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(A) Using the section formula,the coordinates of the point $P$ that divides the line segment joining $A(1,0)$ and $B(2,3)$ in the ratio $1:n$ are:$P =

Vedclass · Accepted Answer

$(1+n)x + 3(1+n)y = n+11$

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(A) Using the section formula,the coordinates of the point $P$ that divides the line segment joining $A(1,0)$ and $B(2,3)$ in the ratio $1:n$ are:$P = \left(\frac{n(1)+1(2)}{1+n}, \frac{n(0)+1(3)}{1+n}\right) = \left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$The slope of the line segment $AB$ is $m = \frac{3-0}{2-1} = 3$.The slope of the line perpendicular to $AB$ is $m' = -\frac{1}{m} = -\frac{1}{3}$.The equation of the line passing through $P$ with slope $m'$ is:$y - \frac{3}{n+1} = -\frac{1}{3} \left(x - \frac{n+2}{n+1}\right)$Multiplying by $3(n+1)$:$3(n+1)y - 9 = -(x(n+1) - (n+2))$$3(n+1)y - 9 = -(n+1)x + n + 2$$(n+1)x + 3(n+1)y = n + 11$

$A$ line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$. Find the equation of the line.

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