[24.2(ax-b)+(a+4b),=0],[2(bx+ay)+(b-4a),=0]

2(ax-by)+(a+4b)=0,2(bx+ay)+(b-4a)=0

Solve the following pair of linear equations 2(ax-by)+(a+4b)=02(bx+ay)+(b-40=0

A circle with center (a,b) passes through the origin.The equation of the tangent to the circle at the origin is ax-by=0 (b) ax+by=0bx-ay=0 (d) bx+ay=0

The straight line x-y-2=0 cuts the axis of x at A. It is rotated about A in such a manner that it is perpendicular to ax+by+c=0. Its equation is: (a) bx-ay-2b=0 (b) ax-by-2a=0 (c) bx+ay-2b=0 (d) ax+by+2a=0

ax+by=a+b bx+ay=b^2

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

If the roots of ax^2+bx+c=0 are real and equal then.......a) b^2-4aclt0 b) b^2-4ac=0 c) b^2-4acgt0 d) Cannot say

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1. Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1 . Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.