Show that the straight lines given by `x(a+2b)+y(a+3b)=a+b`
for different values of `a and b`
pass through a fixed point.
Text Solution
Verified by Experts
Given equation of variable line is x(a+2b)+y(a+3b) =a+b `"or " a(x+y-1)+b(2x+3y-1) =0 " "(1)` At least one of a and b will be non zero. `"Let " a ne 0.` So, equation (1) reduces to `(x+y-1)+(b)/(a)(2x+3y-1)=0` `"or " x+y-1+lambda(2x+3y-1)=0, "where" lambda=(b)/(a) " " (2)` This is the equation of family of straight lines concurrent at point of intersection of lines `x+y-1 = 0 " " (3)` `"and " 2x+3y-1 = 0 " " (4)` Solving (3) and (4), we get x=2, y=-1. Hence, given variable lines pass through the fixed point (2,-1) for all values of a and b.
Show that the straight lines given by x(a+2b)+y(a+3b)=a for different values of aa n db pass through a fixed point.
The lines (a +2b)x +(a – 3b)y= a - b for different values of a and b pass through the fixed point whose coordinates are
Consider the equation y-y_1=m(x-x_1) . If ma n dx_1 are fixed and different lines are drawn for different values of y_1, then (a)the lines will pass through a fixed point (b)there will be a set of parallel lines (c)all the lines intersect the line x=x_1 (d)all the lines will be parallel to the line y=x_1
If aa n db are two arbitrary constants, then prove that the straight line (a-2b)x+(a+3b)y+3a+4b=0 will pass through a fixed point. Find that point.
If non-zero numbers a, b, c are in H.P., then the straight line (x)/(a)+(y)/(b)+(1)/(c)=0 always passes through a fixed point. That point is
If a , b ,c are in harmonic progression, then the straight line ((x/a))+(y/b)+(1/c)=0 always passes through a fixed point. Find that point.
consider a family of circles passing through two fixed points S(3,7) and B(6,5) . If the common chords of the circle x^(2)+y^(2)-4x-6y-3=0 and the members of the family of circles pass through a fixed point (a,b), then
ax + by - a^(2) = 0 , where a, b non-zero, is the equation to the straight line perpendicular to a line l and passing through the point where I crosses the x-axis. Then equation to the line / is
If the parabola y=(a-b)x^2+(b-c)x+(c-a) touches x- axis then the line ax+by+c=0 passes through a fixed point
show that the straight line passing through the points A (6,7,5) and B (8,10,6) is perpendicular to the straight line passing through the points C (10, 2,-5) and D (8,3,-4).
