The line x $$+$$ $$3y$$ $$-$$ $$2$$ $$=$$ $$0$$ bisects the angle between a pair of straight lines of which one has equation x $$-$$ $$7y$$ $$+$$ $$5$$ $$=$$ $$0$$.The equation of the other line is

Question

Infinity Learn · Accepted Answer

The family of line through the given lines is L≡x−$$7y+5+$$λ(x+3y$$−$$2)=0$$ $$=x(1+$$λ$$)+y(-7+3$$λ$$)+5-2$$λ$$-----(1)let$$  $$(2$$, $$0) be a point on $$x+3y-2=0$$ which is equidistant from  x $$-$$ $$7y$$ $$+$$ $$5$$ $$=$$ $$0$$ and the line L $$=$$ $$0Therefore$$,$$2-0+512+72=2(1+$$λ$$)+0(-7+3$$λ$$)+5-2$$λ$$1+$$λ$$2+(-7+3$$λ$$)22+550=2+2$$λ$$+5$$−$$2$$λ$$(1+$$λ$$)2+(3$$λ−$$7)2$$ or $$10$$λ$$2$$−$$40$$λ$$=0$$ i.e., λ$$=4$$ or $$0$$ Hence, $$L=0$$,λ$$=4$$. Therefore, the required line is $$5x+5y$$−$$3=0$$ .

The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0.The equation of the other line is

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