
How do you find the slope, Y-intercept and X-intercept of the line \[x-3y+7=0\]?
Answer
548.4k+ views
Hint: In order to solve the solve question, first we need to convert the given linear equation in the standard slope intercept form of a linear equation by simplifying the given equation. The slope intercept form of a linear equation is,\[y=mx+b\], where ‘m’ is the slope of the line and ‘b’ is the y-intercept. Then for x-intercept, we need to put the value of y = 0 and solve for the value of ‘x’. in this way we will get all required values.
Complete step by step solution:
We have given that,
\[x-3y+7=0\]
As we know that the slope intercept form of a linear equation is,
\[y=mx+b\], where ‘m’ is the slope of the line and ‘b’ is the y-intercept.
Converting the given equation in slope intercept form of equation;
\[x-3y+7=0\]
Adding 3y to both the sides of the equation, we get
\[x-3y+7+3y=0+3y\]
Combining the like terms, we get
\[x+7=3y\]
Rewrite the above equation as,
\[3y=x+7\]
Multiplying both the sides of the equation by 3, we get
\[y=\dfrac{1}{3}x+\dfrac{7}{3}\]
Comparing it with the slope intercept form of a linear equation i.e. \[y=mx+b\]
Thus,
Slope = m = \[\dfrac{1}{3}\]
y-intercept = b = \[\dfrac{7}{3}\]
Now,
Finding the x-intercept,
We need to put the value of y = 0,
We have,
\[y=\dfrac{1}{3}x+\dfrac{7}{3}\]
\[0=\dfrac{1}{3}x+\dfrac{7}{3}\]
\[0=x+7\]
Subtracting 7 from both the side of the equation we get
\[x=-7\]
Therefore, the x-intercept = -7.
Note: While solving these types of questions, students need to know the concept of slope intercept form of linear equation. Solve the equation very carefully and do the calculation part very explicitly to avoid making any errors. For a straight line, if two points $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$ are situated on the line, then by using the slope formula we can calculate the slope (m) as, $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Complete step by step solution:
We have given that,
\[x-3y+7=0\]
As we know that the slope intercept form of a linear equation is,
\[y=mx+b\], where ‘m’ is the slope of the line and ‘b’ is the y-intercept.
Converting the given equation in slope intercept form of equation;
\[x-3y+7=0\]
Adding 3y to both the sides of the equation, we get
\[x-3y+7+3y=0+3y\]
Combining the like terms, we get
\[x+7=3y\]
Rewrite the above equation as,
\[3y=x+7\]
Multiplying both the sides of the equation by 3, we get
\[y=\dfrac{1}{3}x+\dfrac{7}{3}\]
Comparing it with the slope intercept form of a linear equation i.e. \[y=mx+b\]
Thus,
Slope = m = \[\dfrac{1}{3}\]
y-intercept = b = \[\dfrac{7}{3}\]
Now,
Finding the x-intercept,
We need to put the value of y = 0,
We have,
\[y=\dfrac{1}{3}x+\dfrac{7}{3}\]
\[0=\dfrac{1}{3}x+\dfrac{7}{3}\]
\[0=x+7\]
Subtracting 7 from both the side of the equation we get
\[x=-7\]
Therefore, the x-intercept = -7.
Note: While solving these types of questions, students need to know the concept of slope intercept form of linear equation. Solve the equation very carefully and do the calculation part very explicitly to avoid making any errors. For a straight line, if two points $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$ are situated on the line, then by using the slope formula we can calculate the slope (m) as, $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

