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

Hint: 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 \[\left( {{y}_{1}},{{y}_{2}} \right).\]