Answer
The value of A for the line whose equation is $Ax+y-2=0$ and is perpendicular to the line containing the points $\left( 1,-3 \right)\text{ and }\left( -2,4 \right)$ is$-\frac{3}{7}$.
Work Step by Step
The slope of the line for the points $\left( 1,-3 \right)\text{ and }\left( -2,4 \right)$ is
$m=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}$
Substitute the value of ${{x}_{1}}=1\text{ and }{{x}_{1}}=-2$ in the above formula
$\begin{align}
& m=\frac{f\left( -2 \right)-f\left( -1 \right)}{-2-1} \\
& =\frac{4-\left( -3 \right)}{-2-1} \\
& =\frac{7}{-3} \\
& =-\frac{7}{3}
\end{align}$
Solve $Ax+y-2=0$ for y to obtain slope intercept form. Isolate y on one side of the equation.
$\begin{align}
& Ax+y-2=0 \\
& y=-Ax+2
\end{align}$
The line given by the equation $Ax+y-2=0$ is perpendicular to the line containing two points $\left( 1,-3 \right)\text{ and }\left( -2,4 \right)$.
The slope of the line containing ${{x}_{1}}=1\text{ and }{{x}_{1}}=-2$ is $m=-\frac{7}{3}$.
The slope of the perpendicular line is the negative reciprocal of $-\frac{7}{3}$.
Thus, the slope of the perpendicular line is $\frac{3}{7}$.
The line given by the equation $Ax+y-2=0$ is perpendicular to the line containing the two points $\left( 1,-3 \right)\text{ and }\left( -2,4 \right)$.
Thus, the value of $-A$ is $\frac{3}{7}$.
