Question

Find the length of perpendiculars drawn from the origin on the sides of the triangle whose
vertices are A (2, 1), B (3, 2) and (-1, -1).​

Answer 1

Here, a and b are the intercepts on the axes. So, equation of line will be,

xa+yb=1

Length of the perpendicular from origin this line =∣∣∣∣∣a1(0)+b1(0)+c1a21+b21−−−−−−√∣∣∣∣∣

Here, a1=1a,b1=1bandc1=−1

So, ∣∣∣∣∣−11a2+1b2−−−−−−√∣∣∣∣∣=p

Squaring both sides,

11a2+1b2=p2

1p2=1a2+1b2