Question

How do you find slope and intercept of the given linear equation, the equation is \[y = 6x + 1\]?

Answer 1

Hint: Slope is the angle from the positive axis of the graph to the line drawn; it gives the angle of inclination of the graph. To find the slope of any line you can use trigonometric identity of “tanx” where “x” is the ratio of perpendicular length and base length, now to find the intercept which means the points on which the curve is cutting the axis of graph can be found out by some simplification in the equation.

Complete step by step solution:
The given equation is \[y = 6x + 1\]
According to the question we have to find the slope and intercept of the curve given, lets first find the slope of the graph:
For finding slope we are using here the differentiation method of finding slope, in which we have to differentiate the equation as \[\dfrac{{dy}}{{dx}}\]
Now finding the value of “y” from the equation by rearranging the term we get:
\[ \Rightarrow y = 6x + 1\]
Now on differentiating the term we get:
\[
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {6x + 1} \right) \\
\Rightarrow \dfrac{{dy}}{{dx}} = 6 +0
\]
Hence we find the required slope of the given equation.
Now for finding the intercept we have to rearrange the equation as, the general equation which is:
\[ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = 1\,(here\,a\,and\,b\,are\,\operatorname{int}
ercept\,of\,the\,given\,equation)\]
Rearranging our equation we get:
\[
\Rightarrow \dfrac{{ - 6x}}{1} + \dfrac{y}{1} = 1(dividing\,by\,1\,both\,the\,sides\,of\,the\,equation)
\\
\Rightarrow \dfrac{{ - 6(x)}}{1} + \dfrac{{1(y)}}{1} = 1 \\
\Rightarrow \operatorname{int} ercept\,are\,\dfrac{{ -
6}}{1},\dfrac{1}{1}\,for\,x\,and\,y\,respectively \\
\]
This is our final answer.

Note: Here it was easy to rearrange the equation in the general form of the intercept equation but if this rearrangement is not possible then you have to plot the graph and then check the points on which the curve is cutting the both axes and that coordinates are our required intercepts.