Equation of a Straight Line: past paper questions & worked solutions

$m=\dfrac{8-5}{7\sqrt3-(-2\sqrt3)}=\dfrac{3}{9\sqrt3}=\dfrac{1}{3\sqrt3}=\dfrac{\sqrt3}{9}$.

The gradient of $l_2$ is the negative reciprocal of $\dfrac{\sqrt3}{9}$, so $m_2=-\dfrac9{\sqrt3}=-3\sqrt3$.

$l_2$ passes through $A(-2\sqrt3,5)$, so $y-5=-3\sqrt3(x+2\sqrt3)$.

$y-5=-3\sqrt3x-18$, hence $y=-3\sqrt3x-13$.

$m_{PQ}=\dfrac{11-7}{9-(-3)}=\dfrac4{12}=\dfrac13$.

$m_{QR}=\dfrac{2-11}{12-9}=\dfrac{-9}{3}=-3$.

$m_{PQ}m_{QR}=\dfrac13(-3)=-1$, so $PQ$ and $QR$ are perpendicular. Hence angle $PQR=90^\circ$.

For rectangle $PQRS$, $\overrightarrow{PS}=\overrightarrow{QR}=(12-9,2-11)=(3,-9)$.

$S=P+\overrightarrow{QR}=(-3,7)+(3,-9)=(0,-2)$.

The model $V=pt+q$ passes through $(3,1.05)$ and $(5,1.65)$.

$p=\dfrac{1.65-1.05}{5-3}=\dfrac{0.60}{2}=0.30$.

Using $1.05=0.30(3)+q$, we get $q=0.15$.

For $V=2.50$, $2.50=0.30T+0.15$, so $T=\dfrac{2.35}{0.30}=7.833\ldots$.

To one decimal place, $T=7.8$.

Using $H=at+b$: when $t=2$, $2a+b=1.85$; when $t=7$, $7a+b=3.45$.

Subtract the first equation from the second: $5a=1.60$, so $a=0.32$.

Substitute into $2a+b=1.85$: $0.64+b=1.85$, hence $b=1.21$. At planting, $t=0$, so $H=b=1.21\text{ m}$.

At $x=4$ on $l_1$, $y=10-3(4)=-2$, so $P=(4,-2)$. The gradient of $l_1$ is $-3$, so the gradient of the perpendicular line $l_2$ is $\dfrac{1}{3}$.

$y+2=\dfrac{1}{3}(x-4)$. Multiplying by 3 and rearranging gives $x-3y-10=0$.

For $l_1$, setting $x=0$ gives $M=(0,10)$. From $l_2$, $y=\dfrac{x}{3}-\dfrac{10}{3}$, so $N=\left(0,-\dfrac{10}{3}\right)$.

$MN=10-\left(-\dfrac{10}{3}\right)=\dfrac{40}{3}$. The perpendicular distance from $P$ to the $y$-axis is 4, so $\text{Area}=\dfrac12\times\dfrac{40}{3}\times4=\dfrac{80}{3}$.

$m=\dfrac{-4-20}{3-(-5)}=\dfrac{-24}{8}=-3$.

Using $B(3,-4)$, $y=-3x+c$. Then $-4=-9+c$, so $c=5$.

$M=\left(\dfrac{-5+3}{2},\dfrac{20+(-4)}{2}\right)=(-1,8)$.

Since $l_1$ has gradient $-3$, $l_2$ has gradient $\dfrac13$.

$y-8=\dfrac13(x+1)$, so $3y-24=x+1$ and $x-3y+25=0$.

$\dfrac3x>4$ is equivalent to $\dfrac{3-4x}{x}>0$.

The critical values are $x=0$ and $x=\dfrac34$. A sign check gives the positive interval $0<x<\dfrac34$.

The curve $C$ meets the negative x-axis when $2x^2-50=0$, so $x=-5$. Since $l$ has gradient $3$ and passes through $(-5,0)$, $y=3(x+5)=3x+15$.

The shaded region lies above the line, below the curve, and to the left of their negative x-axis intersection. Hence $y>3x+15$, $y<2x^2-50$, and $x<-5$.

$-2x^2+12x+7=-2(x^2-6x)+7$.

$x^2-6x=(x-3)^2-9$, so $-2x^2+12x+7=-2\big((x-3)^2-9\big)+7=-2(x-3)^2+25$.

Because the coefficient of $(x-3)^2$ is negative, the curve has its maximum when $(x-3)^2=0$. Hence $x=3$, and the maximum value is $y=25$, so $P=(3,25)$.

The line through $O=(0,0)$ and $P=(3,25)$ has gradient $\dfrac{25}{3}$, so $l$ has equation $y=\dfrac{25}{3}x$.

Points in $R$ are below the curve, above the line, and on or to the right of the $y$-axis. Therefore $y\leq-2x^2+12x+7$, $y\geq\dfrac{25}{3}x$, and $x\geq0$.

$l_1:x-2y+25=0$ gives $y=\dfrac12x+\dfrac{25}{2}$, so its gradient is $\dfrac12$.

A line perpendicular to $l_1$ has gradient $-2$.

Since $l_2$ passes through the origin, its equation is $y=-2x$.

Substitute $y=-2x$ into $x-2y+25=0$ to get $x+4x+25=0$.

$5x+25=0$, so $x=-5$ and $y=10$. Hence $I=(-5,10)$.

The distance is $\sqrt{(-5)^2+10^2}=\sqrt{125}=5\sqrt5$.

$3y-2x=30$ gives $y=\dfrac23x+10$, so the gradient of $l_1$ is $\dfrac23$.

Since $l_2$ is perpendicular to $l_1$, its gradient is $-\dfrac32$. Through $A(24,0)$, $y=-\dfrac32(x-24)=-\dfrac32x+36$.

At the intersection, $\dfrac23x+10=-\dfrac32x+36$. Multiplying by $6$ gives $4x+60=-9x+216$, so $13x=156$ and $x=12$.

$y=\dfrac23(12)+10=18$, so $P=(12,18)$.

For the x-intercept of $l_1$, set $y=0$: $-2x=30$, so $x=-15$. Hence $B=(-15,0)$.

The base $BA$ has length $24-(-15)=39$, and the height from $P$ to the x-axis is $18$. Area $=\dfrac12(39)(18)=351$.

The gradient is $\dfrac{-5-13}{4-(-2)}=\dfrac{-18}{6}=-3$. Using $P(-2,13)$, $y-13=-3(x+2)$, hence $y=-3x+7$.

The minimum point is $Q(4,-5)$, so the curve has form $y=a(x-4)^2-5$.

Since $P(-2,13)$ lies on the curve, $13=a(-6)^2-5$. Therefore $18=36a$ and $a=\dfrac12$.

$C$ has equation $y=\dfrac12(x-4)^2-5$.

In the shaded second-quadrant region, points are above the line, below the curve, and to the left of $P$. Therefore $y>-3x+7$, $y<\dfrac12(x-4)^2-5$, and $x<-2$.

$3x-4y+20=0$ gives $y=\frac34x+5$, so the gradient is $\frac34$.

Line $l_2$ is parallel to $l_1$ and passes through $R(8,0)$, so $y=\frac34(x-8)=\frac34x-6$. Therefore $3x-4y-24=0$.

For $l_1$, the x-intercept is $P\left(-\frac{20}{3},0\right)$ and the y-intercept is $Q=(0,5)$.

Base $PR=8-\left(-\frac{20}{3}\right)=\frac{44}{3}$ and height $OQ=5$. Area $=\frac{44}{3}\cdot5=\frac{220}{3}$.

In parallelogram $PQRS$, $\vec{PS}=\vec{QR}=(8,-5)$. Hence $S=P+(8,-5)=\left(-\frac{20}{3}+8,-5\right)=\left(\frac43,-5\right)$.

Using $A(-1,8)$ and $B(7,2)$, $m=\dfrac{2-8}{7-(-1)}=-\dfrac68=-\dfrac34$.

$y-2=-\dfrac34(x-7)$.

$4y-8=-3x+21$, hence $3x+4y-29=0$.

Since $L_2$ is perpendicular to $L_1$, its gradient is $\dfrac43$.

$L_2$ passes through $A(-1,8)$, so $y-8=\dfrac43(x+1)$.

At the x-axis, $y=0$, so $-8=\dfrac43(x+1)$. Hence $x+1=-6$ and $x=-7$, giving $C=(-7,0)$.

$AB=\sqrt{8^2+(-6)^2}=10$ and $AC=\sqrt{(-6)^2+(-8)^2}=10$.

Since $AB\perp AC$, $\text{Area}=\dfrac12(10)(10)=50$.

$m=\dfrac{7-2}{15-4}=\dfrac5{11}$.

$y-2=\dfrac5{11}(x-4)$.

$11y-22=5x-20$, so $5x-11y+2=0$.

Since $l_2$ passes through $A(4,2)$ and is parallel to the x-axis, point $C$ has coordinates $(c,2)$.

$(c-15)^2+(2-7)^2=(5\sqrt5)^2$.

$(c-15)^2+25=125$, so $(c-15)^2=100$ and $c=5$ or $25$.

The height from $B$ to $y=2$ is $5$. For $C=(5,2)$, $AC=1$, so the minimum area is $\dfrac12(1)(5)=\dfrac52$.

The maximum is at $x=-2$, and one root is $x=-5$. The other root is the same distance to the right of $-2$, so it is $x=1$.

Let $f(x)=a(x+5)(x-1)$. Since the maximum point is $(-2,12)$, $12=a(3)(-3)=-9a$, so $a=-\dfrac43$.

The gradient of $l_1$ is $\dfrac45$, so the gradient of a perpendicular line is $-\dfrac54$.

Using the point $(-5,0)$, $y=-\dfrac54x+c$. Then $0=\dfrac{25}{4}+c$, so $c=-\dfrac{25}{4}$ and $l_2$ is $y=-\dfrac54x-\dfrac{25}{4}$.

The region is below the curve and above both straight lines, so $y\le f(x)$, $y\ge\dfrac45x$, and $y\ge-\dfrac54x-\dfrac{25}{4}$.

$2x-5y+7=0$ gives $y=\dfrac25x+\dfrac75$, so the gradient is $\dfrac25$.

The gradient perpendicular to $\dfrac25$ is $-\dfrac52$. Through $A(6,-2)$, $y+2=-\dfrac52(x-6)$, so $y=-\dfrac52x+13$.

At the intersection, $\dfrac25x+\dfrac75=-\dfrac52x+13$. Multiplying by $10$ gives $4x+14=-25x+130$, so $x=4$.

$y=\dfrac25(4)+\dfrac75=3$, so $M=(4,3)$.

In a square, the diagonals bisect each other, so $M$ is the midpoint of $AC$. Thus $C=2M-A=(8,6)-(6,-2)=(2,8)$.

$m=\dfrac{6-9}{10-(-2)}=-\dfrac{3}{12}=-\dfrac14$.

Using $P(-2,9)$, $y=-\dfrac14x+c$. Then $9=\dfrac12+c$, so $c=\dfrac{17}{2}$.

A line perpendicular to gradient $-\dfrac14$ has gradient $4$. Since $l_2$ passes through the origin, $l_2$ is $y=4x$.

At the intersection, $4x=-\dfrac14x+\dfrac{17}{2}$. Hence $\dfrac{17}{4}x=\dfrac{17}{2}$, so $x=2$ and $y=8$.

For $O(0,0)$, $P(-2,9)$ and $Q(10,6)$, the area is $\dfrac12\left|(-2)(6)-9(10)\right|=51$.

Line $l_1$ has gradient $\dfrac12$ and passes through the negative x-axis root $(-2,0)$, so $y=\dfrac12(x+2)=\dfrac12x+1$.

Set $-\dfrac14(x+2)(x-b)=\dfrac12(x+2)$.

Multiplying by $-4$ gives $(x+2)(x-b)=-2(x+2)$, so $(x+2)(x-b+2)=0$.

The root $x=-2$ is the x-axis point, so $P$ has $x=b-2$. Then $y=\dfrac12(b-2)+1=\dfrac b2$.

Since $l_2$ is perpendicular to $l_1$, its gradient is $-2$. It passes through the positive x-axis root $(b,0)$, so $y=-2(x-b)=-2x+2b$.

Through $P\left(b-2,\dfrac b2\right)$ with gradient $-2$: $y-\dfrac b2=-2(x-(b-2))$, so $y=-2x+\dfrac{5b}{2}-4$.

Both equations represent $l_2$, so $2b=\dfrac{5b}{2}-4$. Hence $b=8$.

$4+12x-3x^2=-3(x^2-4x)+4=-3\big((x-2)^2-4\big)+4=16-3(x-2)^2$.

The maximum occurs when $(x-2)^2=0$, so $M=(2,16)$.

Line $l_1$ passes through $O(0,0)$ and $M(2,16)$, so its gradient is $\dfrac{16}{2}=8$.

Line $l_2$ is parallel to $l_1$, so its gradient is also $8$. For $y=4+12x-3x^2$, $\dfrac{dy}{dx}=12-6x$.

Set $12-6x=8$, giving $x=\dfrac23$. Then $y=4+12\left(\dfrac23\right)-3\left(\dfrac49\right)=\dfrac{32}{3}$.

Using gradient $8$ through $\left(\dfrac23,\dfrac{32}{3}\right)$, $y-\dfrac{32}{3}=8\left(x-\dfrac23\right)$. Hence $y=8x+\dfrac{16}{3}$.