Question

Find the equation of the line passing through (ab) and parallel to px +qy + 1 = 0.  

Solution

Correct option is

px + qy = pa + qb

Equation of line parallel to px + qy + 1= 0 can be taken as

px + qy + k = 0    (having same slope)

Now, as it passes through (ab):  

          

Substituting for k = –(pa + qb), the required equation of

L:         px + qy = pa + qb  

SIMILAR QUESTIONS

Q1

 be three points. Then the equation of the bisector of angle PQR is

Q2

Find the area of triangle ABC with vertices A (aa2), B (bb2), C (cc2).

Q3

A straight line passes through (2, 3) and the portion of the line intercepted between the axes is bisected at this point. Find its equation

Q4

Find the slope (m), intercepts on X axis, intercept on Y axis of the line 3x+ 2y – 12 = 0. Also trace the line on XY plane.

Q5

Given the triangle A (10, 4), B(–4, 9), C(–2, 1), find the equation of median through B.

Q6

Find the equation of the straight line passing through the points (3, 3) and (7, 6). What is the length of the portion of the line intercepted between the axes of the coordinates.

Q7

Given the triangle with vertices A (–4, 9), B (10, 4), C (–2, –1). Find the equation of the altitude through A.

Q8

Find the equation of perpendicular bisector of the line joining the points (1, 1) and (2, 3). 

Q9

Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line y = 3x + 4.

Q10

Find the equation of the line perpendicular to 3x + 4y + 1= 0 and passing through (1, 1).