slope of line 2 = -2/k2
Therefore X = - 1
That is k2 (k-1) = -2
i.e; k3 - k2 +2 = 0
On factorizing we get (k+1)(k2 -2k - 2) = 0
This imlpies k+1 is a factor, hence k = -1
Hence they are at right angles if k = -1
Since the line is equally inclined the slope of the line should be -1, because it makes 1350 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
Therefore =
This implies c = 2
Hence the equation of the line is x+y=2
Rhombus is a parallelogram in which the opposite sides are equaland parallel.
Therefore the lines y = mx and y = -2x are parallel, similarly y = 2x+ and y = -mx + are parallel.
If two lines are equal, then their slopes are qual
This implies m = -2
The slope of the line joining the points (a,0) and (0,b) is [b-0]/[0-a] = -(b/a)
Hence the equation of the line is y = (-b/a)x +b
i.e; ay = -bx +ab
Substituting the x coordinate 3a in the place of x in the above equation we get y = -2b
Hence (3a,-2b) is another point on the line.
The equation of the line perpendicular to the given line is x - y + k = 0
Since it passes through the origin,
0 - 0 + k = 0
Therefore k = 0
Hence the equation of the line is x - y = 0
On solving these two equations we get x = 1 and y = 1
The point of intersection of these two lines is (1,1)
Hence the coordinates of the foot of the perpendicular is (1,1)
The equation 4x + 5y = 20 can be written as + = 1
This implies the intercepts cut by this line on the X and Y axes are 5 and 4 respectively.
Hence the area of the triangle is 1/2 [ 5 x 4] = 10 square units
The slope of the given line is = = 1
Therefore = 1
That is 4+1 = 3 - x
Therefore x = -2
Equation of the line which is perpendicular to the given line x + y = 6 is x - y + k = 0
Since it passes through the point (4,3)
4 - 3 + k = 0
Therefore k = -1
Hence the equation of the line segment PQ is x - y - 1 = 0
On solving these two lines we get x = 7/2 and y = 5/2
This point of intersection is the midpoint of the line segement PQ
That is = 7/2 Hence x = 3
Similarly = 5/2. Hence y = 2
Hence the coordiates of the point Q is (3,2)
The slope of the first line is -(a/b) and slope of the second line is
Angle between the lines is =
Let m1 =-(a/b) and m2 =
Hence =[
On simplifying we get,
= 1
This implies = 450
Equation of a line passing through the intersection of two lies is given by ax1 +by1 +c1 + k(ax2 +by2 +c2) = 0
Hence x+2y-3 + k(2x+y-3) = 0
Since it passes through (0,0)
-3 -3k = 0
This implies k = -1
Sustituting for k we get,
x+2y-3 +(-1)(2x+y-3) = 0
-x +y =0 or x - y = 0