Find the equation of the line joining A(1, 3) | vedclass

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(B) Let $P(x, y)$ be any point on the line $AB$. Since the points $A, B,$ and $P$ are collinear,the area of triangle $ABP$ is $0$.Using the determinan

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$\mp 2$

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(B) Let $P(x, y)$ be any point on the line $AB$. Since the points $A, B,$ and $P$ are collinear,the area of triangle $ABP$ is $0$.Using the determinant formula for the area of a triangle:$\frac{1}{2} \left| \begin{array}{lll} 0 & 0 & 1 \ 1 & 3 & 1 \ x & y & 1 \end{array} ight| = 0$Expanding along the first row:$\frac{1}{2} [1(y - 3x)] = 0 \implies y - 3x = 0 \implies y = 3x$.This is the equation of the line $AB$.Now,given that the area of triangle $ABD$ is $3 \, 	ext{sq units}$,where $A(1, 3), B(0, 0),$ and $D(k, 0)$:$\frac{1}{2} \left| \begin{array}{lll} 1 & 3 & 1 \ 0 & 0 & 1 \ k & 0 & 1 \end{array} ight| = \pm 3$Expanding along the second row:$\frac{1}{2} [(-1) \cdot (0 - 3k)] = \pm 3$$\frac{1}{2} [3k] = \pm 3$$\frac{3k}{2} = \pm 3 \implies 3k = \pm 6 \implies k = \pm 2$.

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3 \, \text{sq units}$.

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