Question

How do you find the value of $k$ so that the slope of the line through $\left( 2,-k \right)$ and $\left( -1,4 \right)$ is $1?$

Answer 1

We have to find the value of $k$ and the slope if given i.e. $1$ and the line through slope passing is given $\left( 2,-k \right)$ and $\left( -1,4 \right).$ By putting this value in slope formula and solve for $k.$ By putting this we can find the value of $k.$
Slope formula $=m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Complete step by step solution:
The slope can be found by using the formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Where $m$ is the slope and $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are two points on the line.
We have given $\left( {{x}_{1}},{{y}_{1}} \right)$ that is $\left( 2,-k \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\left( -1,4 \right).$
The slope is $1$ as given in the question know substitute the values given in the problem and solve for $k.$
$\Rightarrow m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\Rightarrow 1=\dfrac{4-\left( -k \right)}{-1-2}$
$\Rightarrow 1=\dfrac{4+k}{-3}$
$\Rightarrow -3\times 1=-3\times \dfrac{4+k}{-3}$
$\Rightarrow -3=-3\times \dfrac{4+k}{-3}$
Cancel the same term we will get
$\Rightarrow -3=4+k$
$\Rightarrow -3-4=4+k-4$
$\Rightarrow -3-4=4-4+k$
$\Rightarrow -7=0+k$
$\Rightarrow -7=k$
$\Rightarrow k=-7$

So, the value of the $k$ is $-7.$

Note: Use the correct the slope formula and solve for $k.$ While considering point is $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ students may mistake and consider point as $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right).$