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Analytical Geometry
Q 1 :Find the locus of the middle points of the intercepts made by the axes on the lines drawn thru the pt(a,b)
Q2 : A line forms a ∆ of area 54√3 sq unts wid d coordinate axis.find the eqn of the line if the perpendicular drawn frm the origin makes an angle of 60 wid the + x axisQ3 :In a ∆ABC d(20,25) & e(8,16) are the foot of perpendicular frm A & B to the opp. sides . if c(10,15) find the orthocentre.
Comments (5)
Sol 2---
Let the length of perpendicular drawn from the origin to the line be p...
Then a = p/cos60
b = p/sin60 (where a and b are x- and y- axis intercepts respectively)
Now
Ar(triangle) = 54√3 = 1/2 *base*height = 1/2 * (p/sin60)*(p/cos60)
=> p = 9
=> a= 18/√3 and b= 18
Hence eqn of line is x/18√3 + y/ 18 = 1
Tell me if i'm right or not
hello
we have D ( 20,25) E ( 8,16) C( 10,15 )
slope of AC = -1/2
slope of BC = 1
Thus slope of BE = 2 and that of AD = -1
thus the eq. of AD becomes y +x = 45
eq. of BE is y = 2x
the point of concurrency is x = 15 y=30 ie. coordinate of orthocenter .
thank u sorry if there is a calculation mistake as i m solving in hurry .
one more thing the approach used by the student who have given the second answer is correct .













Let the line intersects axes at (p,0) and (0,q)
equation of line : x/p + y/q = 1
since (a,b) passes through the line : a/p + b/q = 1
mid point of (p,0) and (0,q) is (p/2,q/2) = (h,k) let => p-2h, q=2k
so, a/2h + b/2k = 1
locus of (h,k) is a/x + b/y = 2 => ay+bx = 2xy