Straight Line and Pair of Straight Lines 1 Question 60

60. Let ABC be a triangle with AB=AC. If D is mid point of BC, the foot of the perpendicular drawn from D to AC and F the mid-point of DE. Prove that AF is perpendicular to BE.

(1989,5M)

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Solution:

  1. Let BC be taken as X-axis with origin at D, the mid-point of BC and DA will be Y-axis.

Given, AB=AC

Let BC=2a, then the coordinates of B and C are (a,0) and (a,0) let A(0,h).

Then, equation of AC is

xa+yh=1

and equation of DEAC and passing through origin is

xhya=0x=hya

On solving, Eqs. (i) and (ii), we get the coordinates of point E as follows

hya2+yh=1y=a2ha2+h2

Coordinate of E=ah2a2+h2,a2ha2+h2

Since, F is mid-point of DE.

Coordinate of Fah22(a2+h2),a2h2(a2+h2)

Slope of AF,

m1=ha2h2(a2+h2)0ah22(a2+h2)=2h(a2+h2)a2hah2m1=(a2+2h2)ah

and slope of BE,m2=a2ha2+h20ah2a2+h2+a=a2hah2+a3+ah2

m2=aha2+2h2

From Eqs. (iii) and (iv), m1m2=1AFBE



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