Question

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.

Solution

Correct option is

Given line are

(2 + 6k)x + (3 – k)y + 4 + 12k = 0                            … (i)

and 7x + 5y – 4 = 0                                                  … (ii)  

     

and slope of line (ii) m2 = –7/5   

since line (i) is perpendicular to line (ii) 

or 14 + 42k = 5k – 15 or 37k = –29 or k = –29/37

SIMILAR QUESTIONS

Q1

Find the area of the quadrilateral with vertices (3, 3), (1, 4), (–2, 1), (2, –3).

 

Q2

A straight line segment of length ‘p’ moves with its ends on two mutually perpendicular lines. find the locus of the point which divides the line in the ratio 1:2.

Q3

Find the acute angle between the two lines with slopes 1/5 and 3/2.

Q4

If a line has a slope = ½ and passes through (–1, 2); find its equation. 

Q5

If a line has a slope 1/2 and cuts off along the positive y-axis of length 5/2 find the equation of the line.

Q6

If a line passes through two points (1, 5) and (3, 7) find its equation.

Q7

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.

Q8

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.

Q9

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.

Q10

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.