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 thisÂ 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.

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### 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 (x_{1},y_{1}) 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-y_{1}) = m(x-x_{1})

Now place the given values in the above formula. It will be,

y-(5) = (âˆ’8)(x âˆ’ (-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:Â **Find the equation of a line that passes through the points (4, â€“2) and (1, 4) in point-slope form.

**Solution:**

In the given question,

The points (x_{1}Â ,y_{1}) is (4, -2).

The points of (x_{2} ,y_{2})are (1, 4).

The slope of the line that passes through the points formula is

m = (y_{2} â€“ y_{1})/ (x_{2} â€“ x_{1})

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-y_{1}) = m(x- x_{1})

y -(-2) = -2(x â€“ 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 â€“ 4 = -2(x â€“ 1)

y â€“ 4 = -2x +2

2x + y â€“ 4 -2= 0

2x + y -6 = 0

Therefore, the equation of the line in point slope form isÂ 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 135Â°. Find the equation for a straight line?

**Solution:**

As given in the question,

The points ( x_{1}, y_{1}) are (8,-5)

An angle of 135 Â° with a positive direction of the x-axis line.

The slope of the line m is,

m= tan 135 Â° = tan (90 Â° + 45 Â°) = – cot 45 Â° = -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- y_{1}) = m(x- x_{1})

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 ( x_{1}, y_{1}) 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-y_{1}) = m(x-x_{1})

After placing the given values in the above formula. It will be,

y-(-6) = (âˆ’1/3)(x âˆ’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 60Â° and which passes through the point (0, â€“ 3).

**Solution:**

In the given question,

The equation of a line passes through the points are (0,-3) = ( x_{1}, y_{1})

The inclination of a straight line is 60Â°.

First, find the slope of a line ‘m’ value.

So, the slope of a line is tan 60Â° = âˆš3.

We know the formula of an equation of a line using the point-slope form is,

(y-y_{1}) = m(x-x_{1})

After putting the given values in the above formula. It will be,

y-(-3) = (âˆš3)(x âˆ’0)

y + 3 = âˆš3 (x â€“ 0)

y + 3 = âˆš3x

âˆš3x â€“ y â€“ 3 = 0

Therefore, the required equation of a line is âˆš3x â€“ y â€“ 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 (x_{1}, y_{1}) are (0, 6).

The formula for point-slope of line is (y-y_{1}) = m(x-x_{1}).

Now substitute the values of m, x, and y in the formula.

y â€“ 6 = (-1)(x â€“ 0)

y â€“ 6 = (-1)x

y â€“ 6 = -x

We will get the equation is y + x â€“ 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 isÂ -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 (x_{1}, y_{1}) value is (-7,0).

Now, Put the values of m and (x_{1}, y_{1}) in the point-slope form formula.

The formula is (y-y_{1}) = m(x-x_{1}).

After substituting the value, we get

y-y_{1} = (-3)(x â€“(-7))

y-y_{1} = (-3)(x + 7)

y-0 = -3x â€“ 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, â€“4) and (5, 10) in point-slope form.

**Solution:**

In the given question,

The points (x_{1}Â ,y_{1}) is (8, -4).

The points of (x_{2},y_{2})are (6, 10).

The slope of the line which is passing through the points formula is,

m = (y_{2} â€“ y_{1})/ (x_{2} â€“ x_{1})

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-y_{1}) = m(x- x_{1})

y -(-4) = -7(x â€“ 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 â€“ 10 = -7(x â€“ 6)

y â€“ 10 = -7x +42

7x + y â€“ 10 – 42= 0

7x + y -52 = 0

Therefore, the equation of the line using the point-slope form isÂ 7x + y – 52 = 0.

**Problem 10:** The equation of the line passing through (1, 3) and making an angle of 30Â° in a clockwise direction with the positive direction of the y-axis.

**Solution:**

As given in the question,

The points (x_{1}Â ,y_{1}) is (1,3)

The line makes an angle of 30Â° in a clockwise direction with the positive direction of the y-axis. It will make an angle of 60Â°

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 60Â° = âˆš3.

The equation of a line using point-slope form formula is (y-y_{1}) = m(x- x_{1})

Substitute the values (1,2) and slopeÂ value âˆš3 in formula. Then we get,

yâˆ’3=âˆš3(xâˆ’1)

y-3= âˆš3x-âˆš3

âˆš3xâˆ’y+3-âˆš3 = 0.

Therefore, the required equation of a straight line is âˆš3xâˆ’y+3-âˆš3 = 0.