IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

piyushsahani

1. The straight line 7x -2y +10 = 0 and 7x+2y -10 = 0 forms an isosceles triangle with the line y =2 .find the area of the triangle .

2.The equation of the perpendicular bisector of the sides AB and ACof the triangle ABC are x - y + 5 = 0 and x + 2y =0 respectively . If the point A is (1,-2) then equation of the line BC woud be.

3. A straight line passes through the origin O meets the parallel lines 4x+2y =9 and 2x + y + 6 = 0 at the points P and Q respectively. then the points O divides the the segment PQ in ratio (which ratio).

delhitushar

ans 1:
int. pt. of lines 7x-2y+10=0 & 7x+2y-10=0 is(0,5)
n for y=2 n 7x-2y+10=0 is (-6/7,2) n for y=2 & 7x+2y-10=0 is (6/7,2)
now find the area

netkid07

answer to third ques is 1 ratio 3

tell me if u want detailed solution

delhitushar

ans 3:
3:4 internally
firsty the lines r on opposite side of the orgin
op=9/2root5
oq=6/root5
so op / oq=3/4

delhitushar

ans2:
let E be mid pt of AB
& F be mid pt of AC
B(x1,y1)
C(x2,y2)
E=((x1+1)/2,(y1-2)/2)
F=((x2+1)/2,(y2-2)/2)
putting in perp eq.
x1-y1=-13
x2+2y2-3=0
slope of AB=-1 AC= 2
so x1+y1+1=0
2x2-y2-4=0
so
B=(-7/6)
C(11/5,2/5)
eq of BC 14x+23y-40=0

netkid07

distance of first line from origin=c/root(a^2+b^2)

=9/2root(5)..............................(1)

distance of 2nd line from origin will be

=6/root(5).........................(2)

required ratio is

(1) divided by (2)

we get

3 ratio 4

sorry my first answer was wrong

don't forget to rate

cheers!!!

Ask & Discuss Questions with Community & Experts