Perpendicular gradients and line equations

(a) $-\frac{1}{2}$

1. Perpendicular gradients multiply to $-1$: $2 \times m = -1$, so $m = -\frac{1}{2}$.

• B1 The correct answer, $-\frac{1}{2}$.

(b) $y = -\frac{1}{2}x + 3$

1. Substitute $(4, 1)$ into $y = -\frac{1}{2}x + c$.
   $$1 = -2 + c \Rightarrow c = 3$$
2. $y = -\frac{1}{2}x + 3$.

• M1 Substituting $(4, 1)$ into $y = mx + c$ with their perpendicular gradient.
• A1 $y = -\frac{1}{2}x + 3$ (or $x + 2y = 6$).

(a) $y = -\frac{5}{2}x + 9$

1. $$y+1=(-5/2)(x-4)$$
2. Perpendicular gradients have product $-1$, so the required gradient is $-5$/2.
3. Substitute (4, $-1$) into y = $-$(5/2)x + c: $-1$ = $-10$ + c, so c = 9.
4. The equation is y = $-$(5/2)x + 9.

• P1 Establishing $y+1=(-5/2)(x-4)$ or an equivalent valid method.
• A1 Correct answer: $y = -\frac{5}{2}x + 9$

(a) $k = \frac{2}{5}$

1. Rearrange the second line.
   $$2y = -5x + 8 \Rightarrow y = -\frac{5}{2}x + 4$$
2. Perpendicular gradient.
   $$k = -1 \div \left(-\frac{5}{2}\right) = \frac{2}{5}$$

• P1 Finding the gradient $-\frac{5}{2}$.
• P1 Using the negative reciprocal.
• A1 $k = \frac{2}{5}$.

(a) $y=-\frac{4}{3}x+\frac{14}{3}$

1. Use the negative reciprocal gradient.
   $$(3/4)(-4/3)=-1$$
2. Substitute P to calculate the y-intercept.
   $$2-(-4/3)\times 2$$
3. Therefore $y=-\frac{4}{3}x+\frac{14}{3}$.

• M1 Use the negative reciprocal gradient.
• M1 Substitute P to calculate the y-intercept.
• A1 Correct answer: $y=-\frac{4}{3}x+\frac{14}{3}$

(b) $(0,14/3)$

1. Put x = 0 in the new line equation.
   $$y=14/3$$
2. Therefore $(0,14/3)$.

• M1 Put x = 0 in the new line equation.
• A1 Correct answer: $(0,14/3)$

(a) $y=-\frac{5}{4}x+\frac{41}{4}$

1. Find the midpoint of AB and the negative reciprocal of its gradient.
   $$(4/5)(-5/4)=-1$$
2. Use the midpoint to determine the intercept.
   $$4+\frac{5}{4}\times 5$$
3. Therefore $y=-\frac{5}{4}x+\frac{41}{4}$.

• M1 Find the midpoint of AB and the negative reciprocal of its gradient.
• M1 Use the midpoint to determine the intercept.
• A1 Correct answer: $y=-\frac{5}{4}x+\frac{41}{4}$

(b) $(41/5,0)$

1. Set y equal to zero and solve for x.
   $$x=(41/4)/(5/4)$$
2. Therefore $(41/5,0)$.

• M1 Set y equal to zero and solve for x.
• A1 Correct answer: $(41/5,0)$

(a) $-2$

1. Find the gradient of L first.
   $$\frac{5-3}{4-0}$$
2. Therefore $-2$.

• M1 Find the gradient of L first.
• A1 Correct answer: $-2$

(b) $y=-2x+13$

1. Substitute (4, 5) into y = −2x + c.
   $$5=-2\times 4+c$$
2. Therefore $y=-2x+13$.

• M1 Substitute (4, 5) into y = −2x + c.
• A1 Correct answer: $y=-2x+13$

(a) $\frac{4}{3}$

1. Therefore $\frac{4}{3}$.

• B1 Correct answer: $\frac{4}{3}$

(b) $y=-\frac{3}{4}x+\frac{25}{4}$

1. The tangent is perpendicular to the radius; substitute P into y = −3x/4 + c.
   $$4=-3/4\times 3+c$$
2. Therefore $y=-\frac{3}{4}x+\frac{25}{4}$.

• M1 The tangent is perpendicular to the radius; substitute P into y = −3x/4 + c.
• A1 Correct answer: $y=-\frac{3}{4}x+\frac{25}{4}$

(c) $(25/3,0)$

1. Substitute y = 0 into the tangent equation.
   $$3x=25$$
2. Therefore $(25/3,0)$.

• M1 Substitute y = 0 into the tangent equation.
• A1 Correct answer: $(25/3,0)$

(a) Gradients $\frac{2}{3}$ and $-\frac{3}{2}$ multiply to $-1$.

1. Gradients of $AB$ and $BC$.
   $$m_{AB} = \frac{3 - 1}{4 - 1} = \frac{2}{3}, \qquad m_{BC} = \frac{6 - 3}{2 - 4} = -\frac{3}{2}$$
2. Multiply.
   $$\frac{2}{3} \times \left(-\frac{3}{2}\right) = -1$$
3. So $AB$ is perpendicular to $BC$, and angle $ABC = 90^{\circ}$.

• M1 Finding one gradient correctly.
• M1 Finding both gradients.
• C1 Showing the product is $-1$ and concluding the lines are perpendicular.

(b) $6.5$ square units

1. $AB = \sqrt{3^2 + 2^2} = \sqrt{13}$ and $BC = \sqrt{2^2 + 3^2} = \sqrt{13}$.
2. The right angle is at $B$, so these are base and height.
   $$\frac{1}{2} \times \sqrt{13} \times \sqrt{13} = \frac{13}{2} = 6.5$$

• P1 Finding $AB$ and $BC$ (both $\sqrt{13}$).
• A1 The correct answer, $6.5$.