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Question

# If the line passing through $\left(4,3\right)$ and $\left(2,k\right)$ is perpendicular to $y=2x+3$, then $k=?$

A

$-1$

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B

$1$

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C

$-4$

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D

$4$

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Solution

## The correct option is D $4$Explanation for correct option Step 1: Evaluate the slope of the given pointsGiven points are $\left(4,3\right)$ and $\left(2,k\right)$ Let the slope be ${m}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$So, ${m}_{1}=\frac{k-3}{2-4}=\frac{k-3}{-2}$Step 2: Evaluate the slope of the given line Given equation of the line, $y=2x+3$ Let the slope-intercept form of line be $y={m}_{2}x+b$So, ${m}_{2}=2$ Step 3: Solve for the given variable It is given that the points joining together are perpendicular to the equation of the lineSo, ${m}_{1}Ã—{m}_{2}=-1$ $â‡’\left(\frac{k-3}{-2}\right)Ã—2=-1\phantom{\rule{0ex}{0ex}}â‡’k-3=1\phantom{\rule{0ex}{0ex}}â‡’k=4$Therefore, option (D) is the correct answer.

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