the slope of the line passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is found using the formula…

the slope of the line passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is found using the formula $\frac{y_2 - y_1}{x_2 - x_1}$. the line passing through the points $(1,2)$ and $(x,5)$ is perpendicular to a line that has a slope of $\frac{1}{3}$. what is the value of $x$? a 10 b -2 c -4 d 0
Answer
Explanation:
Step1: Find the slope of the line passing through (1, 2) and (x, 5)
Let $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(x,5)$. Using the slope - formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, the slope $m_1$ of the line passing through these two points is $m_1=\frac{5 - 2}{x - 1}=\frac{3}{x - 1}$.
Step2: Use the perpendicular - slope relationship
If two lines are perpendicular, the product of their slopes is - 1. Given that one line has a slope $m_2=\frac{1}{3}$, and the other line has a slope $m_1$. Then $m_1\times m_2=-1$. Substitute $m_1=\frac{3}{x - 1}$ and $m_2=\frac{1}{3}$ into the equation: $\frac{3}{x - 1}\times\frac{1}{3}=-1$.
Step3: Solve the equation for x
Simplify the left - hand side of the equation $\frac{3}{x - 1}\times\frac{1}{3}=\frac{1}{x - 1}$. So, $\frac{1}{x - 1}=-1$. Cross - multiply to get $1=-(x - 1)$. Expand the right - hand side: $1=-x + 1$. Subtract 1 from both sides: $0=-x$. Then $x = 0$.
Answer:
D. 0