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How do you find slope, point slope, slope intercept, standard form, domain and range of a line for Line \[{\text{J}}\left( { - 3,4} \right){\text{ }}\left( {0,10} \right)?\]

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Last updated date: 13th Jul 2024
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Answer
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Hint: To find the slope of a line whose two points are given is calculated as $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Since we have found the slope so to write the equation of the line, we have to put the value of slope and required value of coordinate of any of point in the equation $y - {y_1} = m\left( {x - {x_1}} \right)$
And to write in slope intercept form, just solve the point slope form for $y$, you will get an equation like this $y = mx + c,\;{\text{where}}\;m\;{\text{and}}\;c$ are slope and intercept of the line.
Standard form of any straight line is given as $ax + by = c$ so kindly shift the variables in the left hand side to get the standard form.
Domain of a function means values of $x$ we can put in the equation to get output, generally polynomials have real numbers as domain, but do check for any discontinuity or critical point in the equation.
Range of a polynomial function can be found by putting the left and right limit of the domain in the function considering critical points of the equation.

Complete step-by-step solution:
Given
\[{\text{J}}\left( { - 3,4} \right){\text{ }}\left( {0,10} \right).........................\left( i \right)\]
Now here we need to find slope, point slope, slope intercept, standard form, domain and range for the given line.
So first let’s find the slope:
1. We know that the equation of slope is given by:
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
On substituting we can write:
$
  m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \\
   = \dfrac{{10 - 4}}{{0 - \left( { - 3} \right)}} \\
   = \dfrac{6}{3} \\
   = 2.............................\left( {ii} \right) \\
 $
Therefore the slope of he given line is$2$.
2. Now we know:
 Point slope form: $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
On substituting we can write:
$
   \Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) \\
   \Rightarrow \left( {y - 4} \right) = 2\left( {x - \left( { - 3} \right)} \right) \\
   \Rightarrow \left( {y - 4} \right) = 2\left( {x + 3} \right) \\
   \Rightarrow \left( {y - 4} \right) = 2x + 6 \\
   \Rightarrow y - 2x = 6 + 4 \\
   \Rightarrow y - 2x = 10..........................\left( {iii} \right) \\
 $
So the point slope form is$y - 2x = 10$.
3. Slope intercept form: Rearrange point slope form and solve for$y$.
Here we have the point slope form as$y - 2x = 10$.
Now rearranging we can write:
$
   \Rightarrow y - 2x = 10 \\
   \Rightarrow y = 2x + 10.........................\left( {iv} \right) \\
 $
So the slope intercept form is$y = 2x + 10$.
4. Standard form:$Ax + By = C$ A, B and C are integers.
So we have the slope intercept form in (iv) as$y = 2x + 10$.
Now to get the standard form let’s express (iv) in terms of the standard equation$Ax + By = C$ A, B and C are integers by rearranging the terms.
 So we can write:
$
  y = 2x + 10 \\
  Ax + By = C \\
   \Rightarrow 2x - y = - 10.......................\left( v \right) \\
 $
So the standard form is$2x - y = - 10$.
5. Domain:
As we can see from (v) in the standard form there are no fractions or restrictions for the variable$x$, such that we can write:
 Domain of the given line:$\left( { - \infty ,\infty } \right)..................................\left( {vi} \right)$
6. Range
As we can see from (v) in the standard form that$x$and $y$are proportional, it also implies that $y$can also take any value.
Such that:
Range of the given line:$\left( { - \infty ,\infty } \right)............................\left( {vii} \right)$

Note: While doing similar questions one should have a thorough knowledge about all the formulas and identities.
Also properties shown below are also useful:
x intercept of \[a\] is the point $\left( {a,0} \right)$
y intercept of \[b\] is the point $\left( {0,b} \right)$
Point slope formula mainly connects the slope of a line and the point on the line.