Practice Questions with Solution

Q1: Determine if the lines y = 2x + 3 and y = 2x − 4 are parallel, perpendicular, or neither.

Solution:

To determine if two lines are parallel, perpendicular, or neither, we can look at their slopes.

The slope of a line in the form y = mx + b is m.

For the line y = 2x + 3, the slope is m=2.

For the line y = 2x − 4, the slope is also m=2.

Since both lines have the same slope of 2, they are parallel.

Q2: Find the equation of the line passing through the point (3, 5) and parallel to the line 2x − 3y = 6.

Solution:

To find the equation of a line passing through the point (3, 5) and parallel to the line 2x − 3y = 6, we need to find the slope of the given line and then use this slope to construct the equation of the parallel line.

First, we rearrange the equation of the given line into slope-intercept form (y = mx + b)

2x − 3y = 6

−3y = −2x + 6

y = 2/3x − 2

The slope of this line is m= 2/3.

Since the line we want to find is parallel to this line, it will have the same slope. So, the slope of the parallel line is also m = 2/3.

Now, we can use the point-slope form of the equation of a line to find the equation of the parallel line:

y − y₁ = m(x − x₁)

where (x1, y1) is the given point (3, 5) and m is the slope 2/3

Substituting

x1 = 3, y1 = 5, and m = 2/3 , we get:

y − 5 = 2/3(x − 3)

Now, we can simplify this equation to obtain the final form:

y − 5 = 2/3(x − 3)

y = 2/3x – 2 + 5

So, the equation of the line passing through the point (3, 5) and parallel to the line 2x − 3y = 6 is y = 2/3x + 3.

Q3: Determine if the lines 3x + 2y = 8 and 6x + 4y = 10 are parallel, perpendicular, or neither.

Solution:

To determine if two lines are parallel, perpendicular, or neither, we can compare their slopes.

First, let’s rearrange both equations into slope-intercept form (y = mx + b):

For 3x + 2y = 8

2y = −3x + 8

y = −3/2x + 4

For 6x + 4y = 10

4y = −6x + 10

y = −3/2x + 5/2

Comparing the slopes of both lines, we see that both lines have the same slope, which is −3/2.

Since both lines have the same slope, they are parallel.

Q4: Find the equation of the line perpendicular to y = 4x − 1 passing through the point (2, −3).

Solution:

To find the equation of the line perpendicular to

y = 4x − 1 passing through the point (2, −3), we first need to determine the slope of the given line.

The given line is in the slope-intercept form

y = mx + b, where

m is the slope. In this case, the slope of the given line is m=4.

The slope of any line perpendicular to this line will be the negative reciprocal of 4. So, the slope of the perpendicular line will be -1/4

Now, we can use the point-slope form of a line to find the equation of the perpendicular line. The point-slope form of a line is

y − y₁ = m(x − x₁),

where (x1, y1) is a point on the line and m is the slope.

We are given a point (2, −3) on the perpendicular line, and the slope m = −1/4. Substituting these values into the point-slope form, we get: y − (−3) = −1/4(x − 2)

Simplify: y + 3 = − 1/4x + 1/2

Now, we can rewrite this equation in slope-intercept form by isolating y:

y = −1/4x + 1/2 − 3

y = −1/4x − 5/2

So, the equation of the line perpendicular to y = 4x − 1 passing through the point (2, −3) is y = −1/4x − 5/2.

Q5: Determine if the lines 2x − 3y = 5 and 6x − 9y = 15 are parallel, perpendicular, or neither.

Solution:

To determine if two lines are parallel, perpendicular, or neither, we can examine their slopes.

The slope-intercept form of a line is y = mx + b, where m is the slope of the line.

For the first line 2x − 3y = 5, let’s rewrite it in slope-intercept form:

3y = 2x − 5

y= 2/3x – 5/3

So, the slope of the first line is m1 = 2/3

For the second line 6x − 9y = 15, let’s rewrite it in slope-intercept form:

9y = 6x − 15

y = 2/3x − 5/3

So, the slope of the second line is m2 = 2/3

Since the slopes of both lines are equal (m1 = m2 = 2/3), the lines are parallel.

So, the lines 2x − 3y = 5 and 6x − 9y = 15 are parallel.

Q6: Find the point of intersection of the lines y = 2x − 1 and y = −3x + 7.

Solution:

To find the point of intersection of the two lines, we need to set their equations equal to each other and solve for the value of x.

So, we have: 2x − 1 = −3x + 7

Now, let’s solve for x:

2x + 3x = 7 + 1

5x = 8

x = 8/5

Now that we have found x, let’s substitute it back into one of the equations to find the corresponding y value. Let’s use the first equation:

y = 2(8/5) − 1

y = 16/5 −1

y = 16/5 – 5/5

y = 16−5/5

y = 11/5

So, the point of intersection of the lines y = 2x − 1 and y = −3x + 7 is (8/5, 11/5).

Q7. Determine if the lines 4x + 2y = 6 and 8x + 4y = 12 are parallel, perpendicular, or neither.

Solution:

To determine if two lines are parallel, perpendicular, or neither, we can examine their slopes.

The slope-intercept form of a line is y = mx + b, where m is the slope of the line.

For the first line 4x + 2y = 6, let’s rewrite it in slope-intercept form:

2y = −4x + 6

y = −2x + 3

So, the slope of the first line is m1 =−2.

For the second line 8x + 4y = 12, let’s rewrite it in slope-intercept form:

4y = −8x + 12

y = −2x + 3

So, the slope of the second line is m2 = −2.

Since the slopes of both lines are equal (m1 = m2 = −2), the lines are parallel.

So, the lines 4x + 2y = 6 and 8x + 4y = 12 are parallel.

Q8. Find the intersection point of the lines 3x + 2y = 8 and 2x − 4y = 5.

Solution:

To find the intersection point of two lines, you need to solve the system of equations formed by the equations of the lines. Let’s solve the system of equations

3x + 2y = 8

2x – 4y = 5

We can solve this system by either substitution or elimination method. Let’s use the elimination method.

Multiply the first equation by 2 and the second equation by 3 to eliminate y:

6x + 4y = 16

6x − 12y = 15

Now, subtract the second equation from the first equation:

(6x + 4y) − (6x − 12y) = 16 − 15

6x + 4y − 6x + 12y = 1

16y = 1

Now, solve for y: y = 1/16

Now, substitute y back into one of the original equations to solve for x. Let’s use the first equation

3x + 2(1/16) = 8

3x + 1/8 = 8

3x = 8 − 1/8

3x = 64/8 – 1/8

3x= 63/8

x = 63/24

So, the intersection point of the lines 3x + 2y = 8 and 2x − 4y = 5 is (x, y) = (63/24, 1/16)

Q9. Determine the intersection point of the lines y = 2x + 1 and y = −3x + 5.

Solution:

To find the intersection point of two lines, you need to set their equations equal to each other and solve for the values of x and y.

Given the equations:

y = 2x + 1

y = −3x + 5

Setting them equal to each other:

2x + 1 = −3x + 5

Now, solve for x:

2x + 3x = 5−1

5x = 4

x= 4/5

Now that we have found the value of x, we can substitute it into either equation to find the corresponding y-coordinate. Let’s use the first equation:

y = 2(4/5) + 1

y = 8/5 + 1

y = 8/5 + 5/5

y = 13/5

So, the intersection point of the lines y = 2x + 1 and y = −3x + 5 is (4/5, 13/5).

Q10. Determine if the lines 2x + 3y = 7 and 4x + 6y = 14 are parallel, perpendicular, or neither.

Solution:

To determine if the lines 2x + 3y = 7 and 4x + 6y = 14 are parallel, perpendicular, or neither, we can compare their slopes.

First, we’ll rewrite the equations in slope-intercept form (y = mx + b) where m represents the slope:

For 2x + 3y = 7

3y = −2x + 7

y = −2/3x + 7/3

So, the slope of this line is m1 = −2/3

For 4x + 6y = 14

6y = −4x + 14

y = −4/6x + 14/6

y = −2/3x + 7/3

So, the slope of this line is m2 = −2/3.

Since the slopes of both lines are identical (m1 = m2 = −2/3), the lines are parallel.

Therefore, the lines 2x + 3y = 7 and 4x + 6y = 14 are parallel.

Parallel, Perpendicular and Intersecting lines are three types of lines based on their properties. This article provide worksheet for practice of questions based on parallel, perpendicular and intersecting lines. This article also contain, formula and solved examples to make you understand how to solve the worksheet