The straight lines x+3y-9=0,4x+5y-1=0,and px+ | vedclass

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(A) Given that the lines $x+3y-9=0$,$4x+5y-1=0$,and $px+qy+10=0$ are concurrent,the determinant of their coefficients must be zero: $\left|\begin{arra

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$(q, -p)$

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(A) Given that the lines $x+3y-9=0$,$4x+5y-1=0$,and $px+qy+10=0$ are concurrent,the determinant of their coefficients must be zero: $\left|\begin{array}{ccc} 1 & 3 & -9 \ 4 & 5 & -1 \ p & q & 10 \end{array}ight| = 0$ Expanding the determinant: $1(50 + q) - 3(40 + p) - 9(4q - 5p) = 0$ $50 + q - 120 - 3p - 36q + 45p = 0$ $42p - 35q - 70 = 0$ Dividing by $7$: $6p - 5q - 10 = 0$ $\Rightarrow 6p - 5q = 10$ The line $5x + 6y + 10 = 0$ passes through the point $(q, -p)$ if $5(q) + 6(-p) + 10 = 0$,which simplifies to $5q - 6p + 10 = 0$,or $6p - 5q = 10$. This matches our derived condition. Thus,the line passes through $(q, -p)$.

The straight lines $x+3y-9=0$,$4x+5y-1=0$,and $px+qy+10=0$ are concurrent. If the line $5x+6y+10=0$ passes through the point $(a, b)$,find the point.

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