Practice Problem on Distance Formula
Distance Formula is an important concept in coordinate geometry to find distance between two points or a point and a line or between two lines. This article will explain concepts related to Distance Formula and presents solved and unsolved questions based on them. These questions are essential for students for better clarity and excel in their exam
Important Concepts Related to Distance Formula
Following are some important concepts related to distance formula
Distance between two points (x1, y1) and (x2, y2) is
- d = β(x2 β x1)2 + (y2 β y1)2
- Midpoint = ((x1 + x2)/2 , (y1 + y2)/2)
Distance between a point and a line:
The distance between a point (x0, y0) and a line Ax + By + C = 0 is:
- Distance= β£Ax0 + By0 + Cβ£/βA2 + B2
Distance between parallel lines:
If two lines have equations Ax + By + C1 = 0 and Ax + By + C2 = 0, then the distance between them is:
- Distance= β£C1 β C2β£/βA2 + B2
Practice Questions on Distance Formula with Solution
Example 1. Given two points A(3, 4) and B(7, 9), find the distance between them.
Solution:
To find the distance between two points A(3, 4) and B(7, 9), we use the distance formula:
d = β(7 β 3)2 + (9 β 4)2
= β16+25
= β41
So, the distance between two points A(3, 4) and B(7, 9) is β41.
Example 2. Find the midpoint of the line segment joining the points P(2, 5) and Q(8, -3).
Solution:
The midpoint of a line segment PQ with endpoints P(xβ, yβ) and Q(xβ, yβ) is given by the midpoint formula:
Midpoint = (x1 + x2)/2 , (y1 + y2)/2
= (2 + 8)/2, (5 + (-3))/2
= 5, 1
So, the midpoint of the line segment joining the points P(2, 5) and Q(8, -3) is (5, 1).
Example 3. Determine the distance between the point (4, -1) and the line 3x + 4y β 7 = 0.
Solution:
The distance between a point (xβ, yβ) and a line Ax + By + C = 0 is given by:
distance = β£Ax0 +By0 + Cβ£/βA2 + B2
= β£3(4) + 4(β1) β 7β£/β32 + 42
= β£12β4β7β£ / β9 + 16
= β£1β£ / β25
= 1/5
So, the distance between the point (4, -1) and the line 3x + 4y β 7 = 0 is 1/5.
Example 4. What is the distance between the point (-1, 6) and the line 2x β 3y + 5 = 0?
Solution:
The distance between a point (xβ, yβ) and a line Ax + By + C = 0 is given by:
distance = β£Ax0 +By0 + Cβ£/βA2 + B2
= β£2(β1) β3(6) + 5β£/β22 + (-3)2
= β£β2 β 18 + 5β£ / β4 + 9
= β£β15β£ / β13
= 15/β13
So, the distance between the point (-1, 6) and the line 3x + 4y β 7 = 0 is 15/β13.
Example 5. Find the distance between the parallel lines 2x + 3y β 4 = 0 and 2x + 3y + 6 = 0.
Solution:
To find the distance between the parallel lines 2x + 3y β 4 = 0 and 2x + 3y + 6 = 0, we use the formula:
distance = β£C2 β C1β£/βA2 + B2
Plugging in the values, we get:
distance = β£6 β (β4)β£/β22 + 32
β= β£10β£/β13
= 10/β13
So, the distance between the two parallel lines is 10/β13.
Example 6. Calculate the distance between the parallel lines 4x β 3y β 9 = 0 and 4x β 3y + 7 = 0.
Solution:
To find the distance between the parallel lines 4x β 3y β 9 = 0 and 4x + 3y + 7 = 0, we use the formula:
distance = β£C2 β C1β£/βA2 + B2
Plugging in the values, we get:
distance = β£7 β (-9)β£/β42+ 32
β= β£16β£/β25
= 16/5
So, the distance between the two parallel lines is 16/5.
Example 7. If A(-2, 1) and B(3, -4) are two points, find the distance between them.
Solution:
To find the distance between two points A(-2, 1) and B(3, -4), we use the distance formula:
d = β(3 β (-2))2 + (-4 β 1)2
= β25 + 25
= 5β2
So, the distance between two points A(-2, 1) and B(3, -4) is 5β2.
Example 8. Determine the midpoint of the line segment joining the points C(5, -2) and D(-3, 7).
Solution:
The midpoint of a line segment PQ with endpoints P(xβ, yβ) and Q(xβ, yβ) is given by the midpoint formula:
Midpoint = (x1 + x2)/2 , (y1 + y2)/2
= (5 + (-3))/2, (-2 + 7)/2
= 2, 5
So, the midpoint of the line segment joining the points P(5, -2) and Q(-3, 7) is (2, 5).
Example 9. What is the distance between the point (1, 3) and the line 5x β 2y + 8 = 0?
Solution:
The distance between a point (xβ, yβ) and a line Ax + By + C = 0 is given by:
distance = β£Ax0 +By0 + Cβ£/βA2 + B2
= β£5(1) β 2(3) + 8β£/β52 + (-2)2
= β£5 β 6 + 8β£ / β25 + 4
= β£7β£ / β29
= 7/β29
So, the distance between the point (4, -1) and the line 3x + 4y β 7 = 0 is 7/β29.
Example 10. Find the distance between the parallel lines 3x β 4y + 6 = 0 and 3x β 4y β 2 = 0.
Solution:
To find the distance between the parallel lines 3x β 4y + 6 = 0 and 3x β 4y β 2 = 0, we use the formula:
distance = β£C2 β C1β£/βA2 + B2
Plugging in the values, we get:
distance = β£-2 β (6)β£/β32 + 42
β= β£-8β£/β25
= 8/5
So, the distance between the two parallel lines is 8/5.
Practice Problem on Distance Formula
Q1. Find the distance between the points (3, 4) and (-1, 2).
Q2. Determine the midpoint of the line segment with endpoints (5, -3) and (-7, 8).
Q3. Calculate the distance between the point (2, -1) and the line 3x + 4y β 5 = 0.
Q4. Find the distance between the parallel lines 2x + 3y β 7 = 0 and 2x + 3y + 9 = 0.
Q5. Determine the distance between the points (-2, 5) and (1, -3).
Q6. Calculate the midpoint of the line segment with endpoints (-4, 6) and (8, -2).
Q7. Find the distance between the point (3, 7) and the line 4x β 2y + 10 = 0.
Q8. Determine the distance between the parallel lines 3x + 2y β 6 = 0 and 3x + 2y + 12 = 0.
Q9. Calculate the distance between the points (0, -1) and (5, 4).
Q10. Find the midpoint of the line segment with endpoints (-3, 2) and (7, -6).
FAQs on Practice Problem on Distance Formula
Does distance have a negative value?
No, distance does not have a negative value. Itβs value is always positive or zero.
Why do we need to use Distance Formula?
We need to use Distance Formula to measure the distance between two points.
Can we calculate speed with the help of Distance?
Yes, we can calculate speed with the help of Distance by applying the following formula, Speed = Distance/Time
What is the SI unit used for distance?
The SI unit used for distance is metre (m).