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1 The line / passes through the coordinates (2, 1) and (4, —5). [ Find an equation for /. XY Le Ya —— M = Ar ~My Arv-% = -$-\ yr - -6 - -2 7 r yu-Yy, = M(x -@,) Y Ug yo l = -3(x-2) -| = -3x +6 u J Yy =-3x +7 = 2 The line /; has the equation 2x + 3y +5=0 The line /; passes through the coordinates (1, 7) and (5, 3 | __ BY, 22.92 Determine, giving full reasons for your answer, whether /; and /; are parallel, perpendicular or neither. L- 34 =-a2x -5 Ug -2 4 = Qo we ms - % x: uM = 4.- 1%) a1 -| = 1-7 §-/ -_ —& = -3 4 2 Ati aes Wor parallé) os art the some grediact -1 === tok perpendicular os “Axe Xi 4 Ae Je 3 (a) Find an equation of the straight line passing through the points (2, 5) and (5, —1). Give your answer in the form ax + by + c = 0, where a, b and c are integers. (3) The line crosses the x axis at point A, the y axis at point B and O is the origin. (b) Find the area of triangle AOB. (3) -[| -—-S —~ -—-6 a/ me ole = = 7 5S --2 7 6z t 74-23 =0 rv b/ Crosses 2% wher yoo 6z - 23=6 6x = 23 x = *% crosses y when x=0 74-23 =0 v Ty 7-3 y = 234 Nn a2 IN Area = > 1 \ = S21 units Pp 34 5 The straight line | has equation 2x — 3y + 24 = 0 and meets the coordinate axis at the points A and B. Find the distance of the midpoint of AB from the origin. - Give your answer in the form k/13 Crosses 2% whea y=o 22+ 24=0 a =-I2 cross€s YY wher x =o —3y +24 =0 y 8 (-)2, 0) and (0, €) X 7 uC 4 {= 64) X aN = > ari= yt ret x* = $d x = aie 6 The line /; has gradient 2 and passes through (5, Hu} Qi 1 (a) Find an equation for /; in the form y = mx + ¢ 2) /: is perpendicular to /; and passes through (0, 1) @ (b) Find an equation for /,. Q) a./ y-7 = 2(x-s) 949-7 = Q2x--t6 YT 2x -3 l b/ perp m= --> cz! The line /, has the equation Sx + 2y-4=0 The line /; has the equation x—4y + 1=0 Find the coordinates of the point where /; and /, intersect. Sa t2y = 4 an ty =-/ (use caleutater ) \ (2,5) ME 4 oe) The line /; has the equation 2x — 3y—4=0 OS The line /> is perpendicular to /; and passes through the point (4, —-1) 2,4 Find an equation for /; in the form ax + by +c=0 Qxn—-4Y = 3y - 2 4 >- Sax -_L JS 3 ms > perp mM = - + 3 i e 2(yti)=- 3% -#) ay +2>-~Sx 7/2 =o ou + ay-lo 10 The line /; has gradient 3 and passes through (-2, 5). (a) Find an equation for /, in the form y = mx + ¢ Q) - his perpendicular to /; and passes through (0, 4) [———— (b) Find an equation for /,. Q) fo (c) Find the coordinates of the point where /; and /; intersect. QB) a/ 4-5 = 3(x+2) ( Uv Y-S = Bx +6 y= Sx tl —————<——— b/ perp Ma _ + T v if 3 c/ Tutersectton where : $a +l) = “7X + loy4 = -7 3 x = cet [oO u= 3( 224) +01 U Sle = ¢ 10 { ~2 47 | YK 710 % 16 / ——— a0] The line /; has the equation Sy — 10 = 2x The point P with x coordinate 4 lies on /. The line /; is perpendicular to /; and passes through the point P. (a) Find an equation for /, in the form ax + by+c=0 The lines /; and /2 cross the x axis at the points Q and R respectively. (4) (b) Calculate the area of the triangle OPR. (4) UL: oy -/0 =< 2z Sy = 2x +/0 . > y = 2xu¢2 (m= % ) J S \ 7 when 2w=¢y y= =(4) +2 7 Uv S 7 = 1T8 - <= cep mae (4 '&%) rt 2 Ca af Dy bo y- 2 = f(x -4) J 5 2° a(y -tb\)= -5(2-+) tw) gS 7 7 Qn -36 =~ ~-Sa +20 ¥ 5 lou —~34 = -25x + /06 J = 0 e5x+10¥g - 1/34 Cross 2 wher yrd Vv i Sx -134 =6 man a = '88& ===" ond —(0 = dx —5* a, 7 x =-8 yer. 18 Area= 3 S565 = 2341 = I, a ae [2S === 14 The point A has the coordinates S. 3 point B has the coordinates 5 (7, 7 v Find the equation the perpendicular bisector of AB Midpoint = (-3+7 ji 442) N 2 ° 2 = (2, -1) Sy m= at%e - & = 3 7+ 3 10 5 perp m =- 5 rent 3 ytl= ~3(x£- 2) 3( y+!) = -5(x-2) by = -Sxer7 iw 35+ U 3 3 |__15 — The line /; has equation 4y — 3x = 11 The line /; passes through the points (3, 5) and (—5, 1). oe a. Ay 2 a . Determine, giving full reasons for your answer, Whether lines /; and /; are parallel, perpendicular or neither. Q Gy = 3x + Il ys nt 4 m= Ve NS 3 ~ -l1-5 = -& 2 3 ——— =8 oF s-3 = Same gradient “. porallel 2, uy Lz 4 16 The point A has the coordinates (-2, 3), point B has the coordinates (4, -7). The perpendicular bisector of AB intersects the line y = 2x + | at the point P. Find the coordinates of P. Midpoint TAtt, B= Ldpoin \ aay 51) (1 } -2) x a Crrodient of AB = 2-3 = 12 => 4-72 6 $ 3 Perp mata | 1 a Perpendicular bisector: Jgr725 = (x - .) 5(y+2) = 30> —\) 54t+10 = Sx - 3 Sy = Sx -13 z 13 y= doo ps ee ee Ax +l = #2-15 =) S \iOxn+ S = Sx -!13 lx = -1% ~ = - |8 “7 y= 2(-t®\ + | [v) Xx av, The line /; has equation 2y + 4x +7=0 The line /, has equation y = mx + 4, where m is a constant. Given that /; and /2 are perpendicular. (a) Find the value of m. (b) Find the coordinates of the point where /; and /; meet. Q) QB) a4-<-4x% -7 cs ey = -ax-%&% perp endicular rls gradient = Uv a) uy az 19 In 1960 the life expectancy in the UK was 71 years {__ In 1975 the life expectancy in the UK was 73 years Given that x years is the life expectancy n years after 1960. (a) Using a linear model, form an equation linking x with n. In 2020 the life expectancy in the UK was 81.8 years. -—— (b) Comment on the suitability of your model in light of this information QB) GB) (0,11) (15, 73) n, x, Nz 22 m= 18-71 ~ 1S 19 = 2 n-+T7/ 15 bl n= £0 wr 2-(60) TTI ] 78 ~7 The model underestimated tne life Rxpectucy Tre model moy vot be switable. 22 La Zt Point C has coordinates (2, c) and point D has coordinates (d, 8) The perpendicular bisector of CD has equation 3y + x = 10 Find c and d. B4u+*+x =10 YY Bu = -x +10 ~ ( 4 = sax + le vw rf) J Pecpeadi Ane Qradi eat = 3 (v) 32 = 38 -c o-2 3(d-2) = B-C Ba -6 = ¥-C C+ 3d = 14 Midpoint = (2rd, crs | 34 +z =/od : VT 2 7 2 J Y af cts) + 2t4 = so iS 27 2 3(c+e) + gto = 2o Set ay t2¢d = 20 dc tad = —6 c+ bd =/4 Bc tad = -6 c= -4 a=6 
