The diagram below shows a3zf/ey)3fw+( d) :bcob/qs mmzjwv;y straight line $L_1$​ which passes through A(0,−2) and B(8,0).

1.Write down the coordinates of the midpoint of line segment [AB]. (a,b) a=   b=  
Another line, $L_2$​ , intersects the $y$-axis at C(0,3) and is parallel to $L_1$​​.
2.Find the gradient of $L_2$​​. $L_2$ =   
3.Find the equation of $L_2$​​​, giving your answer in the form y=mx+c. m =    c =   

  A small town is planning the construction of a new road. The new relx)m, 5lmspb01 zf ,en5to/k oad will pass through the point A(0,5) and will be perpendicular to the road cb p),nleo5mle 1 0m5z sxf,kt/onnecting the points B(3,0) and C(6,6). This information is shown in the following diagram.





1.Find the gradient of the line through points B and C. $m_{BC}$ =   
2.Hence, state the gradient of the line through points A and D. $m_{AD}$ =   
3.Find the equation of the line through A and D. Give your answer in the form y=mx+c. m =    c =   
4.Point D lies on the x-axis. Find the coordinates of point D. (a,b) a=   b=  

  The equation of a lin ihzi+0q5hjat+ wa9.m e $L_1$​ is y=0.5x+p. The point A(2,−1) lies on $L_1$
1.Find the value of p. p =   
A second line, $L_2$​ , is perpendicular to $L_1$​ and intersects $L_1$​ at point A.
2.Find the gradient of $L_2$​. $L_2$ =   
3.Find the equation of $L_2$​. Give your answer in the form y=mx+c. m =    c =   
4.Write your answer to part (c) in the form ax+by+d=0, where a,b,d∈Z. a =    b =    c =   

  The equation of a lin/j1r/ae k02*u vmz2 bsve0n ipe $L_1$​ is y−3x+5=0.
1.For the line $L_1$​​, find:
(1)the $x$-intercept;x =   
(2)the gradient. y = ax+b a =    b =   
A second line, $L_2$​​, intersects the $y$-axis at P(0,2)(0,2) and is parallel to $L_1$​.
2.Find the equation of $L_2$​. Give your answer in the form y=mx+c. m =    c =   
A third line, $L_3$​, passes through the point Q(3,1) and is perpendicular to $L_1$​​​.
3.Find the gradient of the line $L_3$. $m_{L_3}$ =   
4.Find the equation of $L_3$, giving your answer in the form ax+by+d=0,
where a,b,d∈Z. a =    b =    d =   

  The diagram shows the straig-. suho.hanico1yi5m; q(6nh ht line $L_1$​, which intersects the $x$-axis at A(−8,0) and the $y$-axis at B(0,4).


1.Write down the coordinates of M, the midpoint of line segment [AB]. (a,b) a=   b=  
2.Calculate the gradient of $L_1$​.
The line $L_2$​ is perpendicular to $L_1$​ and passes through the point P(1,2). $L_1$ =   
3.Find the equation of $L_2$​. Give your answer in the form y=mx+c. m =    c =   

  The straight line, $L_1$​, has the equation y=$\frac{1}{3}$x−5.
1.Write down the $y$-intercept of $L_1$​​. (a,b) a=   b=  
2.Write down the gradient of $L_1$​​. $mL_1$ =   
The line $L_2$​​ is perpendicular to $L_1$​ and passes through the point A(2,4).
3.Find the gradient of $L_2$​. $mL_2$ =   
4.Find the equation of $L_2$, giving your answer in the form ax+by+d=0,
where a,b,d∈Z. a =    b =    d =   

  The coordinates of point P are(−4,6) and to7.oi9wb 4umv yd: xo,he coordinates of point Q are(5,1). M is the midpou4i,dy:o wxbo 7m. vo9int of [PQ].
1.Find the coordinates of M.
$L_1$​ is the line which passes through P and Q. (a,b) a=   b=  
2.Find the gradient of $L_1$.
A new line, $L_2$, is perpendicular to $L_1$ and passes through M. $m_{L_1}$ =   
3.(1)Write down the gradient of $L_2$​. $m_{L_2}$ =   
(2)Write down the equation of $L_2$ in the form y=mx+c. m =    c =   

  The coordinates of point K are (−2,−3) and the coordinates of ,d0x 3t 4.okh,azcm fzpoint N are (8,6). M is the midpoint of [KN.o0fzzhdtmc, a34, x k].
1.Find the coordinates of M.
$L_1$​ is the line which passes through K and N. (a,b) a=   b=  
2.Find the gradient of $L_1$.
A new line, $L_2$​, is perpendicular to $L_1$ and passes through M. $L_1$ =   
3.(1)Write down the gradient of $L_2$​.$L_2$ =   
(2)Write down the equation of $L_2$ in the form y=mx+c. m =    c =   

  The equation of a li2cyw t rxx5.d+rjek0g 8qm;d1ne $L_1$​ is 2x+3y=−5.
1.Find the gradient of $L_1$​.
A second line, $L_2$, is perpendicular to $L_1$​. $mL_1$ =   
2.Find the gradient of $L_2$​.
The point P(4,0)(4,0) lies on $L_2$. $mL_1$ =   
3.Find the equation of $L_2$, giving your answer in the form ax+by+d=0, where a,b,d∈Z. a =    b =    d =   
The point Q is the intersection of $L_1$​ and $L_2$.
4.Find the coordinates of (a,b) a=   b=  

  The coordinates of point A are (6, 03/ b-jtwaq4p-amvawi0kj 5 s−3) and the coordinates of point B are (−2,−1)aa5iv-wsj 4bq0/3j0k -patwm . $L_1$ is the line which passes through A and B.
1.Find the equation of $L_1$.
Point M is the midpoint of [AB]. The line $L_2$​ is perpendicular to $L_1$​and passes through M. y = ax+b a =    b =   
2.Find the equation of$L_2$. y = ax+b a =    b =   

  The coordinates of point A a pwtmdu3.ud b8i 8;wn w)ru(:bre (−1,−7) and the coordinates of point B are (5,2).nd wd8wubp(8umr w; i) :b.3tu $L_1$ is the line which passes through A and B.
1.Find the equation of $L_1$​.
Point M is the midpoint of [AB]. The line $L_2$ is perpendicular to $L_1$​ and passes through M. y = ax+b a=    b =   
2.Find the equation of $L_2$​. y = ax+b a=    b =   

  The diagram below shows the straigh vifa- /52lml8cw0mh bt lines $L_1$1​ and $L_2$.

1.Find:
(1)the gradient of $L_1$​; $mL_1$ =   
(2)the equation of $L_1$​, giving your answer in the form y=mx+c. m =    c =   
The equation of $L_2$​ is x−2y=0. y =   
2.Find the area of the shaded triangle. A =   

  Bruce goes into a car dealership to purchase a new vehicle. The one he wants to83 (n:+xyvc7 dkx1k6.ayf ixnw1rr q x buy costs 16000 dollars , however he doesn't have that much money in his bank. The salesman offers him a finax:xf(i13v +r 18xy ckk.ynqxw rnda 76ncing option of a 30 % deposit followed by 12 monthly payments of 1150 dollars.
1.Find the amount of the deposit. PV =   
2.Calculate the total cost of the loan under this financing option.
Bruce's father generously offers him an interest free loan of 16000 dollars to buy the car to avoid the expensive loan repayments. They agree that Bruce will repay the loan by paying his father $x$ in the first month and $y$ every following month until the 16000 dollars is repaid.
The total amount Bruce's father receives after 12 months is 5200 dollars. This can be expressed by the equation x+11y=5200. The total amount that Bruce's father receives after 24 months is 10600dollars.FV =   
3.Write down a second equation involving x and y. ax+by=c a =    b =    c =   
4.Determine the value of x and the value of y. x =    y =   
5.Calculate the number of months it will take Bruce's father to receive
the 16000 dollars. n =   
Bruce decides to buy a cheaper car for 12000 dollars and invest the remaining 4000 dollars. He is considering two investment options over four years.
Option A: Compound interest at an annual rate of 6.5 %.
Option B: Compound interest at a nominal annual rate of 6 %, compounded monthly.
Express each answer in part (f) to the nearest dollar.
6.Calculate the value of each investment option after four years.
(1)Option A. ≈   
(2)Option B. ≈   


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  The diagram below shows the triangle ABC. i rmr8l3wb l ,y(gi8c -8j)dneThe point C has coordinates (9,5) and tb (r8)ir8el dwnl 3m-8jc,gyihe equation of the line (AB) is x+4y=12.

1.Find the coordinates of:
(1)A; y =   
(2)B. x =   
2.Show that the length of [AB] is 12.412.4 correct to three significant figures. AB ≈   
The point D lies on the line (AB). The line (CD) is perpendicular to the line (AB).
3.Find:
(1)the gradient of (CD); $m_{CD}$ =   
(2)the equation of (CD). y =ax+b a =    b =   
4.Find the coordinates of D. (a,b) a=   b=  
5.Calculate the length of [CD] correct to three significant figures. CD ≈   
6.Calculate the area of triangle ABC. A ≈   

  The equation of the lineyk1c zi 1vm4fk.50ht d5p ovq6 $L_1$​ is 3y−x−5=0. The line $L_1$ is shown on the diagram below.

The point K has coordinates(4,3). The point T has coordinates (2,−3). The point M is the midpoint of [TK].
1.lculate the coordinates of the point M. (a,b) a=   b=  
2.Show that the point K lies on the line $L_1$​.
3.lculate the length of [TK]. TK ≈   
The line $L_2$​ passes through the point M and is perpendicular to $L_1$​.
4.Find the gradient of the line $L_2$. gradients is   
5.Find the equation of $L_2$​. Give your answer in the form ax+by+d=0, where a,b and d are integers.
The point N is the intersection of $L_1$​ and $L_2$. ax+by+c=0 a =    b =    c =   
6.lculate the coordinates of N. (a,b) a=   b=  


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  The line $L_1$​ has equation 2y−x−10=0 and is shown on the diagram.

The point A has coordinates (2,6)(2,6).
1.Show that A lies on $L_1$​.
The point C has coordinates (8,18)(8,18). M is the midpoint of [AC].
2.Find the coordinates of M. (a,b) a=   b=  
3.Find the length of [AC]. AC ≈   
The straight line, $L_2$​, is perpendicular to [AC] and passes through M.
4.Show that the equation of $L_2$ is 2y+x−29=0.
The point D is the intersection of $L_1$ and $L_2$.
5.Find the coordinates of D.
The length of [MD] is $\frac{5\sqrt{5}}{4}$​​. (a,b) a=   b=  
6.Write down the length of [MD] correct to three significant figures.
The point B is such that ABCD is a rhombus. MD ≈   
7.Find the area of ABCD. the area ≈