Q49 of 51 Page 10

The perpendicular from the origin to the line y = mx+c meets it at the point (-1,2). Find the values of m and c.


Now(-1,2) lies on y=mx+c

2 = -m+c

m-c = -2

Slope of the perpendicular =

Therefore the slope of the required line perpendicular the line from the origin m =

Substituting in the equation m-c = -2

-c=-2

c=

m=, c=

More from this chapter

All 51 →
47
Find the equation of the right bisector of the line segment joining the points (3,4) and (-1,2).

48
(-1,3) to the line 3x-4y-16=0

50
If p and q are the lengths of perpendiculars from the origin to the lines xcosθ -ysinθ =kcos2θ and xsecθ +ycosθ =k, respectively, prove that p2+4q2=k2

51
In a triangle ABC with vertices A(2,3) B(4,-1) and C(-1,2), find the equation and length of the altitude from the vertex A.