Worksheet on Equation of a Straight line is helpful for the students who are willing to solve the problems based on the Equation of a Straight Line. If you are preparing for any exam, you can begin by referring to the Equation of a Straight Line Worksheet. Make use of these Equations of a Straight Line Question and Answers during practice and perform well in the test. We have covered various types of questions in our Finding the Equation of a Straight Line Given Two Points Worksheet. So, Practice all the questions from the 10th Grade Math Worksheet as often as possible to understand the concept easily.
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Equation of a Straight Line Worksheet with Answers PDF
Problem 1: What is the equation of a straight line that passes through the points (3,2) and (-5, 4)?
Solution:
Given that, the points are (3,1) and (-2, 4).
So, we will use the point-slope form formula to find the equation.
First, we need to find out the slope of the line.
The Slope is (4-2)/(-5-3) = 2/-8.
Therefore, the equation of the line passing through (3, 2) and (-5, 4) is y – 4 = (2/-8) (x + 5)
⇒ y – 4 = 2x/-8 + 10/-8
⇒ y + 2x/-8 = 4 + 10/-8
⇒ 2x – 8y = -32+10
2x-8y = -22
2x-8y+22=0
So, the required equation of a straight line is 2x – 8y+22=0.
Problem 2: The equation of a line is given by 3x – 9y + 4 = 0. Find the slope of both intercepts?
Solution:
As given in the question, the equation is 3x – 9y + 4 = 0.
Now, it can be represented in slope-intercept form as:
y = x/4 + 1/3
Now compare the equation with y = mx + c,
So, the slope of the line is m = 1/4.
Now, the above equation will be rewrite in the intercept form as,
x/a + y/b = 1
3x – 9y = -4
x/(-4/3) – y/(-9/-4) = 1
Thus, the x-intercept is given as a = -4/3 and the y-intercept is b = 9/4.
Problem 3: The equation of a line is given by, 12x – y + 10 = 0. Find the value of the slope and both the values of the intercepts?
Solution:
The given equation is 12x – y + 10 = 0
The equation is represented in slope-intercept form as,
y = 12x + 10
Now, comparing it with y = mx + c,
The slope of the line m is 12.
Rewrite the above equation in the intercept form as,
x/a + y/b = 1
12x – y = -10
x/(-10/12) + y/10 = 0
Hence, the x-intercept is given as a = -10/12 and the y-intercept as b = 10, and the value of the slope is 12.
Problem 4: Find the equation of the line whose gradient is 3/2 and which passes through P, where P divides the line segment joining A(-4, 12) and B (6, -5) in the ratio 2 : 3 internally.
Solution:
Given that, the line segment joining at A(-4,12) and B(6,-5)
Now, we need to find out point P.
So, the formula is P = (lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)
Substitute the values in the formula. We get,
P = (2(6) + 3(-4))/(2+3), (2(-5) + 3(10))/(2+3)
= (12 -12)/5, (-10 + 30)/5
= 0/5, 20/5
P = (0, 4)
Next, the (x₁, y₁) is (0, 4)
That is m = 3/2.
(y – y₁) = m(x – x₁)
Place the y₁ value in the above equation.
(y – 4) = (3/2)(x – 0)
2(y – 4) = 3x
2y – 8 = 3x
3x – 2y + 8 = 0
Thus, the required equation is 3x – 2y + 8 = 0.
Problem 5: What is the general equation of the straight line whose angle of inclination is 45° and the y-intercept is 3/6?
Solution:
In the given question, the angle of inclination is 45°
We can get the slope from the angle of inclination of 45°
So, the slope of the line m is tan45°
So, the value of m is 1.
Next, the given y-intercept b is 3/6.
We know that, the Equation of a straight line in the slope-intercept form as,
y = mx + b
Substitute the value of m = 1 and b = 3/6
Then the value of y is x + 3/6.
Now, multiply each side by 6.
6y = 6x + 3
Subtract 6y from each side,
0 = 6x – 6y + 3
6x – 6y + 3 = 0
Hence, the general equation of straight line is 6x – 6y + 3 = 0.
Problem 6: If the straight line x-intercept and y-intercept are 2/3 and 3/4 respectively. Find the general equation of the straight line.
Solution:
Given that,
The x-intercept a is 2/3.
The y-intercept b is 3/4.
Now, we will find the equation of a straight line.
So, the equation of the straight line in intercept form is
x/a + y/b = 1
Now, substitute the value of a is 2/3 and b is 3/4 in the above equation.
We get, x/(2/3) + y/(3/4) = 1
3x / 2 + 4y / 3 = 1
(9x + 8y) / 6 = 1
Now, multiply by 6 on both sides.
9x + 8y = 6
Subtract 6 on both sides. Then we get,
9x + 8y – 6 = 0
Therefore, the general equation of straight line is 9x + 8y – 6 = 0.
Problem 7: If the straight line has a slope of 5. If the line cuts the y-axis at -4. Find the general equation of the straight line?
Solution:
As given in the question,
The slope of a straight line, m is 5.
The line cut at the y-axis is -4. So the y-intercept is -4.
We know the formula of the Equation of a straight line in slope-intercept form is,
y = mx + b
Now, substitute the value of m and b in the above equation.
y = 5x – 4
Subtract y from each side.
0 = 5x – y – 4
Thus, the required general equation of the straight line is 5x-y-4=0.
Problem 8: Find the equation of the line which intersects the y-axis at a distance of 4 units above the origin and makes an angle of 30° with the positive direction of the x-axis.
Solution:
Given that,
Line AB intersects the y-axis 4 units of the origin.
As we know, the y-axis and x-coordinate will be 0 always.
Therefore, Line AB cuts the y-axis at P(0,4)
Next, line AB makes an angle of 30° with the x-axis.
So, the Slope of the line = tan30°= 1/√3
As we know the equation of a line passing through (x1,y1) and having slope m is (y-y1) = m(x-x1)
Substitute the values in the formula. We get the equation of the line AB is,
(y−4)=1/√3(x−0)
x−√3y+4=0
Hence, the required equation of a line is x−√3y+4=0.
Problem 9: Find the value of x. If the slope is 2 and the points are (2,2) and (x,6)?
Solution:
In the given question, the values are
The slope of a line m is 2.
The points of the line are (2,2) and (x, 6)
Now, we will find the value of the x.
We all know the formula of the slope of a line m = (y2 – y1)/(x2 – x1)
Substitute the values in the formula. We get,
2 = (6 – 2)/(x – 2)
2(x-2) =4
x-2 = 2
x = 4
So, the value of the x is 4.
Problem 10: Find the equations of the straight lines each passing through the point (6, -2), and whose sum of the intercepts is 5?
Solution:
Given that,
The sum of the intercept is 5.
The lines (x,y) are passing through the points (6, -2).
So ‘a’ and ‘b’ be the x-intercept and y-intercept of the required straight line respectively.
So, the value of a+b is 5.
Now, subtract ‘a’ from both sides. Then it will be,
a+b-a = 5-a
b = 5-a
Next, the equation of the straight line in intercept form is,
x/a + y/b = 1
Now, place the value of b in the above formula. We get,
x / a + y / (5-a) = 1
Simplify the equation,
[(5-a)x + ay ] / a(5-a) = 1
i.e., (5-a)x + ay = a(5-a) —–(1)
While, the straight line is passing through (6, -2).
So, (x, y) = (6, -2).The values of the x and y are substituted in the equation 1.
Then the equation is (5 – a)6 – 2a = a(5 – a)
i.e., 30 – 6a – 2a = 5a – a²
a² – 13a + 30 = 0
a² – 13a + 30 = 0
(a – 10)(a – 3) = 0
Thusn the value of the a is10 and 3.
When a = 10,
Put the value of a in equation(1)
(5 – 10)x + 10y = 10(5 – 10)
– 5x + 10y = – 50
5x – 10y – 50 = 0
x – 2y – 10 = 0
Now, put the value of a is 3, in equation 1.
(5 – 3)x + 3y = 3(5 – 3)
2x + 3y = 6
Therefore, the general equations of the required straight line is x – 2y – 10 = 0 and 2x + 3y – 6 = 0.