Straight lines x-y=7 and x+4y=2 intersect at | vedclass

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(B) The given equations of the lines are:$x-y=7$ ...$(i)$$x+4y=2$ ...(ii)Solving equations $(i)$ and (ii),we find the intersection point $B(6,-1)$.Let

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$23x+7y+3=0$

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(B) The given equations of the lines are:$x-y=7$ ...$(i)$$x+4y=2$ ...(ii)Solving equations $(i)$ and (ii),we find the intersection point $B(6,-1)$.Let the slope of line $AC$ be $m$. Since $A(2,-7)$ lies on $x-y=7$,line $AC$ passes through $A(2,-7)$.The angle between line $AC$ and line $AB$ (which is $x-y=7$,slope $m_1=1$) must be equal to the angle between line $AC$ and line $BC$ (which is $x+4y=2$,slope $m_2=-1/4$) because $AB=AC$ implies $	riangle ABC$ is isosceles with $\angle ABC = \angle ACB$.Using the formula for the angle between two lines $	an 	heta = \left| \frac{m_1-m_2}{1+m_1m_2} ight|$,we equate the tangents of the angles:$\left| \frac{m-1}{1+m} ight| = \left| \frac{m-(-1/4)}{1+m(-1/4)} ight| = \left| \frac{4m+1}{4-m} ight|$.Solving this gives $m=1$ or $m=-23/7$.For $m=1$,the line equation is $y-(-7)=1(x-2) \Rightarrow x-y-9=0$.For $m=-23/7$,the line equation is $y-(-7)=-\frac{23}{7}(x-2) \Rightarrow 23x+7y+3=0$.

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

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