Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Tier A

Three points are
O(0,0),A(4,2) and
B(1,2). Find the equation of the line which is parallel to
OA,C is the point on this line whose
x-coordinate is
-(11)/(5)
y-coordinate of
C and show that
OB=OC.
In the trapezium
ABCD,AD is parallel to
BC and the points
A,B and
C are
(-2,-2),(2,5) and
(8,3) respectively. The perpendicular from
B to
AC meets
AC at
M. Find
(i) the equations of
AC and
BM,
(ii) the coordinates of
M,
(iii) the value of
k, where
k > 0, given that
D is the point
(k,-k),
(iv) the coordinates of
P, on
BM produced such that
PM=BM,
The points
A,B and
C have coordinates
(0,1),(1,3) and
(-1,-1) respectively.
(i) Calculate the gradient of the line
AB.
(ii) Find the equation of the line
AB.
(iii) The point
(10,k) lies on the line
AB produced. Find the value of
k.

Tier A $\newline$ $1$ . Three points are $O(0,0), A(4,2)$ and $B(1,2)$ . Find the equation of the line which is parallel to $O A, C$ is the point on this line whose $x$ -coordinate is $-\frac{11}{5}$ $y$ -coordinate of $C$ and show that $O B=O C$ . $\newline$ $2$ . In the trapezium $A B C D, A D$ is parallel to $B C$ and the points $B(1,2)$ $0$ and $C$ are $B(1,2)$ $2$ and $B(1,2)$ $3$ respectively. The perpendicular from $B(1,2)$ $4$ to $B(1,2)$ $5$ meets $B(1,2)$ $5$ at $B(1,2)$ $7$ . Find $\newline$ (i) the equations of $B(1,2)$ $5$ and $B(1,2)$ $9$ , $\newline$ (ii) the coordinates of $B(1,2)$ $7$ , $\newline$ (iii) the value of $O A, C$ $1$ , where $O A, C$ $2$ , given that $O A, C$ $3$ is the point $O A, C$ $4$ , $\newline$ (iv) the coordinates of $O A, C$ $5$ , on $B(1,2)$ $9$ produced such that $O A, C$ $7$ , $\newline$ $3$ . The points $B(1,2)$ $0$ and $C$ have coordinates $x$ $0$ and $x$ $1$ respectively. $\newline$ (i) Calculate the gradient of the line $x$ $2$ . $\newline$ (ii) Find the equation of the line $x$ $2$ . $\newline$ (iii) The point $x$ $4$ lies on the line $x$ $2$ produced. Find the value of $O A, C$ $1$ .

View step-by-step help

Home

Math Problems

Algebra 2

Write equations of cosine functions using properties

Full solution

Q. Tier A

\newline

1

. Three points are

O(0,0), A(4,2)

and

B(1,2)

. Find the equation of the line which is parallel to

O A, C

is the point on this line whose

x

-coordinate is

-\frac{11}{5}

y

-coordinate of

C

and show that

O B=O C

\newline

2

. In the trapezium

A B C D, A D

is parallel to

B C

and the points

B(1,2)

0

and

C

are

B(1,2)

2

and

B(1,2)

3

respectively. The perpendicular from

B(1,2)

4

B(1,2)

5

meets

B(1,2)

5

B(1,2)

7

. Find

\newline

(i) the equations of

B(1,2)

5

and

B(1,2)

9

\newline

(ii) the coordinates of

B(1,2)

7

\newline

(iii) the value of

O A, C

1

, where

O A, C

2

, given that

O A, C

3

is the point

O A, C

4

\newline

(iv) the coordinates of

O A, C

5

, on

B(1,2)

9

produced such that

O A, C

7

\newline

3

. The points

B(1,2)

0

and

C

have coordinates

x

0

and

x

1

respectively.

\newline

(i) Calculate the gradient of the line

x

2

\newline

(ii) Find the equation of the line

x

2

\newline

(iii) The point

x

4

lies on the line

x

2

produced. Find the value of

O A, C

1

Find Slope of OA: To find the equation of the line parallel to OA, we first find the slope of OA using points $O(0,0)$ and $A(4,2)$ . Slope of OA = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{4 - 0} = \frac{1}{2}$ .
Equation of Line Parallel to OA: The line parallel to OA will have the same slope, $\frac{1}{2}$ . Using point-slope form $y - y_1 = m(x - x_1)$ with point C's x-coordinate $-\left(\frac{11}{5}\right)$ , we find the y-coordinate of C. $\newline$ Let $y_C$ be the y-coordinate of C. Then $y_C - 0 = \left(\frac{1}{2}\right)\left(-\left(\frac{11}{5}\right) - 0\right)$ , $y_C = -\left(\frac{11}{10}\right)$ .
Length of OB and OC: The equation of the line parallel to OA passing through C( $-(11/5)$ , $-(11/10)$ ) is $y = (1/2)x - (11/10)$ .
Equation of AC: To show $OB=OC$ , find the length of $OB$ using points $O(0,0)$ and $B(1,2)$ . $\newline$ $OB = \sqrt{((2 - 0)^2 + (1 - 0)^2)} = \sqrt{(4 + 1)} = \sqrt{5}.$
Equation of BM: Find the length of OC using points $O(0,0)$ and $C\left(-\left(\frac{11}{5}\right), -\left(\frac{11}{10}\right)\right)$ . $OC = \sqrt{\left(-\left(\frac{11}{10}\right) - 0\right)^2 + \left(-\left(\frac{11}{5}\right) - 0\right)^2} = \sqrt{\left(\frac{121}{100}\right) + \left(\frac{121}{25}\right)} = \sqrt{\frac{121}{100} + \frac{484}{100}} = \sqrt{\frac{605}{100}} = \sqrt{\frac{121}{20}} = \sqrt{5}.$
Coordinates of M: Since $OB = OC = \sqrt{5}$ , $OB=OC$ is shown to be true.
Value of k for AD: For the trapezium ABCD, to find the equation of AC, use points $A(-2,-2)$ and $C(8,3)$ . $\newline$ Slope of AC = $\frac{3 - (-2)}{8 - (-2)} = \frac{5}{10} = \frac{1}{2}$ .
Coordinates of P: Using point-slope form with point $A(-2,-2)$ , the equation of $AC$ is $y + 2 = \frac{1}{2}(x + 2)$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for x: $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $x = 4$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\newline$ $\frac{1}{2}x - 1 = -2x + 9$ .Solve for x: $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $x = 4$ .Substitute $x = 4$ into the equation of AC to find y: $\frac{1}{2}$ $1$ , $\frac{1}{2}$ $2$ , $\frac{1}{2}$ $3$ .The coordinates of M are $\frac{1}{2}$ $4$ .For part (iii), to find the value of $\frac{1}{2}$ $5$ where $\frac{1}{2}$ $6$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $4$ units away from M in the direction opposite to B.
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for x: $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $x = 4$ .Substitute $x = 4$ into the equation of AC to find y: $\frac{1}{2}$ $1$ , $\frac{1}{2}$ $2$ , $\frac{1}{2}$ $3$ .The coordinates of M are $\frac{1}{2}$ $4$ .For part (iii), to find the value of $\frac{1}{2}$ $5$ where $\frac{1}{2}$ $6$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $\frac{1}{2}$ $9$ , the equation of AD is $-2$ $0$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $2$ into the equation of AD to find $\frac{1}{2}$ $5$ : $-2$ $4$ .Solve for $\frac{1}{2}$ $5$ : $-2$ $6$ , $-2$ $7$ , $-2$ $8$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $4$ .BM = $(2,5)$ $1$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $2$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $3$ . Normalize this to get the unit vector in the direction from M to B.
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for x: $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $x = 4$ .Substitute $x = 4$ into the equation of AC to find y: $\frac{1}{2}$ $1$ , $\frac{1}{2}$ $2$ , $\frac{1}{2}$ $3$ .The coordinates of M are $\frac{1}{2}$ $4$ .For part (iii), to find the value of $\frac{1}{2}$ $5$ where $\frac{1}{2}$ $6$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $\frac{1}{2}$ $9$ , the equation of AD is $-2$ $0$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $2$ into the equation of AD to find $\frac{1}{2}$ $5$ : $-2$ $4$ .Solve for $\frac{1}{2}$ $5$ : $-2$ $6$ , $-2$ $7$ , $-2$ $8$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $4$ .BM = $(2,5)$ $1$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $2$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $3$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $4$ . Multiply this by $(2,5)$ $2$ to get the direction from M to P: $(2,5)$ $6$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $4$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $5$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $6$ . Multiply this by $(2,5)$ $4$ to get the direction from M to P: $(2,5)$ $8$ .Add the direction vector to M's coordinates to find P: P = $(2,5)$ $9$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $4$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $5$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $6$ . Multiply this by $(2,5)$ $4$ to get the direction from M to P: $(2,5)$ $8$ .Add the direction vector to M's coordinates to find P: P = $(2,5)$ $9$ .For the points A $y - 5 = -2(x - 2)$ $0$ , B $y - 5 = -2(x - 2)$ $1$ , calculate the gradient of AB. Slope of AB = $y - 5 = -2(x - 2)$ $2$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $4$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $5$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $6$ . Multiply this by $(2,5)$ $4$ to get the direction from M to P: $(2,5)$ $8$ .Add the direction vector to M's coordinates to find P: P = $(2,5)$ $9$ .For the points A $y - 5 = -2(x - 2)$ $0$ , B $y - 5 = -2(x - 2)$ $1$ , calculate the gradient of AB. Slope of AB = $y - 5 = -2(x - 2)$ $2$ .The equation of AB using point-slope form with point A $y - 5 = -2(x - 2)$ $0$ is $y - 5 = -2(x - 2)$ $4$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for x: $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $x = 4$ .Substitute $x = 4$ into the equation of AC to find y: $\frac{1}{2}$ $1$ , $\frac{1}{2}$ $2$ , $\frac{1}{2}$ $3$ .The coordinates of M are $\frac{1}{2}$ $4$ .For part (iii), to find the value of $\frac{1}{2}$ $5$ where $\frac{1}{2}$ $6$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $\frac{1}{2}$ $9$ , the equation of AD is $-2$ $0$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $2$ into the equation of AD to find $\frac{1}{2}$ $5$ : $-2$ $4$ .Solve for $\frac{1}{2}$ $5$ : $-2$ $6$ , $-2$ $7$ , $-2$ $8$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $4$ .BM = $(2,5)$ $1$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $2$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $3$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $4$ . Multiply this by $(2,5)$ $2$ to get the direction from M to P: $(2,5)$ $6$ .Add the direction vector to M's coordinates to find P: P = $(2,5)$ $7$ .For the points A $(2,5)$ $8$ , B $(2,5)$ $9$ , calculate the gradient of AB.Slope of AB = $y - 5 = -2(x - 2)$ $0$ .The equation of AB using point-slope form with point A $(2,5)$ $8$ is $y - 5 = -2(x - 2)$ $2$ .Simplify the equation of AB to get $y - 5 = -2(x - 2)$ $3$ .
Gradient of AB: Simplify the equation of AC to get $y = \frac{1}{2}x - 1$ .To find the equation of BM, we need the slope of AC, which is perpendicular to BM. The slope of BM is the negative reciprocal of $\frac{1}{2}$ , which is $-2$ .Using point-slope form with point B $(2,5)$ , the equation of BM is $y - 5 = -2(x - 2)$ .Simplify the equation of BM to get $y = -2x + 9$ .To find the coordinates of M, solve the system of equations of AC and BM. $\frac{1}{2}x - 1 = -2x + 9$ .Solve for $x$ : $\frac{1}{2}x + 2x = 9 + 1$ , $\frac{5}{2}x = 10$ , $\frac{1}{2}$ $0$ .Substitute $\frac{1}{2}$ $0$ into the equation of AC to find $\frac{1}{2}$ $2$ : $\frac{1}{2}$ $3$ , $\frac{1}{2}$ $4$ , $\frac{1}{2}$ $5$ .The coordinates of M are $\frac{1}{2}$ $6$ .For part (iii), to find the value of $\frac{1}{2}$ $7$ where $\frac{1}{2}$ $8$ , AD must be parallel to BC. Since AC has a slope of $\frac{1}{2}$ , AD will also have a slope of $\frac{1}{2}$ .Using point-slope form with point A $-2$ $1$ , the equation of AD is $-2$ $2$ .Simplify the equation of AD to get $y = \frac{1}{2}x - 1$ .Substitute $-2$ $4$ into the equation of AD to find $\frac{1}{2}$ $7$ : $-2$ $6$ .Solve for $\frac{1}{2}$ $7$ : $-2$ $8$ , $-2$ $9$ , $(2,5)$ $0$ .For part (iv), to find the coordinates of P, we know PM = BM. The length of BM is the distance between B $(2,5)$ and M $\frac{1}{2}$ $6$ .BM = $(2,5)$ $3$ .Since PM = BM, P lies on the line BM and is $(2,5)$ $4$ units away from M in the direction opposite to B.The direction vector from M to B is $(2,5)$ $5$ . Normalize this to get the unit vector in the direction from M to B.The unit vector is $(2,5)$ $6$ . Multiply this by $(2,5)$ $4$ to get the direction from M to P: $(2,5)$ $8$ .Add the direction vector to M's coordinates to find P: P = $(2,5)$ $9$ .For the points A $y - 5 = -2(x - 2)$ $0$ , B $y - 5 = -2(x - 2)$ $1$ , calculate the gradient of AB.Slope of AB = $y - 5 = -2(x - 2)$ $2$ .The equation of AB using point-slope form with point A $y - 5 = -2(x - 2)$ $0$ is $y - 5 = -2(x - 2)$ $4$ .Simplify the equation of AB to get $y - 5 = -2(x - 2)$ $5$ .For the point $y - 5 = -2(x - 2)$ $6$ on AB, substitute $y - 5 = -2(x - 2)$ $7$ into the equation of AB: $y - 5 = -2(x - 2)$ $8$ , $y - 5 = -2(x - 2)$ $9$ .