The diagram below shows a straight lineoro-q e6t +x.p $L_1$​ 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 :z xu lzq,xq*(new road. The new road will pass through the point A(0,5) and will be perpendicular to the road connecting thzxul(xq qz ,:*e 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 line3g5,p6s i0bq1 qju+w h/emtsy $L_1$​ 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 lin2az6wi6rg lnz n/1 s b0d;+xxie $L_1$​ 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 straight linl*+)zm t;b1bdvp x4 kbe $L_1$​, 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) anduuxi: ;xt(jo et/6z 5qgb, /xqfw*-it the coordinates of point Q are(5,1). M is the midpoint of [PQ*j ;otqxq,tgzi(u /5:eiu fxx6-t bw/].
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 pol,w *q)dv3m .anksy4virj- v -int N are (8,6). .k* y3 4ds, wqa-rimjlv)-vv nM is the midpoint of [KN].
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 linbd 6(h+ 8oc htl)xfb4le $L_1$​ 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,−3) and the coordinates of poinic76*)* s-xmn7(lf kwj nnz iy)8xrx ot B are (−2,−1). jxn coif6lwrysx*7()8xi 7)n-*kz m n$L_1$ 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 are (−1,−7) and the coordinates of point B are j09)95b cvts0 oni jsl(5,2). l is5j ts0j )b9no9vc0$L_1$ 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 sho5ro4lue:6+ 1dscn / hxw flk5gws the straight 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(pf; f6fi6 /oj0 hm54 tvvgg7escjd9f a new vehicle. The one he wants to buy costs 16000 dollars , however he doesn't have that much money in his bank. The salesman offers hv4 h 760c9 ;tmj6ofgf/ jdp5siegv(ffim a financing 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 triangl v28i1jhdvfbp5 dh v9.e ABC. The point C has coordinates (9,5) and dhf8v29.hv1pdib5 jv the 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 linef ipvrb-ug;s3vc p*3 d, p0xc+ $L_1$​ 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=  


​

  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 ≈