IB exam questions for topic 4

a)

$$BD=\sqrt{(6-(-2){)}^2+(3-7{)}^2}=\sqrt{64+16}=\sqrt{80}\approx \mathrm{8.94.}$$

Answer: $BD=\sqrt{80}\approx 8.94$

b)

1. The diagonals of a kite are perpendicular: $m_{AC}=\frac{9+11}{4+6}=2$ and $m_{BD}=\frac{3-7}{6+2}=-\frac{1}{2}$, with product $-1$.
2. As $B$ and $D$ also lie on opposite sides of the line $AC$, the area is half the product of the diagonals:

$$\text{Area}=\frac{AC\cdot BD}{2}=\frac{\sqrt{500}\cdot \sqrt{80}}{2}=\frac{\sqrt{40000}}{2}=\frac{200}{2}=100.$$

Answer: area of $ABCD=100$

Answer: a) $\sqrt{80}\approx 8.94$; b) $100$

a) With $A(0,4)$ and $B(6,0)$:

$$m_{AB}=\frac{0-4}{6-0}=-\frac{2}{3}\approx -\mathrm{0.667.}$$

Answer: $m_{AB}=-\frac{2}{3}$

b) The midpoint is found by averaging the coordinates of $A(0,4)$ and $C(8,3)$:

$$M=\left(\frac{0+8}{2},\frac{4+3}{2}\right)=(4;3.5).$$

Answer: $M(4;3.5)$

c) Length of $AC$:

$$AC=\sqrt{(8-0{)}^2+(3-4{)}^2}=\sqrt{64+1}=\sqrt{65}\approx \mathrm{8.06.}$$

Answer: $AC=\sqrt{65}\approx 8.06$

d) Line $BM$ with $B(6,0)$ and $M(4;3.5)$.

1. Gradient: $m_{BM}=\frac{3.5-0}{4-6}=\frac{3.5}{-2}=-\frac{7}{4}$.
2. Point-gradient form using $B$: $y-0=-\frac{7}{4}(x-6)$.
3. Multiplying by $4$: $4y=-7x+42$.
4. Moving all terms to the left-hand side: $7x+4y-42=0$.

Answer: $7x+4y-42=0$

e) Two lines are perpendicular if the product of their gradients is $-1$.

1. Gradient of $AB$: $-\frac{2}{3}$.
2. Gradient of $BC$, with $B(6,0)$ and $C(8,3)$: $\frac{3-0}{8-6}=\frac{3}{2}$.
3. Product: $-\frac{2}{3}\cdot \frac{3}{2}=-1$.

As the product is $-1$, the line $AB$ is perpendicular to the line $BC$.

Answer: a) $-\frac{2}{3}$; b) $M(4;3.5)$; c) $\sqrt{65}\approx 8.06$; d) $7x+4y-42=0$; e) they are perpendicular, because $m_{AB}\cdot m_{BC}=-1$

a) The coordinates of the midpoint are the average of those of the endpoints.

1. $x$-coordinate: $\frac{s+(-2)}{2}=2\Rightarrow s-2=4\Rightarrow s=6$.
2. $y$-coordinate: $\frac{8+t}{2}=3\Rightarrow 8+t=6\Rightarrow t=-2$.

Answer: $s=6$, $t=-2$

b) Now $A(6,8)$ and $B(-2,-2)$.

1. Gradient of $AB$: $m_{AB}=\frac{-2-8}{-2-6}=\frac{-10}{-8}=\frac{5}{4}$.
2. Gradient of the perpendicular: $-\frac{4}{5}$.
3. Passing through $M(2,3)$: $y-3=-\frac{4}{5}(x-2)$.
4. Rearranging: $y=-\frac{4}{5}x+\frac{8}{5}+3=-\frac{4}{5}x+\frac{23}{5}$.

Answer: a) $s=6$, $t=-2$; b) $y=-\frac{4}{5}x+\frac{23}{5}$

a) Plot the point $P$ two units to the left of the $y$-axis and six units up, and the point $Q$ six units to the right and two units up.

P(−2,6) plotted in the second quadrant with its label and point $Q(6,2)$ plotted in the first quadrant with its label."> b)

1. (i) $PQ=\sqrt{(6-(-2){)}^2+(2-6{)}^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}\approx 8.94$.
2. (ii) $m_{PQ}=\frac{2-6}{6-(-2)}=\frac{-4}{8}=-\frac{1}{2}$.
3. (iii) If $RPQ$ is a right angle, $PR\perp PQ$, so $m_{PR}\cdot m_{PQ}=-1$ and $m_{PR}=-\frac{1}{-1/2}=2$.
4. (iv) The gradient of $PR$ with $R(-8,a)$ is $\frac{a-6}{-8-(-2)}=\frac{a-6}{-6}$. Equating to 2: $a-6=-12$, so $a=-6$.

Check: with $R(-8,-6)$, $PR=\sqrt{(-6{)}^2+(-12{)}^2}=\sqrt{180}$, which agrees with the value given in part (c).

Answer: (i) $\sqrt{80}\approx 8.94$; (ii) $-\frac{1}{2}$; (iii) $2$; (iv) $a=-6$

c) As the angle at $P$ is a right angle, $PQ$ and $PR$ are the base and height of the triangle:

$$\text{Area}=\frac{1}{2}\cdot PQ\cdot PR=\frac{1}{2}\sqrt{80}\cdot \sqrt{180}=\frac{1}{2}\sqrt{14400}=\frac{1}{2}\cdot 120=60.$$

Answer: a) (graph); b) (i) $\sqrt{80}\approx 8.94$, (ii) $-\frac{1}{2}$, (iii) $2$, (iv) $a=-6$; c) $60$

a)

$$m=\frac{8-4}{5-(-1)}=\frac{4}{6}=\frac{2}{3}\approx \mathrm{0.667.}$$

Answer: $m=\frac{2}{3}\approx 0.667$

b)

1. Point-gradient form with $A(-1,4)$: $y-4=\frac{2}{3}(x+1)$.
2. We solve for $y$: $y=\frac{2}{3}x+\frac{2}{3}+4=\frac{2}{3}x+\frac{14}{3}$.

Answer: $y=\frac{2}{3}x+\frac{14}{3}$

c) We substitute $x=8$ into the equation of $L$:

$$y=\frac{2}{3}\cdot 8+\frac{14}{3}=\frac{16+14}{3}=10.$$

Answer: a) $\frac{2}{3}$; b) $y=\frac{2}{3}x+\frac{14}{3}$; c) $y=10$

a)

1. $P$ is on the $y$-axis, so $x=0$: $2y=12$, $y=6$. Therefore $P(0,6)$.
2. $Q$ is on the $x$-axis, so $y=0$: $x=12$. Therefore $Q(12,0)$.

Answer: $P(0,6)$, $Q(12,0)$

b) We solve the system (by hand or with the GDC): $$\{\begin{array}{l}x+2y=12\\ x-y=3\end{array}$$ Subtracting the second equation from the first: $3y=9$, so $y=3$. Then $x=3+y=6$.

Answer: $A(6,3)$

Answer: a) $P(0,6)$, $Q(12,0)$; b) $A(6,3)$

a) $A$ is on the $y$-axis, so $x=0$: $2y=18\Rightarrow y=9$.

b) $B$ is on the $x$-axis, so $y=0$: $3x=18\Rightarrow x=6$.

c) $M$ is the midpoint of $A(0,9)$ and $B(6,0)$:

$$M=\left(\frac{0+6}{2},\frac{9+0}{2}\right)=(3;4.5).$$

Answer: a) $A(0,9)$; b) $B(6,0)$; c) $M(3;4.5)$

a) Start from the point $(-2,0)$. A gradient of $-3$ means that, when we move $1$ to the right, the line drops by $3$. This gives the points $(-1,-3)$ and $(0,-6)$. Going to the left, when we move back $1$ the line rises by $3$: the point $(-3,3)$. Join them with a straight line.

b) With gradient $m=-3$ and point $(-2,0)$:

1. Point-gradient form: $y-0=-3(x-(-2))$.
2. $y=-3(x+2)=-3x-6$.

Check: the line cuts the $y$-axis at $-6$, as seen in the drawing.

Answer: a) line passing through $(-3,3)$, $(-2,0)$, $(-1,-3)$ and $(0,-6)$; b) $y=-3x-6$

a) From the figure we can read two points on $l$ with clear coordinates: $(0,0)$ and $(5,10)$ (it also passes through $(1,2)$, $(2,4)$, ...).

1. Gradient: $m=\frac{10-0}{5-0}=2$.
2. It passes through the origin, so the $y$-intercept is $0$: $y=2x$.

Answer: $l:y=2x$

b) Parallel lines have the same gradient, $2$. As $n$ passes through $(0,8)$, the $y$-intercept is $8$:

$$n:y=2x+8.$$

Answer: $n:y=2x+8$

c) $P$ is on the horizontal axis, so $y=0$:

$$0=2x+8\Rightarrow x=-4.$$

Answer: a) $y=2x$; b) $y=2x+8$; c) $P(-4,0)$

In $y=mx+c$, the constant $c$ is the $y$-intercept (the same in all four diagrams) and the gradient $m$ determines the slope.

1. $y=c$: gradient $0$, a horizontal line. It is diagram 2.
2. $y=-x+c$: gradient $-1$, a line falling from left to right. It is diagram 3.
3. $y=3x+c$: gradient $3$, positive and large: the steepest line rising to the right. It is diagram 4.
4. $y=\frac{1}{3}x+c$: gradient $\frac{1}{3}$, positive and small: it rises gently. It is diagram 1.

Answer: $y=c$: 2; $y=-x+c$: 3; $y=3x+c$: 4; $y=\frac{1}{3}x+c$: 1

a)

1. Let $OA=t$ with $t>0$. Then $A(t,0)$ and, as $OB=3t$, $B(0,3t)$ (according to the figure, $A$ is on the positive $x$-semi-axis and $B$ on the positive $y$-semi-axis).
2. The gradient is $m=\frac{3t-0}{0-t}=-3$.

Answer: $m=-3$

b)

1. With $m=-3$: $y=-3x+c$.
2. As $(2,6)$ lies on $L$: $6=-3\cdot 2+c$, from which $c=12$.

Answer: $y=-3x+12$

c) At $A$ we have $y=0$: $0=-3x+12$, so $x=4$.

Answer: a) $-3$; b) $y=-3x+12$; c) $x=4$

a) $M$ is the midpoint of $AB$, so its x-coordinate is the average of those of $A$ and $B$: $a=\frac{0+24}{2}=12$. (Check with the y-coordinate: $\frac{26+10}{2}=18$.)

Answer: $a=12$

b)

$$m_{AB}=\frac{10-26}{24-0}=\frac{-16}{24}=-\frac{2}{3}\approx -\mathrm{0.667.}$$

Answer: $m_{AB}=-\frac{2}{3}\approx -0.667$

c)

1. With $a=12$, $M(12,18)$ and $O(0,0)$: gradient of $OM=\frac{18}{12}=\frac{3}{2}$.
2. $AM$ lies on the line $AB$, whose gradient is $-\frac{2}{3}$.
3. The product of the gradients is $\frac{3}{2}\cdot \left(-\frac{2}{3}\right)=-1$, so $OM\perp AM$.
4. Therefore triangle $OAM$ has a right angle at $M$: it is right-angled.

Answer: yes, it is right-angled (right angle at $M$), because $\frac{3}{2}\cdot \left(-\frac{2}{3}\right)=-1$

Answer: a) $a=12$; b) $-\frac{2}{3}$; c) yes, it is right-angled, with the right angle at $M$

a) From the diagram, $L_1$ passes through $(0,2)$ and $(6,0)$.

1. (i) Gradient: $$m_1=\frac{0-2}{6-0}=-\frac{1}{3}\approx -\mathrm{0.333.}$$
2. (ii) The y-intercept is $2$, so $$y=-\frac{1}{3}x+2.$$

Answer: (i) $-\frac{1}{3}$; (ii) $y=-\frac{1}{3}x+2$

b)

1. The shaded triangle has vertices $O(0,0)$, $B(6,0)$ and the point $P$ where $L_1$ and $L_2$ meet.
2. To find $P$ we equate $y=x$ with $y=-\frac{1}{3}x+2$: $$x=-\frac{1}{3}x+2\Rightarrow \frac{4}{3}x=2\Rightarrow x=\frac{3}{2},y=\frac{3}{2}.$$
3. We take as base the side on the $x$-axis, of length $6$, and as height the y-coordinate of $P$, which is $\frac{3}{2}$: $$\text{Area}=\frac{1}{2}\cdot 6\cdot \frac{3}{2}=\frac{9}{2}=\mathrm{4.5.}$$

Answer: Area $=\frac{9}{2}=4.5$

Answer: a) (i) $-\frac{1}{3}$, (ii) $y=-\frac{1}{3}x+2$; b) $\frac{9}{2}=4.5$

In $y=mx+c$, $m$ is the gradient (positive if the line rises from left to right, negative if it falls) and $c$ is the $y$-intercept (positive if the line cuts the $y$-axis above the origin, negative if it cuts below the origin, and zero if it passes through the origin). We classify the six lines:

$L_2$ and $L_6$ do not satisfy any of the four conditions, because $c=0$ for both. We assign the remaining lines:

1. $m>0$ and $c>0$: the line rises and cuts the $y$-axis above the origin, so it is $L_5$.
2. $m<0$ and $c>0$: the line falls and cuts the $y$-axis above the origin, so it is $L_4$.
3. $m<0$ and $c<0$: the line falls and cuts the $y$-axis below the origin, so it is $L_1$.
4. $m>0$ and $c<0$: the line rises and cuts the $y$-axis below the origin, so it is $L_3$.

Answer: $m>0,c>0$: $L_5$; $m<0,c>0$: $L_4$; $m<0,c<0$: $L_1$; $m>0,c<0$: $L_3$

a) From the diagram, the ramp passes through $(0,80)$ and meets the horizontal axis at $(940,0)$:

$$m=\frac{0-80}{940-0}=-\frac{80}{940}=-\frac{4}{47}\approx -\mathrm{0.0851.}$$

Answer: $m\approx -0.0851$

b) The safe gradient is $-\frac{1}{12}\approx -0.0833$. Ramp A has a gradient of $-0.0851$, which is smaller (more negative) than $-0.0833$: it descends faster than is allowed, that is, it is too steep and therefore it is not safe.

Answer: It is not safe, because $-0.0851<-\frac{1}{12}\approx -0.0833$ (it is steeper).

c) (i) We solve for $y$ in $2x+24y-1920=0$:

$$24y=-2x+1920\Rightarrow y=-\frac{1}{12}x+80.$$

The gradient of ramp B is $-\frac{1}{12}$, exactly the safe gradient, so ramp B is safe (and it starts from the same height of 80 cm).

(ii) The ramp reaches the ground when $y=0$:

$$-\frac{1}{12}x+80=0\Rightarrow x=960.$$

The horizontal distance is $960$ cm.

Answer: (i) it is safe, since its gradient is $-\frac{1}{12}$; (ii) $960$ cm

Answer: a) $-0.0851$; b) not safe, because $-0.0851<-\frac{1}{12}$; c) (i) safe (gradient $-\frac{1}{12}$), (ii) $960$ cm

a) $A$ is on the $x$-axis ($y=0$): $3x=24\Rightarrow x=8$, so $A(8,0)$. $B$ is on the $y$-axis ($x=0$): $4y=24\Rightarrow y=6$, so $B(0,6)$.

Answer: $A(8,0)$, $B(0,6)$

b) Midpoint: $M=\left(\frac{8+0}{2},\frac{0+6}{2}\right)=(4,3)$.

Answer: $M(4,3)$

c)

1. Gradient of $L_2$ through $M(4,3)$ and $C(0,-2)$: $m=\frac{3-(-2)}{4-0}=\frac{5}{4}$.
2. It cuts the $y$-axis at $-2$, so $y=\frac{5}{4}x-2$.

Check with $M$: $\frac{5}{4}\cdot 4-2=3$, correct.

Answer: $y=\frac{5}{4}x-2$

d)

1. (i) $MC=\sqrt{(4-0{)}^2+(3-(-2){)}^2}=\sqrt{16+25}=\sqrt{41}\approx 6.40$.
2. (ii) $AC=\sqrt{(8-0{)}^2+(0-(-2){)}^2}=\sqrt{64+4}=\sqrt{68}\approx 8.25$.

Answer: (i) $\sqrt{41}\approx 6.40$; (ii) $\sqrt{68}\approx 8.25$

e) In triangle $CMA$ we know all three sides: $MC=\sqrt{41}$, $MA=5$ and $AC=\sqrt{68}$.

1. (i) Cosine rule for the angle at $M$ (opposite $AC$):

$$\cos C\hat{M}A=\frac{M{C}^2+M{A}^2-A{C}^2}{2\cdot MC\cdot MA}=\frac{41+25-68}{2\sqrt{41}\cdot 5}=\frac{-2}{10\sqrt{41}}\approx -\mathrm{0.0312.}$$

Therefore $C\hat{M}A=\arccos (-0.0312\dots )\approx 91.8^{\circ }$ (using the GDC in degrees).

2. (ii) Area $=\frac{1}{2}\cdot MC\cdot MA\cdot \sin C\hat{M}A=\frac{1}{2}\cdot \sqrt{41}\cdot 5\cdot \sin 91.79^{\circ }\approx 16.0$.

(This can be checked using coordinates: $\overrightarrow{CM}=(4,5)$, $\overrightarrow{CA}=(8,2)$ and the area is $\frac{1}{2}|4\cdot 2-5\cdot 8|=16$.)

Answer: (i) $C\hat{M}A\approx 91.8^{\circ }$; (ii) area $\approx 16.0$

Answer: a) $A(8,0)$, $B(0,6)$; b) $M(4,3)$; c) $y=\frac{5}{4}x-2$; d) (i) $\sqrt{41}\approx 6.40$, (ii) $\sqrt{68}\approx 8.25$; e) (i) $91.8^{\circ }$, (ii) $16.0$

a) The line cuts the $y$-axis at $P(0,2)$, so the $y$-intercept is $2$. The gradient is

$$m_{PQ}=\frac{0-2}{2-0}=-1.$$

Therefore $y=-x+2$.

Answer: $y=-x+2$

b) 1. Midpoint: $M=\left(\frac{0+2}{2},\frac{2+0}{2}\right)=(1,1)$.

2. The gradient of a perpendicular to $PQ$ is the negative reciprocal of $-1$, that is, $1$.

3. Point-gradient form: $y-1=1\cdot (x-1)$, that is, $y=x$.

Answer: a) $y=-x+2$; b) $y=x$

a) $m_{AB}=\frac{9-3}{4-2}=\frac{6}{2}=3$.

Answer: $m_{AB}=3$

b) Two perpendicular lines satisfy $m_1\cdot m_2=-1$, so the gradient of the perpendicular is $m=-\frac{1}{3}$.

Answer: $-\frac{1}{3}$

c) Rearranging $2x+by-12=0$ to make $y$ the subject: $y=-\frac{2}{b}x+\frac{12}{b}$, whose gradient is $-\frac{2}{b}$.

As it is perpendicular to $[AB]$, its gradient is $-\frac{1}{3}$:

$$-\frac{2}{b}=-\frac{1}{3}\Rightarrow b=6.$$

Answer: $b=6$

Answer: a) $3$; b) $-\frac{1}{3}$; c) $b=6$

a) The gradient is the change in $y$ divided by the change in $x$:

$$m_1=\frac{y_B-y_A}{x_B-x_A}=\frac{-3.5-0.5}{9-(-3)}=\frac{-4}{12}=-\frac{1}{3}.$$

Answer: $m_1=-\frac{1}{3}$

b)

1. Parallel lines have the same gradient, so $L_2$ has gradient $-\frac{1}{3}$.
2. We use the point-gradient form with $C(3,1)$: $y-1=-\frac{1}{3}(x-3)$.
3. We multiply by 3 to remove the denominator: $3y-3=-(x-3)=-x+3$.
4. We move everything to the left-hand side: $x+3y-6=0$.

Answer: $x+3y-6=0$

c) The x-intercept occurs when $y=0$:

$$x+3\cdot 0-6=0\Rightarrow x=6.$$

Answer: a) $-\frac{1}{3}$; b) $x+3y-6=0$; c) $(6,0)$

a) The midpoint is the average of the coordinates:

$$M=\left(\frac{6+0}{2},\frac{0+2}{2}\right)=(3,1).$$

Answer: $M(3,1)$

b)

$$m_1=\frac{2-0}{0-6}=-\frac{1}{3}.$$

Answer: $m_1=-\frac{1}{3}$

c)

1. Parallel to $L_1$: $m_2=-\frac{1}{3}$, so $y=-\frac{1}{3}x+c$.
2. It passes through $(3,2)$: $2=-\frac{1}{3}\cdot 3+c=-1+c$, so $c=3$.

Answer: $y=-\frac{1}{3}x+3$

Answer: a) $M(3,1)$; b) $-\frac{1}{3}$; c) $y=-\frac{1}{3}x+3$

a)

$$m_{AC}=\frac{3-9}{5-3}=\frac{-6}{2}=-3.$$

Answer: $m_{AC}=-3$

b)

1. The angle $ACB$ has its vertex at $C$ and its sides are $CA$ and $CB$, which lie on the lines $AC$ and $BC$.
2. $m_{BC}=\frac{3-6}{5-15}=\frac{-3}{-10}=\frac{3}{10}$.
3. The lines are perpendicular if the product of the gradients is $-1$: $m_{AC}\cdot m_{BC}=-3\cdot \frac{3}{10}=-\frac{9}{10}\ne -1$.
4. Therefore $AC$ and $BC$ are not perpendicular and the angle $ACB$ is not a right angle.

Answer: not a right angle, because $-3\cdot \frac{3}{10}=-\frac{9}{10}\ne -1$

c)

1. $L\parallel BC$, so its gradient is $\frac{3}{10}$.
2. It passes through $A(3,9)$: $y-9=\frac{3}{10}(x-3)$.
3. We multiply by 10: $10y-90=3x-9$.
4. We move everything to the left-hand side: $3x-10y+81=0$.

Answer: $3x-10y+81=0$

Answer: a) $-3$; b) not a right angle ($-\frac{9}{10}\ne -1$); c) $3x-10y+81=0$

a) We substitute $A(3,-2)$ into $y=2.5x+k$: $-2=2.5\cdot 3+k=7.5+k$, so $k=-9.5$.

Answer: $k=-9.5$

b) The gradient of $L_1$ is $2.5=\frac{5}{2}$; that of a perpendicular is its reciprocal with the sign changed:

$$m_2=-\frac{1}{2.5}=-0.4=-\frac{2}{5}.$$

Answer: $m_2=-0.4$

c)

1. $y=-0.4x+c$ and it passes through $A(3,-2)$: $-2=-0.4\cdot 3+c=-1.2+c$.
2. Therefore $c=-0.8$.

Answer: $y=-0.4x-0.8$

d)

1. We multiply by 5: $5y=-2x-4$.
2. We move everything to the left-hand side: $2x+5y+4=0$.

Answer: a) $k=-9.5$; b) $-0.4$; c) $y=-0.4x-0.8$; d) $2x+5y+4=0$

a) The y-intercept occurs at $x=0$: $y=-3$, that is, the point $(0,-3)$.

Answer: $(0,-3)$

b) In $y=2x-3$ the coefficient of $x$ is the gradient: $m_1=2$.

Answer: $m_1=2$

c) $L_2$ has gradient 2 and cuts the $y$-axis at 3 (it passes through $(0,3)$), so $y=2x+3$.

Answer: $y=2x+3$

d) Perpendicular to $L_1$: $m_3=-\frac{1}{2}$.

Answer: $m_3=-\frac{1}{2}$

e)

1. Point-gradient form with $(-2,6)$: $y-6=-\frac{1}{2}(x+2)$.
2. We multiply by 2: $2y-12=-x-2$.
3. We move everything to the left-hand side: $x+2y-10=0$.

Answer: $x+2y-10=0$

Answer: a) $(0,-3)$; b) $2$; c) $y=2x+3$; d) $-\frac{1}{2}$; e) $x+2y-10=0$

a) We solve for $y$: $5y=-2x-4$, that is, $y=-\frac{2}{5}x-\frac{4}{5}$. The gradient is $-\frac{2}{5}$.

Answer: $m_1=-\frac{2}{5}$

b) Perpendicular: $m_2=-\frac{1}{m_1}=\frac{5}{2}$.

Answer: $m_2=\frac{5}{2}$

c)

1. Point-gradient form with $(5,3)$: $y-3=\frac{5}{2}(x-5)$.
2. We multiply by 2: $2y-6=5x-25$.
3. We rearrange: $5x-2y=19$.

Answer: $5x-2y=19$

d)

1. We solve the system $2x+5y=-4$ and $5x-2y=19$ (with the GDC or by hand).
2. By hand: we multiply the first equation by 2 and the second by 5, and add: $4x+10y=-8$ and $25x-10y=95$, so $29x=87$ and $x=3$.
3. Then $y=\frac{-4-2\cdot 3}{5}=-2$.

Answer: $P(3,-2)$

Answer: a) $-\frac{2}{5}$; b) $\frac{5}{2}$; c) $5x-2y=19$; d) $P(3,-2)$