Find the equation of the line perpendicular to 2x – 3y = 5 and cutting off an intercept 1 on the x-axis
3x + 2y – 3 = 0
Any line perpendicular to 2x – 3y = 5 is of the form 3x + 2y = k
Putting y = 0, we get x = k/3, the x-intercept.
Since, k/3 = 1 ⇒ k = 3
The required line is 3x + 2y – 3 = 0
If a line passes through two points (1, 5) and (3, 7) find its equation.
A straight line passes through a point A (1, 2) and makes an angle 60owith the x-axis. This line intersects the line x + y = 6 at the point P. find AP.
Find the equation of the straight line, which passes through the point (3, 4) and whose intercept on y-axis is twice that on x-axis.
Find the equation of the straight line upon which the length of perpendicular from origin is units and this perpendicular makes an angle of 75o with the positive direction of x-axis.
Find the value of k so that the straight line 2x + 3y + 4 + k (6x – y + 12) = 0 and 7x + 5y – 4 = 0 are perpendicular to each other.
Show that the lines 2x – y – 12 = 0 and 3x + y – 8 = 0 intersect at a points which is equidistant from both the coordinates areas.
Find the area of triangle formed by the lines x – y + 1 = 0, 2x + y + 4 = 0 and x + 3 = 0.
The line x + λy – 4 = 0 passes through the point of intersection of 4x – y+ 1 = 0 and x + y + 1 = 0. Find the values of λ.
Find the equation of a line parallel to x + 2y = 3 and passing through the point (3, 4).
Find the equation of the straight line passing through (2, –9) and the point of intersection of lines 2x + 5y – 8 = 0,
3x – 4y – 35 = 0