?> Distance and Midpoint Formula Worksheet - Page 3 | Problems & Solutions

# Distance and Midpoint Formula Worksheet - Page 3

Distance and Midpoint Formula Worksheet
• Page 3
21.
One endpoint of a line-segment is (5$x$, 7$y$) and other endpoint is (2, 4). Find the value of $x$ and $y$, if the midpoint of the segment is (3, 5).
 a. $x$ = $\frac{4}{5}$ and $y$ = $\frac{7}{8}$ b. $x$ = $\frac{4}{5}$ and $y$ = $\frac{6}{7}$ c. $x$ = $\frac{5}{6}$ and $y$ = $\frac{6}{7}$ d. $x$ = 5 and $y$ = 7

#### Solution:

Two endpoints are (5x, 7y) and (2, 4).

The midpoint of the segment is (3, 5).

Midpoint of a line-segment with endpoints (x1, y1) and (x2, y2) is (x1+x22, y1+y22).
[Use the midpoint formula.]

Midpoint = (5x+22, 7y+42)
[Replace (x1, y1) with (5x, 7y) and (x2, y2) with (2, 4).]

(5x+22, 7y+42) = (3, 5)
[Equate the midpoints.]

5x+22 = 3
[Equate x-coordinates.]

5x + 2 = 6

5x = 6 - 2 = 4
[Subtract 2 from both sides.]

x = 4 / 5
[Divide by 5 on both sides.]

7y+42 = 5
[Equate y-coordinates.]

7y + 4 = 10

7y = 10 - 4 = 6
[Subtract 4 from both sides.]

y = 6 / 7
[Divide by 7 on both sides.]

x = 4 / 5 and y = 6 / 7

22.
What is the midpoint of the line-segment with one endpoint at origin and other endpoint at ($x$1, $y$1)?
 a. ($x$1, $y$1) b. ($\frac{{x}_{1}}{2}$, $\frac{{y}_{1}}{2}$) c. (2$x$1, $y$1) d. (${x}_{1}$ - 2, ${y}_{1}$ - 2)

#### Solution:

One endpoint of the line-segment is at origin (0, 0) and other endpoint is at (x1,y1).

Midpoint of a line-segment with endpoints (x1, y1) and (x2, y2) is (x1+x22, y1+y22).
[Use the midpoint formula.]

Midpoint = (x1 + 02, y1 + 02)
[Replace (x2, y2) with (0, 0).]

= (x12, y12)
[Simplify the numerators.]

The midpoint of the line-segment with one endpoint at origin and other endpoint at (x1, y1) is (x12, y12).

23.
Find the midpoint of the segment OB if B is at (8, 10) and O is the origin.
 a. (4, - 5) b. (- 4, 5) c. (- 4, - 5) d. (4, 5)

#### Solution:

B is at (8, 10) and O is the origin (0, 0).

Midpoint of a line segment with endpoints (0, 0) and (x1, y1) is (x12, y12).
[Use the midpoint formula.]

Midpoint of OB = (8 / 2 , 10 / 2)
[Replace (x1, y1) with (8, 10).]

= (4, 5)
[Write the fractions in simplest form.]

The midpoint of the line-segment OB = (4, 5).

24.
In the line-segment AB, the coordinates of A are ($x$, $y$) and B are (4$x$, 5$y$) and the midpoint of the segment AB is at (15, 6). Find the coordinates of A and B.
 a. A(- 6, 2) and B(- 24, 10) b. A(6, - 2) and B(24, - 10) c. A(- 6, - 2) and B(- 24, - 10) d. A(6, 2) and B(24, 10)

#### Solution:

The coordinates of A are (x, y), B are (4x, 5y)

Midpoint of a line-segment with endpoints (x1, y1) and (x2, y2) is (x1+x22, y1+y22).
[Use the midpoint formula.]

Midpoint of AB = (x + 4x2, y +   5y2)
[Replace (x1, y1) with (x, y) and (x2, y2) with (4x, 5y).]

= (5x2, 6y2)
[Simplify the numerators.]

(5x2, 6y2) = (15, 6)
[Equate the midpoints.]

5x2 = 15; 6y2 = 6
[Equate x and y coordinates.]

x = 6; y = 2
[Simplify.]

The coordinates of A are (x, y) = (6, 2).

The coordinates of B are (4x, 5y) = (4 × 6, 5 × 2) = (24, 10).

25.
The coordinates of A are (5$p$, 6$p$) and its distance from origin is 5$\sqrt{61}$units. Find the values of $p$.
 a. 25 and - 25 b. 11 and - 11 c. 5 and - 5 d. 6 and - 6

#### Solution:

The coordinates of A are (5p, 6p).

Distance of a point (x, y) from origin is x2+y2
[Write the distance formula.]

Distance of A from origin = (5p)2+(6p)2
[Replace (x, y) with (5p, 6p).]

= 25p2+36p2
[Evaluate powers.]

= 61p2

The distance of point A from origin is 561units.

61p2 = 561
[Equate distances.]

61p2 = 25 × 61
[Squaring on both sides.]

p2 = 25
[Divide each side by 61.]

p = 25

= ± 5
[Find the square root.]

The values of p are 5 and - 5.

26.
The coordinate plane shows a parallelogram ABCD. Find the length of the diagonal AC.

 a. 4.5 units b. 8.6 units c. 6.4 units d. 5.8 units

#### Solution:

From the graph, the coordinates of A are (- 3, 4) and C are (4, - 1).

Distance between the line segment with endpoints (x1, y1) and (x2, y2) is (x2-x1)2+(y2-y1)2.

Distance between A and C = (4 - (- 3))2+(- 1 - 4)2
[Replace (x1, y1) with (- 3, 4) and (x2, y2) with (4, - 1).]

= 72+(- 5)2
[Subtract.]

= 49 + 25
[Evaluate powers.]

= 74

= 8.6

Length of the line segment = Distance between the endpoints of the segment.

The length of the diagonal AC is 8.6 units.

27.
The coordinates of three points are A(12, 0), B(24, 0) and C(12, 9). What kind of figure is ABC?
 a. scalene triangle b. equilateral triangle c. right triangle d. isosceles triangle

#### Solution:

The coordinates of three points are A(12, 0), B(24, 0) and C(12, 9).

The distance between (x1, y1) and (x2, y2) = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

AB = (24-12)2+(0-0)2
[Replace (x1, y1) with (12, 0) and (x2, y2) with (24, 0).]

= 122+0=144 = 12
[Simplify.]

BC = (12-24)2+(9-0)2
[Replace (x1, y1) with (24, 0) and (x2, y2) with (12, 9).]

= (- 12)2+92=144+81=225 = 15
[Simplify.]

CA = (12-12)2+(0-9)2
[Replace (x1, y1) with (12, 9) and (x2, y2) with (12, 0).]

= 02+(- 9)2=81 = 9
[Simplify.]

AB = 12 units, BC = 15 units and CA = 9 units

BC2 = 152 = 225
AB2 = 122 = 144
AC2 = 92 = 81

BC2 = AB2 + AC2
[Since 225 = 144 + 81.]

ABC is a right triangle, since it satisfies the Pythagorean theorem.

28.
What type of triangle is PQR, if P(0, 12), Q(12, - 4) and R(12, 4)?
 a. isosceles triangle b. right triangle c. equilateral triangle d. scalene triangle

#### Solution:

P(0, 12), Q(12, - 4) and R(12, 4).

The distance between (x1, y1) and (x2, y2) = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

PQ = (12 - 0)2+(-4 - 12)2
[Replace (x1, y1) with (0, 12) and (x2, y2) with (12, - 4).]

= 122+( - 16)2=144+256=400 = 20
[Simplify.]

QR = (12 - 12)2+[4 - (- 4)]2
[Replace (x1, y1) with (12, - 4) and (x2, y2) with (12, 4).]

= 02+82=0+64 = 8
[Simplify.]

RP = (0 - 12)2+(12 - 4)2
[Replace (x1, y1) with (12, 4) and (x2, y2) with (0, 12).]

= ( - 12)2+82=144+64=413
[Simplify.]

PQ = 20, QR = 8 and RP = 413

RP2 + QR2 = (413)2 + 82 = 272 and PQ2 = 202 = 400

RP2 + QR2 ≠ PQ2. So, triangle PQR is not a right triangle.

PQ ≠ QR ≠ RP. So PQR is a scalene triangle.

29.
The coordinates of B are formed by interchanging the coordinates of A. If the coordinates of A are (7, 11), then find the distance between A and B.
 a. 8 units b. 4$\sqrt{2}$ units c. 20 units d. 5$\sqrt{2}$ units

#### Solution:

The coordinates of A are (7, 11).

The coordinates of B are formed by interchanging the coordinates of A.

The coordinates of B are (11, 7).

Distance between the line-segment with endpoints (x1, y1) and (x2, y2) is (x2-x1)2+(y2-y1)2.

Distance between A and B = (11 - 7)2+(7 - 11)2
[Replace (x1, y1) with (7, 11) and (x2, y2) with (11, 7).]

= 42+(- 4)2
[Subtract.]

= 16 + 16

= 32

= 42
[Simplify.]

The distance between A and B is 42units.

30.
Find the distance between the center of the circle(C) and origin(0, 0) of the coordinate axes.

 a. 6$\sqrt{2}$units b. 8$\sqrt{2}$units c. 4$\sqrt{2}$units d. 3$\sqrt{2}$ units

#### Solution:

From the graph, the center of the circle C is at (4, 4) and the origin is O(0, 0).

Distance of a point (x, y) from origin is x2+y2
[Write the distance formula.]

Distance between the center of the circle C and origin = 42+42
[Replace (x, y) with (4, 4).]

= 16 + 16
[Substitute the values.]

= 32