Consider the family of lines $$(x$$−y−$$6)+$$λ$$(2x+y+3)=0$$ and $$(x+2y$$−$$4)+$$μ$$(3x$$−$$2y$$−$$4)=0$$.If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

Question

Consider the family of lines $$(x$$−y−$$6)+$$λ$$(2x+y+3)=0$$ and $$(x+2y$$−$$4)+$$μ$$(3x$$−$$2y$$−$$4)=0$$.If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

Infinity Learn · Accepted Answer

solving x−y−$$6=0$$ and $$2x+y+3=0$$ we get (1$$,$$-5)  and  $$x+2y$$−$$4=0$$ , $$3x$$−$$2y$$−$$4=0$$ we get (2$$,$$1) Given family of lines are concurrent at $$A(1$$, $$-5)$$ and $$B(2$$, $$1)$$. Now point $$P(h$$, $$k)$$ is such that AP ⊥ BP ⇒$$k+5h$$−$$1$$⋅k−$$1h$$−$$2=$$−$$1$$∴  Locus of (h$$,$$k) is $$x2+y2$$−$$3x+4y$$−$$3=0$$.

$Consider the family of lines (x - y - 6) + λ (2 x + y + 3) = 0 and (x + 2 y - 4) + μ (3 x - 2 y - 4) = 0$ .If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

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Consider the family of lines (x−y−6)+λ(2x+y+3)=0 and (x+2y−4)+μ(3x−2y−4)=0.If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

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$Consider the family of lines (x - y - 6) + λ (2 x + y + 3) = 0 and (x + 2 y - 4) + μ (3 x - 2 y - 4) = 0$ .If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is