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Slope of a line or straight line describes the direction (using the variables or axis X and Y) as well as the steepness of the line (means angle made at x-axis or theta. Hence, slope intercept form is one the ways to describe the slope of a straight line in mathematics. The slope of the straight line is represented by the variable ‘m’ and used in the equation y = mx+b.
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Key Takeaways: Slope, Slope intercept form, One point slope form, Two point slope form, Straight line, Slope intercept
What is Slope Intercept Form?
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Intercept means the straight line intersects a segment on the axis.
Slope intercept form uses an equation to represent two things about a straight line passing through the X and Y axis:
- Direction of the straight line with the help of x and y variables
- Steepness of the straight line by using its slope ‘m.’
The equation of the slope intercept form is y = mx + b, where, m is the slope of the line and b is the y-intercept.
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Slope Intercept Form: Formula Derivation
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For finding the slope of a straight line, m, you have to divide the coordinate of the y-axis by the x-axis. Normally, while comparing two collinear points on a line, considering x1,y1 and x2,y2 is according to convention of how the equation has to be formed.
m = (y2 – y1)/(x2 – x1)
Slope Intercept Form Formula Derivation
Hence, let’s consider that the straight line intersects the y-axis at (0,b). It means that the 'b' or y-intercept in the equation of the line y = mx + b is the y-axis coordinate.
y – b = m (x – 0)
m = (y - b) / (x – 0)
⇒ m = (y – b) / (x)
Hence,
mx = y – b
And, by rearranging we get,
y = mx + b
Slope Intercept Form: General Equation of Straight Line
While the slope intercept form has the equation of y = mx + b and also called the general equation of straight line which is usually used, there is another method of representation that can be commonly found:
Ax + By + C = 0
Slope Intercept Form For A Straight Line: Intercept Form
Let’s take an example: Consider that a straight line passing through the x and y axis makes two intercepts, one on x-axis (a,0) and another on y-axis (0,b).
Slope Intercept Form For A Straight Line Intercept Form
Thus, the intercept form of the straight line will be:
x/a + y/b = 1
Slope Intercept Form For A Straight Line: Normal Form
Let’s consider that there is a straight line. Now, if the length of the perpendicular drawn from the straight line to the origin is ‘p,’ and the angle made by the perpendicular with respect to the positive x-axis is α, therefore:
x cos É + y sinα = p
Slope Intercept Form For A Straight Line Normal Form
Slope Intercept Form For A Straight Line: One Point Form
The one point form or point form of slope of a straight line involves picking two points lying on the straight with coordinates (x1,y1) and (x2,y2). Then, instead of y = mx + c where the coordinates of only one point was used, replace them with the coordinates of two points like this:
y = y2 – y1 and x = x2 – x1
Hence, the equation of the slope of a straight line according to one point form is:
y2 – y1 = m (x2 – x1)
Slope Intercept Form For A Straight Line: Two Point Form
The two point form of slope of a line is quite simple. You only have to insert the equation of ‘m’ in the linear equation of one point form. The linear equation of a straight line according to one point form is y2 – y1 = m (x2 – x1).
y2 – y1 = {(y2 – y1) / (x2 – x1)} (x2 – x1)
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Things to Remember
Following are some important points:
- The equation of a line represents the slope of the line and direction via x as well as y axis.
- The equation of a straight line for slope intercept form is: y = mx + b.
- The general equation of a straight line that can be rearranged is Ax + By + C = 0, where A, B, and C are real constants (A and B are non-zero).
- Some more important formulas are:
- m = (y2 – y1)/(x2 – x1)
- One point form = y2 – y1 = m (x2 – x1)
- Intercept form = x/a + y/b = 1
- Normal form = x cosÉ\(\Box\) + y sinα = p
- Two point form = y2 – y1 = {(y2 – y1)/(x2 – x1)} (x2 – x1)
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Sample Questions
Ques: What will be the equation in the slope intercept form of a straight line passing through the points with coordinates (2,3) and (4,6) respectively? (3 Marks)
Ans: Consider (x1,y1) as (2,3) and (x2,y2) as (4,6). Also, for slope, m = (y2 – y1)/(x2 – x1).
So, m = 6-3 / 4-2 = 3/2
Now, using the coordinate (2,3) as per the y = mx+b equation,
3 = 3/2 (2) + b
3 = 3 + b
3 – 3 = b
b=0.
Hence, y = 3/2 (x) + 0
So, y = 3/2 (x)
Ques: What is the equation of a straight line in slope intercept form that has slope as -2 and passes through a point of coordinate (-4,-5)? (3 Marks)
Ans: According to the question, m = -2 and (x,y) coordinates are (-4,-5). Hence, putting it in the equation y = mx +b will be:
-5 = (-2) (-4) + b
-5 = 8 + b
b = -13.
Therefore, the equation of the straight line is:
y = (-2)x + (-13)
y = -2x - 13
Ques: Form the equation of the straight line that runs parallel to another straight line of equation 2x + 3y + 11 = 0. The sum of the intercepts intercepted by the axis is 15. (3 Marks)
Ans: If 2x + 3y + 11 is the equation of the line, then the equation parallel to this line will be 2x + 3y + c = 0
According to the question, a + b = 15 if the equation x/a + y/b =1 is considered, where a and b are intercepts cut on x as well as y axis.
Let’s rearrange 2x + 3y + c = 0 to bring it in x/a + y/b = 1 form.
2x + 3y = -c
x /(-c/2) + y/(-c/3) = 1
Hence, a = -c/2 and b = – c/3
Therefore,
-c/2 + (-c/3) = 15
-3c/6 - 2c/6 = 15
-5c/6 = 15
c = -18
So, the equation of the line is: 2x + 3y - 18 = 0
Ques: What is the equation of the line which is parallel to the x-axis and drawing an intercept -3 on the y-axis? (2 Marks)
Ans: As the intercept is -3, it means the intercept is on the negative y-axis.
We know that y = mx + b
So, y = mx + (-3),
y = mx -3
As the line is parallel to the x-axis, therefore m is zero.
Ques: What is the equation of the line running parallel to 3x + 6y = 5 and having coordinates (1,3) by using the slope intercept form? (3 Marks)
Ans: According to the question, 3x + 6y = 5 runs parallel to the line. Therefore, the line represented in the slope intercept form will be:
y = -3/6 (x) + 5
y = -½ (x) + 5
Hence, slope is -½ and b is 5. So, now using the one point form with the help of the coordinates (1,3), the slope of the straight line parallel to the given straight line will be:
y - 3 = -½ {x - 1}
x+2y = 5
Ques: Reduce the given equation 3x + 2y -12 into slope intercept form and find: (i) slope of the equation, (ii) x-intercept and y-intercept formed. (3 Marks)
Ans: Rearrange the equation 3x + 2y - 12y = 0 in the following way:
3x + 2y = 12
2y = 12 - 3x
y = -3x/2 + 12/2
y = -3/2(x) + 6
So, (i) slope of the equation after comparing the reduced equation with the slope intercept form of y = mx + b is -3/2.
(ii) 6 is the y-intercept, also represented as ‘b.’
For x-intercept, rearrange 3x + 2y = -12 in the following way:
x/(â) + y/(½) = 1/(12)
Hence, x/4 + y(6) = 1
So, x-intercept is 4.
Ques: Reduce the equation: x + √3y + 4 = 0 into: (i) slope intercept form and find axis intercepts, (ii) slope of the straight line. (3 Marks)
Ans: (i) reduce into slope intercept form by rearranging the equation in the following way:
x + √3y + 4 = 0
y = (-x - 4)/√3
y = -x/√3 - 4/√3
The y-intercept is -4/√3 = B. For finding the x-intercept, rearrange again in the following format:
x + √3y + 4 = 0
x + y/(1/√3) + (¼) = 0
x/(-4) + y(-4/√3) = 1
So, x-intercept = A = -4.
(ii) the slope of the line is -1/√3
Ques: What is the general solution of a straight line that has been reduced into slope intercept form as well as normal form? Also, find the slope of the line and its intercepts on each axis. (3 Marks)
Ans: Slope intercept form can be formed by reducing the general equation of Ax + By + C = 0, where A, B and C are real constants (A and B are non-zero).
Hence, for reducing to slope intercept form, follow the given steps:
Ax + By + C = 0
By = -Ax -C
y = -A/B (x) - C/B
For reducing to normal form, follow the given steps:
Ax + By + C = 0
Ax + By = -C
As, x cosÉ + y sinα = p
Divide x cosÉ/p + y sinα/p = -C/p
It can also be represented as x cosÉ/√(A2 + B2) + y sinα/√(A2 + B2) = 1
Ques: If p is the length of the perpendicular from the origin to the straight line whose intercepts are: a and b on x and y axis, prove that 1/p^2 = 1/a^2 + 1/b^2. (2 Marks)
Ans: Intercepts are a and b. Equation of the line is x/a + y/b = 1.
Perpendicular distance origin is (x1,y1) from the line ax + by + c = 0 is:
\(= | {ax_1 + by_1 + c \over \sqrt{a^2 + b^2}}|\)
\(= p = { |{0 \over a} + {0 \over b} - 1| \over \sqrt {{1 \over a^2} + {1 \over b^2}}}\)
\(= p = { 1 \over \sqrt {{1 \over a^2} + {1 \over b^2}}} = {1 \over p^2} ={1 \over a^2} + {1 \over b^2}\)
Ques: What are the coordinates of the foot of the perpendicular to the point with coordinate (-1,3) to the line with equation 3x - 4y - 16 = 0. (2 Marks)
Ans: Let us assume that (a,b) are the coordinates of the foot of the perpendicular from the point (-1,3) to the line with equation 3x - 4y - 16 = 0.
Slope of the line (m1) is (b-3) / (a+1)
Slope of the line (m2) is ¾
As these lines are perpendicular, m1.m2 = -1
{(b-3) / (a+1)} . (¾) = -1
=> 4a + 3b = 5 …(1)
(a,b) lies on 3x - 4y = 16
Therefore, 3a - 4b = 16 …(2)
a = 68/25
b = -49/25
Thus, the coordinates are {68/25 , 49/25}
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