The worksheet on Point-Slope Form consists of all Point Slope Form Problems with Answers related to finding the equation of a line using the point-slope form method. Solve all the Point-Slope problems with Solutions on your own to know how to find the equation of a line. Different tricks and tips are also included in this Point Slope Form Practice Worksheet PDF for easy solving of questions.
So, first, quickly go through the entire article and gain a high knowledge of point-slope form word problems as well. The complete information about finding the equation of a line using a point-slope form is given in this Worksheet for 10th Grade Math PDF articles. This Point Slope Form Worksheet with Answers PDF will help students to solve problems easily.
Also, Read:
Point-Slope Form Quiz PDF
Check out the below problems to learn the Point-slope form concepts deeply. The different problems along with their answers are given below.
Problem 1: Find the equation of a line that passes through (-2, 5) and with a slope is -8.
Solution:
Given that,
The points (x1,y1) are (-2,5).
The slope of a line i.e., m = -8.
We know the formula of the equation of a line using a point-slope form is,
(y-y1) = m(x-x1)
Now place the given values in the above formula. It will be,
y-(5) = (−8)(x − (-2))
i.e., y-5 = -8x -16
8x+y-5 +16 =0
8x+y+11 =0
Hence, the equation of the line is 8x+y+11 =0.
Problem 2: Find the equation of a line that passes through the points (4, –2) and (1, 4) in point-slope form.
Solution:
In the given question,
The points (x1Â ,y1) is (4, -2).
The points of (x2 ,y2)are (1, 4).
The slope of the line that passes through the points formula is
m = (y2 – y1)/ (x2 – x1)
Now place the given values in the above formula. Then it will be
m = (4-(-2))/(1-4)
m= 6/-3
So, the slope of the line m is -2.
Next, the equation of a line which is passing through the point (4, -2) with a slope is -2.
We all know the formula for equation of a line using point-slope form is,
(y-y1) = m(x- x1)
y -(-2) = -2(x – 4)
y + 2 = -2x +8
2x + y + 2 – 8 = 0
2x + y -6 = 0
Similarly, the equation of a line passing through the point (1, 4) with slope of a line -2.
Place these value in a point slope from formula. Then we get,
y – 4 = -2(x – 1)
y – 4 = -2x +2
2x + y – 4 -2= 0
2x + y -6 = 0
Therefore, the equation of the line in point slope form is 2x + y – 6 = 0.
Problem 3: A straight line that passes through the point (8, -5), and the positive direction of the x-axis gives an angle of 135°. Find the equation for a straight line?
Solution:
As given in the question,
The points ( x1, y1) are (8,-5)
An angle of 135 ° with a positive direction of the x-axis line.
The slope of the line m is,
m= tan 135 ° = tan (90 ° + 45 °) = – cot 45 ° = -1.
The line that is needed to pass through the point (8, -5).
We know the equation of a line using the point-slope formula.
So, the formula is (y- y1) = m(x- x1)
Substitute the values in the above formula. We get,
y – (-5) = -1 (x -8)
y + 5 = -x + 8
x + y + 5- 8 = 0
x + y -3 = 0
Thus, the equation of a straight line is x + y – 3 = 0.
Problem 4: What is the value of the slope of the line when the equation of the line is 5x-7y +1.
Solution:
Given that, the equation of a line is 5x-7y+1
Now, we will find the value of the slope of a line.
The given equation of a line is in the form of ax + by + c is -a/b
So, the value of a is 5 and the value of b is -7.
Then the value of the m is -(5/-7)
Hence, the slope of a line is 5/7.
Problem 5: Find the line equation with a slope of -1/3 and go through the points (4, -6)?
Solution:
In the given question,
The points ( x1, y1) are (4,-6)
The slope of the line m is -1/3.
We know the formula of the equation of a line using a point-slope form is,
(y-y1) = m(x-x1)
After placing the given values in the above formula. It will be,
y-(-6) = (−1/3)(x −4)
i.e., 3(y+6) =(-1)(x -4)
3y+18=-x+4
x+3y-4 +18 =0
x+3y+14 =0
Hence, the equation of the line according to the point-slope form is x+3y+14 =0.
Problem 6: Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).
Solution:
In the given question,
The equation of a line passes through the points are (0,-3) = ( x1, y1)
The inclination of a straight line is 60°.
First, find the slope of a line ‘m’ value.
So, the slope of a line is tan 60° = √3.
We know the formula of an equation of a line using the point-slope form is,
(y-y1) = m(x-x1)
After putting the given values in the above formula. It will be,
y-(-3) = (√3)(x −0)
y + 3 = √3 (x – 0)
y + 3 = √3x
√3x – y – 3 = 0
Therefore, the required equation of a line is √3x – y – 3 = 0.
Problem 7:Â Find the equation of the straight line with slope -1 and the y-intercept is 6?
Solution:
Given that,
The slope of the line is -1 i.e., m = -1.
The y-intercept value is 3. This means the line cuts the y-axis on the fixed point (0, 3).
So, the points (x1, y1) are (0, 6).
The formula for point-slope of line is (y-y1) = m(x-x1).
Now substitute the values of m, x, and y in the formula.
y – 6 = (-1)(x – 0)
y – 6 = (-1)x
y – 6 = -x
We will get the equation is y + x – 6 = 0
So, the required equation of a line is x+y-6 = 0.
Problem 8: What is the equation of the straight line whose slope of a line is -3 and the x-intercept is -8?
Solution:
As given in the question,
The slope of the line is -3 i.e., m = -3.
The x-intercept is -7. This means the (x1, y1) value is (-7,0).
Now, Put the values of m and (x1, y1) in the point-slope form formula.
The formula is (y-y1) = m(x-x1).
After substituting the value, we get
y-y1 = (-3)(x –(-7))
y-y1 = (-3)(x + 7)
y-0 = -3x – 21
y + 3x + 21 = 0
Hence, the required equation is 3x+y+21= 0.
Problem 9: What is the equation of a line which is passing through the points (8, –4) and (5, 10) in point-slope form.
Solution:
In the given question,
The points (x1Â ,y1) is (8, -4).
The points of (x2,y2)are (6, 10).
The slope of the line which is passing through the points formula is,
m = (y2 – y1)/ (x2 – x1)
After placing the given values in the above formula. Then we get,
m = (10-(-4))/(6-8)
m= 14/-2
So, the slope of the line m is -7.
The equation of a line passing through the point (8, -4) with slope -7.
The equation of a line using point-slope form formula is,
(y-y1) = m(x- x1)
y -(-4) = -7(x – 8)
y + 4 = -7x +56
7x + y + 4 – 56= 0
7x + y -52 = 0
Similarly, the equation of a line passing through the point (6, 10) with slope of a line -7.
Substitute the value in a point-slope formula. Then we get,
y – 10 = -7(x – 6)
y – 10 = -7x +42
7x + y – 10 – 42= 0
7x + y -52 = 0
Therefore, the equation of the line using the point-slope form is 7x + y – 52 = 0.
Problem 10: The equation of the line passing through (1, 3) and making an angle of 30° in a clockwise direction with the positive direction of the y-axis.
Solution:
As given in the question,
The points (x1Â ,y1) is (1,3)
The line makes an angle of 30° in a clockwise direction with the positive direction of the y-axis. It will make an angle of 60°
in an anti-clockwise direction with the positive direction of the x-axis.
Now, we need to find out the value of the slope of a line and the equation of a line.
The slope of the line m = tan 60° = √3.
The equation of a line using point-slope form formula is (y-y1) = m(x- x1)
Substitute the values (1,2) and slope value √3 in formula. Then we get,
y−3=√3(x−1)
y-3= √3x-√3
√3x−y+3-√3 = 0.
Therefore, the required equation of a straight line is √3x−y+3-√3 = 0.