Let the equation of the line which has equal intercepts be
+ =1
Substituting the values for x and y
1 -2 =a
Therefore a = -1
Hence the equation of the line is x+y+1=0
If the line is parallel to the given line x-2y + 5 =0,
then the required line will have same slope
Hence the equaion of the given line is x-2y+k=0
since it passes through the origin,
0+0+k=0
Therefore k=0
Hence the equation of the required line is x-2y=0 or x=2y
The required line is parallel to the given line, hence it has same slope
Therefore the equation of the required line is 2x-3y+k=0
Since it passes through (1,1)
2(1) - 3(1) +k = 0
Therefore k=1
Hence the equation of the required line is 2x - 3y + 1=0
The equation of the line which cuts equal intercepts is
x+y=a
Since it passes through (2,-3),
2 - 3 =a
Hence a = -1
Hence the equation of the required line is x + y +1 = 0
The triangle formed by these lines is a right angled triangle
If the lines are perpendicular to each other, then the product of their slopes is -1
The slope of lines 3x + y – 4 = 0 , x - 3y – 4 = 0 are -3 and 1/3 respectively.
The product of the slopes is -1
Hence these two lines are perpendicular to each other
This infers that the triangle formed by these lines is a right angled triangle.
Distance between two parallel lines is given by
c1 = 2 x 11 = 22 and c2 = 15 and A = 8 and B = 6
Now substituting the values we get
= 7/10
The line which passes through the point ( 0 , 1 ) and perpendicular to the line x – 2y + 11 = 0 is
The line which is perpendicular to the given line is 2x + y + k = 0
Since it passes through (0,1)
2(0) + 1 + k = 0
This implies k = -1
Hence the equation of the required line is 2x + y - 1 = 0
The perpendicular distance from the origin to the line is given by
For the given line c = 1, A = 3 and B = 4 and since it passes through the origin,
Sustituting the values we get,
= 1/5
Slope of the line x - y = 0 is 1. This imlpies the tan = 1 Hence the angle made by the line with X- axis is 450
The angle made by the line y = 0 with the Y axis is 900
Therefore the acute angle between the lines is 450
From the given points it is clear that these points form a right angled isosceles triangle.
The orthocentre of all right angle lies on its right angle vertex.
Here the right angle vertex is (0,3)
Hence the orthocentre is (0,3)