The diagram below shows a rxkgk56ueb*ztajmo 8a4 lx i859j 74a 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 rosd2zk c7 nx,)zvs c(5tad will pass through the point A(0,5) and will be perpendicular to the road connecting the points B(3,0) anzv5cksc szd,)xt7 2( nd 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 line bvv+x46cv9f3:r b aqk$L_1$​ 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 line,)rjc kasi0 ny4mw 98g $L_1$​ 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 stra61)wjp0xt xyu4z-dnl ight 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- spass6bi f32a6yzr ( the coordinates of point Q are(5,1). M is the midpoi(b-p iy3sa6 s62sarfznt 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 coordinate39*mcac q9natdyvc2cb yd7y5/ a s0i/s of point N are (8,6). M is the midp3dcby i/a920q 5cv9 y7cystm/a*dn a coint 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 lin 5tv9u w6q.apem(e vl;)vqo-xe $L_1$​ 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 point B a6)t m9f2zh z*:6s 9iyvsk2sx 3af azhwre (−2,−1) *39m2x:9t) 6iywfszhfa ah6kssz2 zv. $L_1$ 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 /q nhcd-dx lyu39.li0 are (5,2).xycdd 0i9n u/.lq3h- l $L_1$ 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 straight lines irm1l) jrb(rb. e 5f8x$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 0nd;vu .4xm f5b3myqp new vehicle. The one he wants to buy costs 16000 dollars , however he doesn't have pf4q vd50.3ybmu n ;mxthat much money in his bank. The salesman offers him 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 triangle ABC. The point C has cos n5rhkgg55w67bh dmqy(26f6o1wo5 x 4 basnd ordinates (9,5) and the equation of the line (AB) is x+4y=1x aw5 n46s1ydmbhgfnbkdq h5(r27 6s 6oog55w2.

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 lo b /.;9s t1xt0kyvdrsnu)/+ tefw o.cine $L_1$​ 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 ≈