The angle between lines joining the origin to the point of intersection of the line x + y = 2 and the curve y2 - x2 = 4 is
A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is
3x - 4y + 7 = 0
4x + 3y = 24
3x + 4y = 25
x + y = 7
The equation of straight line through the intersection of the lines x - 2y = 1 and x + 3y = 2 and parallel to 3x + 4y = 0 is
3x + 4y + 5 = 0
3x + 4y - 10 = 0
3x + 4y - 5 = 0
3x + 4y + 6 = 0
To the lines ax2 + 2hxy + by2 = 0, the lines a2x2 + 2h(a + b) xy + b2y2 = 0 are
equally inclined
perpendicular
bisector of the angle
None of these
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. What is its y intercept
1/3
2/3
1
4/3
The equation of straight line through the intersection of line 2x +y = 1 and 3x + 2y = 5 and passing through the origin is
7x + 3y = 0
7x - y = 0
3x + 2y = 0
x + y = 0
The condition that the line lx + my = 1 may be normal to the curve y2 = 4ax, is
al3 - 2alm2 = m2
al2 + 2alm3 = m2
al3 + 2alm2 = m3
al3 + 2alm2 = m2
If the lines and intersect, then the value of k, is
-
B.
We have the lines
...(i)
...(ii)
Let a point (2r + 1, 3r - 1, 4r + 1) be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also
...(iii)
Taking first and third part of Eq. (iii), we get
2r - 2 = 4r + 1
Taking second and third part of Eq. (iii), we get
3r - 1 - k = 8r + 2
3r - 1 - k - 8r - 2 = 0