5. Find the equation of the straight lines passing through (-2,-7) &having an intercept of length 3 between the straight lines 4x+3y=12, 4x+3y=3.
7. Two sides of a rhombous ABCD are parallel to the lines Y=x+2 & y=7x+3. If the diagonals of the rhombous intersect at the point (1,2) & the vertex A is on the y-axis, find the possible coordinates of A.
10. A point P is such that its perpendicular distance from the line y-2x+1=0 is equal to its distance from the origin. Find the equation of the locus of the point P. Prove that the line y=2x meets the locus in two points Q & R, such that the origin is the mid point of Q R.
11. Atraingle has two sides y=m1x and y=m2x where m1 and m2 are the roots of the equation bα2+2hα+a=0. If (a,b) be the orthocentre of the triangle, then find the euation of the third side in terms of a,b and h.
627. Two sides of a rhombous ABCD are parallel to the lines Y=x+2 & y=7x+3. If the diagonals of the rhombous intersect at the point (1,2) & the vertex A is on the y-axis, find the possible coordinates of A.
lines parallel to Y=x+2 & y=7x+3
will be y=x+a, y=7x+b
the point of intersection lies on the y axis... ie x=0
point of intersection is 6x+b-a=0... ie x=(a-b)/6
y = (7a-b)/6
here, x=0. hence, a=b
hence the point of intersection is (0, a)
also, the line passign thorugh (0,a) and (1,2) is the angular bisector of y=x+a and y=7x+a
which is something you need to verify :)
Can you finish it now?
15.take the st line as x = - 2 + r cos θ
y = -7 +r sin θ,where r is the distance of any point on this st line from (-2,7).
insert the above equation in the given two st lines n den u will get 2 eqn in r and θ....r1+r2 = given intercept....hence u will get the vlue of θ...hence u will get the st line...
the process looks tedius here but is better done than said...so try it...
110.take a poin p(x,y)
then √x2+y2= (y-2x+1)/√5
then insert y =2x and u will get a quadratic...DONT EVER NEVER SOLVE IT!!! let the quad be Ax2+Bx+C=0...then the sum of root is -B/A
hence to get the last part -B/2A=0.....hope u get it????
1wells yaar..but i cant solve the next one coz i have an inbuilt fear for the centers(orthocentre,centriod n circum centre n incentre..:P)
