EQUATION OF A STRAIGHT LINE WORKSHEET
About "Equation of a Straight Line Worksheet"
Equation of a Straight Line Worksheet :
Worksheet given in this section is much useful to the students who would like to practice problems on finding the equation of a line.
Before look at the worksheet, if you would like to know the basic stuff about finding the equation of a line,
Equation of a Straight Line Worksheet - Problems
Problem 1 :
Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, -4).
Problem 2 :
Find the general equation of the straight line whose angle of inclination is 45° and y-intercept is 2/5.
Problem 3 :
Find the general equation of the straight line passing through the point (-2, 3) with slope 1/3.
Problem 4 :
Find the general equation of the straight line passing through the points (-1, 1) and (2, -4).
Problem 5 :
The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.
Problem 6 :
If the x-intercept and y-intercept of a straight line are 2/3 and 3/4 respectively, find the general equation of the straight line.
Problem 7 :
Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.
Problem 8 :
Find the general equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis.
Problem 9 :
Find the slope and y-intercept of the straight line whose equation is 4x - 2y + 1 = 0.
Problem 10 :
A straight line has the slope 5. If the line cuts y-axis at -2, find the general equation of the straight line.
Equation of a Straight Line Worksheet - Solutions
Problem 1 :
Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, -4)
Solution :
Let L and L' be the straight lines passing through the point (3, -4) and parallel to x-axis and y-axis respectively.
The y-coordinate of every point on the line L is – 4.
Hence, the equation of the line L is
y = - 4
Similarly, the x-coordinate of every point on the straight line L' is 3
Hence, the equation of the line L' is
x = 3
Problem 2 :
Find the general equation of the straight line whose angle of inclination is 45° and y-intercept is 2/5.
Solution :
From the angle of inclination 45°, we can get the slope.
Slope of the line m = tan45° = 1
Given : y-intercept b = 2/5
So, the equation of the straight line in slope-intercept form is
y = mx + b
Substitute m = 1 and b = 2/5.
y = 1x + 2/5
Multiply each side by 5.
5y = 5x + 2
Subtract 5y from each side.
0 = 5x - 5y + 2
or
5x - 5y + 2 = 0
Hence the general equation of straight line is
5x - 5y + 2 = 0
Problem 3 :
Find the general equation of the straight line passing through the point (-2, 3) with slope 1/3.
Solution :
Given : Point = (-2, 3) and slope m = 1/3
So, the equation of the straight line in point-slope form is
y - y1 = m(x - x1)
Substitute (x1 , y1) = (-2 , 3) and m = 1/3.
y - 3 = 1/3 ⋅ (x + 2)
Multiply each side by 3.
3(y - 3) = x + 2
Simplify.
3y - 9 = x + 2
Subtract 3y from each side.
-9 = x - 3y + 2
Add 9 to each side.
0 = x - 3y + 11
So, the general equation of straight line is
x - 3y + 11 = 0
Problem 4 :
Find the general equation of the straight line passing through the points (-1, 1) and (2, -4).
Solution :
Given : Two points on the straight line : (-1, 1) and (2, -4).
So, the equation of the straight line in two-points form is
(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)
Substitute (x1 , y1) = (-1, 1) and (x2, y2) = (2, -4).
(y - 1) / (-4 - 1) = (x + 1) / (2 + 1)
Simplify.
(y - 1) / (-5) = (x + 1) / 3
Cross multiply.
3(y - 1) = -5(x + 1)
3y - 3 = -5x - 5
5x + 3y + 2 = 0
Hence the general equation of straight line is
5x + 3y + 2 = 0
Problem 5 :
The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.
Solution :
Median is a straight line joining a vertex and the midpoint of the opposite side.
Let D be the midpoint of BC.
The median through A is nothing but the line joining two points A (2,1) and D(1, 4).
So, the equation of the median through A is
(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)
Substitute (x1 , y1) = (2, 1) and (x2, y2) = (1, 4).
(y - 1) / (4 - 1) = (x - 2) / (1 - 2)
Simplify.
(y - 1) / 3 = (x - 2) / (-1)
Cross multiply.
-1(y - 1) = 3(x - 1)
Simplify.
-y + 1 = 3x - 3
3x + y - 4 = 0
Hence, the equation of the median through A is
3x + y - 4 = 0
Problem 6 :
If the x-intercept and y-intercept of a straight line are 2/3 and 3/4 respectively, find the general equation of the straight line.
Solution :
Given :
x- intercept "a" = 2/3
y-intercept "b" = 3/4
So, the equation of the straight line in intercept form is
x/a + y/b = 1
Substitute a = 2/3 and b = 3/4.
x / (2/3) + y / (3/4) = 1
Simplify.
3x / 2 + 4y / 3 = 1
(9x + 8y) / 6 = 1
Multiply each side by 6.
9x + 8y = 6
Subtract 6 from each side from 6.
9x + 8y - 6 = 0
Hence the general equation of straight line is
9x + 8y - 6 = 0
Problem 7 :
Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.
Solution :
Let "a" and "b" be the x-intercept and y-intercept of the required straight line respectively.
Given : Sum of the intercepts = 5
So, we have
a + b = 5
Subtract a from each side.
b = 5 - a
Now, equation of the straight line in intercept form is
x/a + y/b = 1
Plugging b = 5 - a, we get
x / a + y / (5-a) = 1
Simplify.
[(5-a)x + ay ] / a(5-a) = 1
(5-a)x + ay = a(5-a) -----(1)
The straight line is passing through (6, -2).
So, substitute (x, y) = (6, -2).
(5 - a)6 - 2a = a(5 - a)
30 - 6a - 2a = 5a - a²
a² - 13a + 30 = 0
a² - 13a + 30 = 0
(a - 10)(a - 3) = 0
a = 10 and a = 3
When a = 10,
(1)-----> (5 - 10)x + 10y = 10(5 - 10)
- 5x + 10y = - 50
5x - 10y - 50 = 0
x - 2y - 10 = 0
When a = 3,
(1)----->(5 - 3)x + 3y = 3(5 - 3)
2x + 3y = 6
Hence, x - 2y - 10 = 0 and 2x + 3y - 6 = 0 are the general equations of the required straight lines.
Problem 8 :
Find the general equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis.
Solution :
From the given information, we can sketch the two lines as given below.
One line is above the x-axis at a distance of 5 units. And another line is below the x-axis at a distance of 5 units.
Hence, y = 5 and y = -5 are the required straight lines.
Problem 9 :
Find the slope and y-intercept of the straight line whose equation is 4x - 2y + 1 = 0.
Solution :
Since we want to find the slope and y-intercept, let us write the given equation 4x - 2y + 1 = 0 in slope-intercept form.
4x - 2y + 1 = 0
4x + 1 = 2y
Divide each side by 2.
(4x + 1)/2 = y
2x + 1/2 = y
or
y = 2x + 1/2
The above form is slope intercept form.
If we compare y = 2x + 1/2 and y = mx + b, we get
m = 2 and b = 1/2
Hence, the slope is 2 and y-intercept is 1/2.
Problem 10 :
A straight line has the slope 5. If the line cuts y-axis at -2, find the general equation of the straight line.
Solution :
Since the line cuts y-axis at "-2", clearly y-intercept is "-2"
So, the slope m = 5 and y-intercept b = -2.
Equation of a straight line in slope-intercept form :
y = mx + b
Substitute m = 5 and b = -2
y = 5x - 2
Subtract y from each side.
5x - y - 2 = 0
Hence, the general equation of the required line is
5x - y - 2 = 0
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