Question

How do you find the slope of the equation \[y=3+2\left( x-1 \right)\]?

Answer 1

Hint: To find the slope of the equation of type \[y=c+a\left( x+b \right)\], we will first convert the equation into the form of \[y=ax+b\]. Then, we will try to find the coordinates of the point for that we will put x = 0 in the equation and then we will solve for y and then we will put y = 0 in the equation, and will solve for x and write it in the form of \[B\left( {{x}_{2}},{{y}_{2}} \right)\]. Therefore, we will have two points and their coordinates which will satisfy the equation \[y=3+2\left( x-1 \right)\]. Then, we will find the slope (m) of the equation using the formula \[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].

Complete step-by-step solution:
It is given in the question that our equation is \[y=3+2\left( x-1 \right)\] and we have been asked to find the slope of the equation.
Now, we will apply the distributive property \[a\left( b+c \right)=ab+ac\], in the term \[2\left( x-1 \right)\]. Therefore, we get
\[\Rightarrow y=3+2x-2\]
And on further simplifications, we get
\[\Rightarrow y=2x+1\]
Now, we will try to find the y coordinate of a point on the equation \[y=2x+1\] by putting x = 0.
\[\Rightarrow y=2\left( 0 \right)+1\]
Therefore, we get
\[\Rightarrow y=0+1\]
\[\Rightarrow y=1\]
So, the coordinates of the point are (0, 1). Let us call this point A.
So we get, \[{{x}_{1}}=0\] and \[{{y}_{1}}=1\]
Now, we will put y = 0 in the equation \[y=2x+1\] and then find the x coordinate of the second point.
\[\Rightarrow 0=2x+1\]
\[\Rightarrow 2x+1=0\]
And on further simplication, we get
\[\Rightarrow 2x=-1\]
\[\Rightarrow x=\dfrac{-1}{2}\]
So, the coordinates of the point are \[\left( -\dfrac{1}{2},0 \right)\]. Let us call this point B.
So we get, \[{{x}_{2}}=-\dfrac{1}{2}\] and \[{{y}_{2}}=0\]
Now, we will calculate the slope of a line crossing from points A (0,1) and B \[\left( -\dfrac{1}{2},0 \right)\].
We know that slope of line with two given points can be calculated using \[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Now, we will put the values as \[{{x}_{1}}=0\] and \[{{y}_{1}}=1\], and \[{{x}_{2}}=-\dfrac{1}{2}\] and \[{{y}_{2}}=0\] for points A (0,1) and B \[\left( -\dfrac{1}{2},0 \right)\]. We get
\[\Rightarrow m=\dfrac{0-\left( 1 \right)}{-\dfrac{1}{2}-0}\]
\[\Rightarrow m=\dfrac{-1}{\left( -\dfrac{1}{2} \right)}\]
And on further simplification, we get
\[\Rightarrow m=2\]
So, we have found the slope of the equation \[y=3+2\left( x-1 \right)\].
Hence, the slope of the equation \[y=3+2\left( x-1 \right)\] is 2.

Note: Whenever we get this type of problem, just put x=0 and then find y and after that put y = 0 and find x. Therefore, we have two points to find the slope of the equation. We should not make mistakes in the calculation part because one mistake and the slope of the equation will be wrong. We can also solve this question by converting the equation in the form y = mx + c where m is slope and hence we have our answer.