23. Show that the parallelogram for...
Question

# 23. Show that the parallelogram formed by ax + by+C =0, aix+by+ c = 0, ax + by + 1 = 0 and dix + b1y + C1 = 0 will be rhombus if a2 + b2 = a 2 + 6,2 [B. U. 67

JEE/Engineering Exams
Maths
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Let A be the intersection of (1) and (2) ( a x+b y+c+mleft(a^{prime} x+b^{prime} y+c^{prime}right)=0 ldots ldots .(7) ) The orientation of line depends on value of ( mathbf{m} ). Equation of a line passing through ( mathrm{A} ) is given ( quad ) Let ( mathbf{D} ) be the intersection of (2) and (3) ( mathrm{for} ) by [ text { which } a x+b y=-c text { ' and } a^{prime} x+b^{prime} y=-c ] ( a x+b y+c+1left(a^{prime} x+b^{prime} y+cright)=0 ldots ldots .(5) ) The orientation of line depends on 1. plugging these in ( (5) mathrm{c}-mathrm{c}^{prime}+1left(mathrm{c}^{prime}-mathrm{c}right)=mathrm{O} Rightarrow mathrm{m}=1 ) and (5) becomes Let ( mathbf{C} ) be the intersection of (3) and (4) for ( left(a+a^{prime}right) x+left(b+b^{prime}right) y+c+c^{prime}=0 ldots ldots ldots ldots ldots ldots(8) . ) this is which ( a x+b y=-c^{prime} ) and ( a^{prime} x+b^{prime} y=-c^{prime} ) plugging these in ( (5) mathrm{c}-mathrm{c}^{prime}+1left(mathrm{c}-mathrm{c}^{prime}right)=mathrm{O} ) as D, therefore of diagonal BD. Now slope of ( mathrm{AC} ) from (6) is ( -frac{b-b^{prime}}{a-a^{prime}} ) ( left(a-a^{prime}right) x+left(b-b^{prime}right) y=0 ldots ldots ldots ldots ldots ldots .(6) . ) this is the equation of line passing through ( mathbf{A} ) as well as slope of ( mathrm{BD} ) from (8) is ( -frac{b+b^{prime}}{a+a^{prime}} ) C, therefore of diagonal AC. Let B be the intersection of (1) and (4) Product of slopes ( =frac{left(b-b^{prime}right)left(b+b^{prime}right)}{left(a-a^{prime}right)left(a+a^{prime}right)}=frac{b^{2}-b^{2}}{a^{2}-a^{prime 2}} ) Equation of a line passing through ( mathrm{B} ) is given If they are perpendicular to each other by product must be - 1 ( a x+b y+c+mleft(a^{prime} x+b^{prime} y+c^{prime}right)=0 ldots ldots .(7) ) The hence orientation of line depends on value of ( mathbf{m} ). ( frac{b^{2}-b^{2}}{a^{2}-a^{12}}=-1 Rightarrow a^{2}+b^{2}=a^{2}+b^{2} )