Diagram for an analytic geometry problem asking students to find coordinates, gradients, and angles for triangle AVC given vertex V and midpoint T.

Let’s solve each part step by step. We are given:

- V(-1, -1)
- T(3, 2) is the midpoint of CV → so we can find C using midpoint formula.
- Equation of AB: y + k + 3x = 0 → rewrite as y = -3x - k
- θ = angle of inclination of AB → slope of AB is tanθ = -3 (from equation)
- β = angle of inclination of CV
- ∠AVC = α
- AV produced cuts y-axis at B → so line AV extended meets y-axis at point B.

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(a) Determine the value of k.

We know that point A lies on line AB: y + k + 3x = 0 → y = -3x - k

Also, point A lies on line AV — which goes from A to V(-1,-1). But we don’t yet have coordinates of A.

Wait — perhaps we can use the fact that B is where AV meets the y-axis, and also lies on AB? Let’s think.

Actually, let’s look again: “AV produced cuts the y-axis at B” → meaning if you extend line AV beyond V, it hits the y-axis at B.

But also, AB is a line with equation y + k + 3x = 0 → so points A and B both lie on this line.

So: Points A, B, and V are colinear? Not necessarily — wait: AV is a segment from A to V; extending it past V hits y-axis at B. So A-V-B are colinear, in that order? Or A-B-V? The phrase “AV produced” usually means extend beyond V, so order is A → V → B.

But then B is on the y-axis AND on line AB? That would mean line AB contains points A and B, and since A-V-B are colinear, then V must also be on line AB!

Is that possible?

Check: If V(-1,-1) lies on line AB: y + k + 3x = 0

Plug in x=-1, y=-1:

→ (-1) + k + 3*(-1) = 0
→ -1 + k - 3 = 0
→ k - 4 = 0
→ k = 4

That works! And makes sense because if AV extended meets y-axis at B, and AB is the same line, then V must lie on AB.

✔ So answer for (a): k = 4

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(b) Determine the coordinates of C.

T(3,2) is the midpoint of CV.

V is (-1, -1)

Let C = (x, y)

Midpoint formula:

T_x = (x_C + x_V)/2 → 3 = (x + (-1))/2 → 6 = x - 1 → x = 7

T_y = (y_C + y_V)/2 → 2 = (y + (-1))/2 → 4 = y - 1 → y = 5

✔ So C = (7, 5)

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(c) Determine the gradient of CV.

C(7,5), V(-1,-1)

Gradient m = (y2 - y1)/(x2 - x1) = (5 - (-1)) / (7 - (-1)) = 6 / 8 = 3/4

✔ Gradient of CV = 3/4

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(d) Calculate the size of α.

α = ∠AVC → angle at V between points A, V, C.

So vectors VA and VC.

First, we need coordinates of A.

We know:

- Line AB: y + 4 + 3x = 0 → y = -3x - 4 (since k=4)
- Point A is on this line.
- Also, A, V, B are colinear, and B is where AV meets y-axis.

Since V(-1,-1) is on line AB, and line AB has slope -3 (from y = -3x -4), then line AV is the same as line AB.

So parametric or vector approach:

From V(-1,-1), moving along direction of line AB (slope -3) to reach A and B.

But we need another condition to find A.

Wait — we’re told later in (f): AV = 2 BV → but that’s for part (f). Maybe we can use geometry now.

Alternatively, perhaps we can find point B first.

B is intersection of line AV (which is line AB) with y-axis.

Line AB: y = -3x - 4

Set x=0 → y = -4 → so B = (0, -4)

Now, we have points:

- B(0, -4)
- V(-1, -1)
- A is somewhere on the line beyond V from B? Wait: “AV produced cuts y-axis at B” → so starting at A, go through V, then hit B on y-axis.

So order is A → V → B.

So vector from V to B: B - V = (0 - (-1), -4 - (-1)) = (1, -3)

Then vector from V to A should be opposite direction? Since A is before V.

If AV = some length, and VB is from V to B.

Actually, since A-V-B are colinear, and we know V and B, we can express A as lying on the line beyond V away from B.

Parametrize line: from B to V to A.

Direction vector from B to V: V - B = (-1 - 0, -1 - (-4)) = (-1, 3)

So from V, going further in same direction to get to A: A = V + t*(-1, 3) for some t > 0? Wait no.

If order is A → V → B, then from A to V to B, so vector AV and VB are in same direction.

Vector VB = B - V = (1, -3) as above.

Then vector AV should be in same direction? No: from A to V is toward V, then V to B is same direction.

So let’s say position vector:

Let’s define parameter s such that any point on line is:

P(s) = V + s * direction_vector

Direction vector of line: since slope is -3, direction vector can be (1, -3) [because dy/dx = -3]

So line: P(s) = (-1, -1) + s*(1, -3)

When s=0 → V

When does it hit y-axis? When x=0.

x = -1 + s*1 = 0 → s=1 → then y = -1 + 1*(-3) = -4 → that’s point B → so B corresponds to s=1

Then A must correspond to s < 0, since A is before V.

But we don’t have more info yet... unless we use part (f) early? But part (f) says "if AV=2BV", which is conditional.

Wait — actually, looking back, parts (a)-(e) don't depend on (f). So maybe we can find A without (f)?

Another idea: perhaps triangle AVC is drawn, and we can compute angles using gradients.

We have:

- Line AV: same as AB, slope = -3 → angle of inclination θ = arctan(-3). But since it's going down to right, θ is obtuse? Actually, angle of inclination is measured from positive x-axis counterclockwise, so for negative slope, it's between 90° and 180°.

tanθ = -3 → θ = 180° + arctan(-3) = 180° - arctan(3) ≈ 180° - 71.565° = 108.435°

But maybe better to keep exact.

Similarly, line CV: from C(7,5) to V(-1,-1), gradient = 3/4 → angle β = arctan(3/4)

Now, angle α = ∠AVC → angle between lines VA and VC at point V.

Vectors:

From V to A: let’s denote vector u = A - V

From V to C: vector v = C - V = (7 - (-1), 5 - (-1)) = (8, 6)

We need vector u. But we don’t have A yet.

However, since A lies on line with direction vector (1, -3) from V (as established), and since the line goes through V with slope -3, then vector VA is parallel to (1, -3) or (-1, 3)?

From earlier: when we go from V to B, we add (1, -3) to get to B.

Since A is on the other side of V from B, then vector VA should be in direction opposite to VB.

VB = (1, -3), so VA = t*(-1, 3) for some t>0.

So let’s set A = V + t*(-1, 3) = (-1 -t, -1 + 3t)

Now, we can compute angle between vectors VA and VC.

Vector VA = A - V = (-t, 3t)

Vector VC = C - V = (8, 6)

Angle α between them:

cosα = (VA • VC) / (|VA| |VC|)

VA • VC = (-t)(8) + (3t)(6) = -8t + 18t = 10t

|VA| = sqrt((-t)^2 + (3t)^2) = sqrt(t² + 9t²) = sqrt(10t²) = t√10

|VC| = sqrt(8² + 6²) = sqrt(64+36) = sqrt(100) = 10

So cosα = (10t) / (t√10 * 10) = 10t / (10t √10) = 1/√10

Thus α = arccos(1/√10)

We can leave it like that, or compute degree measure.

arccos(1/√10) ≈ arccos(0.3162) ≈ 71.565°

But let’s see if we can find exact value or if problem expects decimal.

Note: tanβ = 3/4, so β = arctan(3/4) ≈ 36.87°

And θ = arctan(-3) but as inclination, it's 180° + arctan(-3) = 180° - arctan(3) ≈ 108.435°

At point V, the angle between VA and VC.

The direction of VA: since it's along the line with slope -3, but from V to A is towards decreasing x and increasing y? From our parametrization, A = (-1-t, -1+3t), so for t>0, x decreases, y increases → so direction is left and up, slope = rise/run = 3t / (-t) = -3, same line.

The angle that VA makes with positive x-axis: since it's going left and up, second quadrant.

The vector VA = (-t, 3t), so angle φ = arctan(3t / -t) = arctan(-3), but in second quadrant, so φ = 180° + arctan(-3) = 180° - arctan(3) ≈ 108.435° — same as θ.

Vector VC = (8,6), first quadrant, angle β = arctan(6/8) = arctan(3/4) ≈ 36.87°

So angle between them is difference: 108.435° - 36.87° = 71.565°, which matches arccos(1/√10)

So α = arccos(1/√10) or approximately 71.57°

But perhaps they want exact expression or degrees.

Looking at context, probably acceptable to give as arccos(1/√10) or compute.

But let's see part (g) asks for area, so maybe we need numerical.

For now, we'll box the exact value.

Actually, note that cosα = 1/√10, so α = \cos^{-1}(1/\sqrt{10})

But perhaps simplify.

We can write α = \tan^{-1}(3) ? Let's check.

If cosα = 1/√10, then sinα = sqrt(1 - 1/10) = sqrt(9/10) = 3/√10, so tanα = sin/cos = 3, so α = arctan(3)

Oh! Nice!

Because tanα = 3, so α = arctan(3)

Is that correct?

From above: sinα = 3/√10, cosα = 1/√10, so yes, tanα = 3.

And since both components positive in the dot product sense? Vector VA is (-t,3t), VC is (8,6), dot product positive, so acute angle, and arctan(3) is about 71.57°, which is acute, good.

So α = \tan^{-1}(3)

But typically we say the angle whose tangent is 3.

In many contexts, they might expect the degree measure, but since not specified, we can leave as \tan^{-1}(3) or compute.

Looking at the diagram, it shows α at V, and it looks acute, so arctan(3) is fine.

But let's confirm with another method.

The angle between two lines with slopes m1 and m2 is given by tanφ = |(m2 - m1)/(1 + m1 m2)|

Here, line VA has slope m1 = -3 (same as AB)

Line VC has slope m2 = 3/4

So tanα = |(m2 - m1)/(1 + m1 m2)| = |(3/4 - (-3)) / (1 + (-3)(3/4))| = |(3/4 + 12/4) / (1 - 9/4)| = |(15/4) / (-5/4)| = |15/4 * 4/-5| = | -3 | = 3

So tanα = 3 → α = arctan(3)

Perfect.

So size of α is \tan^{-1}(3) radians or degrees? Usually in such problems, if not specified, we can leave as inverse tan, or compute approximate.

But since part (d) says "calculate the size", probably expects numerical value.

arctan(3) ≈ 71.565° , so perhaps 71.6° or exact.

But let's see other parts.

I think for accuracy, we'll state α = \tan^{-1}(3) , but since it's "calculate", maybe compute.

Note that in part (g) we need area, so perhaps keep symbolic.

But for now, we'll put α = \tan^{-1}(3)

To be precise, let's calculate it.

Using calculator, arctan(3) = approximately 71.565051177 degrees.

Often rounded to one decimal: 71.6°

But let's see if exact fraction.

No, so perhaps leave as is, but I think for school level, they might expect the expression or the value.

Another way: since tanα=3, and it's an angle in triangle, we can use it.

But for answer, I'll box the exact form.

Wait, the problem says "calculate the size", so likely numerical.

Perhaps in degrees.

Let me assume we need to compute.

So α ≈ 71.6°

But let's hold off and do other parts.

Actually, for consistency, since other answers are exact, perhaps keep α = \tan^{-1}(3)

But let's move on and come back.

(e) Calculate the length of BV.

B is (0, -4), V is (-1, -1)

Distance BV = sqrt[ (0 - (-1))^2 + (-4 - (-1))^2 ] = sqrt[ (1)^2 + (-3)^2 ] = sqrt[1 + 9] = sqrt(10)

✔ So BV = √10

(f) If AV = 2 BV and CV = √111, calculate the length of AC.

Given: AV = 2 * BV

From (e), BV = √10, so AV = 2√10

CV = √111 (given)

We need AC.

Points A, V, C.

We have distances AV and CV, and angle at V is α, with cosα = 1/√10 as before.

In triangle AVC, we have sides AV, CV, and included angle α, so we can find AC by law of cosines.

Law of cosines: AC² = AV² + CV² - 2 * AV * CV * cos(angle at V)

Angle at V is α, and cosα = 1/√10

AV = 2√10, CV = √111

So AC² = (2√10)^2 + (√111)^2 - 2 * (2√10) * (√111) * (1/√10)

Simplify:

= 4*10 + 111 - 2 * 2√10 * √111 / √10

= 40 + 111 - 4 * √111 [since √10 / √10 = 1]

= 151 - 4√111

Is that simplifiable? Probably not.

But let's check calculation.

2 * AV * CV * cosα = 2 * 2√10 * √111 * (1/√10) = 4 * √111 * (√10 / √10) = 4√111

Yes.

So AC² = 40 + 111 - 4√111 = 151 - 4√111

Then AC = sqrt(151 - 4√111)

But is this simplified? Perhaps not, but maybe we can leave it.

Note: earlier we had coordinates, but now with AV=2BV, we can find coordinates of A.

From earlier, we had A = V + t*(-1, 3) = (-1 -t, -1 + 3t)

Distance AV = distance from A to V = sqrt[ (-t)^2 + (3t)^2 ] = sqrt(t² + 9t²) = sqrt(10t²) = t√10

Set equal to 2√10 → t√10 = 2√10 → t=2

So A = (-1 -2, -1 + 3*2) = (-3, 5)

Now, C is (7,5) from (b)

So AC = distance between A(-3,5) and C(7,5) = |7 - (-3)| = 10, since same y-coordinate.

Oh! Much simpler!

Why didn't I think of that earlier.

With t=2, A=(-3,5), C=(7,5), so AC = 10 units.

And CV = distance from C(7,5) to V(-1,-1) = sqrt((8)^2 + (6)^2)=sqrt(64+36)=sqrt(100)=10, but the problem says CV=√111? Contradiction?

Problem says: "if AV=2BV and CV=√111"

But from earlier, with C(7,5), V(-1,-1), CV=sqrt(8^2+6^2)=sqrt(64+36)=sqrt(100)=10, not √111.

Inconsistency.

What's wrong?

Ah, I see. In part (b), we found C using T as midpoint of CV, with V(-1,-1), T(3,2), so C(7,5), and CV=10.

But in part (f), it says "if CV=√111", which contradicts.

Unless... perhaps the "if" means we ignore previous and use new values? But that doesn't make sense because C is defined by T being midpoint.

Perhaps I misinterpreted.

Let me read carefully.

Part (f): "If AV=2BV and CV=√111, calculate the length of AC."

But from earlier, CV is fixed as 10, since C and V are fixed by T being midpoint.

Unless the "if" is hypothetical, but that would change everything.

Perhaps there's a mistake in my assumption.

Another possibility: in part (b), we assumed T is midpoint of CV, which is given, so C is fixed.

But then CV is fixed at 10, but part (f) says "if CV=√111", which is different, so perhaps it's a separate scenario, but that doesn't make sense for a single diagram.

Perhaps "CV=√111" is a typo, or perhaps I miscalculated CV.

C(7,5), V(-1,-1): delta x=8, delta y=6, distance sqrt(64+36)=sqrt(100)=10, yes.

√111 is approximately 10.535, close but not same.

Perhaps the "if" is for this part only, and we should use the given CV=√111, but then C is not (7,5)? But that contradicts part (b).

This is confusing.

Let me read the problem again.

"T (3,2) is the midpoint of CV" — this is given at the beginning, so for all parts, C is determined as (7,5), so CV=10.

But part (f) says "if CV=√111", which is inconsistent.

Unless "CV=√111" is a mistake, and it's supposed to be something else.

Perhaps "CV" in part (f) refers to something else, but unlikely.

Another idea: perhaps "CV" in part (f) is a typo, and it's "AC" or something, but it says CV.

Or perhaps it's "the length of CV is given as √111 for this part", but that would mean we ignore the midpoint, which is not reasonable.

Perhaps in part (f), the "if" includes changing C, but that seems odd.

Let's look at the diagram description. It says "in the diagram below", and gives V, T, etc., so probably the diagram is fixed, so C is fixed.

But then why say "if CV=√111"?

Perhaps it's "and CV = sqrt(111)" but that can't be.

Another thought: perhaps "CV" in part (f) is not the same as in the diagram, but that doesn't make sense.

Maybe I miscalculated C.

T(3,2) midpoint of C and V(-1,-1)

So (x_c + (-1))/2 = 3 => x_c -1 = 6 => x_c=7

(y_c + (-1))/2 = 2 => y_c -1 = 4 => y_c=5, yes.

Distance CV = sqrt((7+1)^2 + (5+1)^2) = sqrt(64+36)=10, yes.

Perhaps "CV=√111" is for a different purpose, or perhaps it's a red herring, but unlikely.

Another idea: perhaps "CV" in part (f) means the length from C to V, but in the context, with AV=2BV, and we have A from that, but C is still (7,5), so CV is 10, not 111.

Unless the "if" is additional condition, but then it's over-constrained.

Perhaps for part (f), we are to assume that CV=√111, and ignore the midpoint, but that would contradict part (b).

This is problematic.

Let's calculate what CV would be if we use the A from AV=2BV.

From earlier, with t=2, A= (-3,5)

V= (-1,-1)

C is not known yet, but T is midpoint of C and V, so if V is fixed, T is fixed, C is fixed, so CV is fixed.

Perhaps the "if" is only for AV=2BV, and CV=√111 is given to find something else, but it's stated as "and CV=√111".

Perhaps it's "and the length of CV is 111", but that can't be with the diagram.

Another possibility: perhaps "CV" in part (f) is a typo, and it's "AC" or "BC", but let's see.

Perhaps in the diagram, CV is not from C to V, but that doesn't make sense.

Let's calculate AC with A(-3,5), C(7,5) = 10, as above.

And if CV=10, then in triangle AVC, AV=2√10, CV=10, angle at V with cosα=1/√10, then AC^2 = (2√10)^2 + 10^2 - 2*2√10*10*(1/√10) = 40 + 100 - 40 = 100, so AC=10, which matches.

But the problem says "if CV=√111", which is not 10.

Perhaps it's "if AV=2BV and the length of AC is to be found, and CV is given as 111 for verification", but that doesn't help.

Maybe "CV=√111" is a mistake, and it's supposed to be "AC= something", but let's assume that for part (f), we use the coordinates we have.

With A(-3,5), C(7,5), AC=10.

And CV=10, not √111, so perhaps the "CV=√111" is irrelevant or typo.

Perhaps in some versions, T is different, but here it's given.

Another idea: perhaps "T(3,2) is the midpoint of CV" is for the diagram, but in part (f), it's a different configuration, but that seems unlikely.

Perhaps for part (f), we are to use the given CV=√111, and find C accordingly, but then T is not midpoint, which contradicts.

I think there might be a typo in the problem, and "CV=√111" should be "AC= something" or omitted.

Perhaps "CV" is " the length from C to V is √111", but then we can find C from V and T, but T is midpoint, so if CV=√111, and T is midpoint, then VT = CV/2 = √111 /2, but T is at (3,2), V at (-1,-1), distance VT = sqrt((3+1)^2 + (2+1)^2) = sqrt(16+9)=5, so CV=2*VT=10, as before.

So VT=5, CV=10.

So "CV=√111" must be a mistake.

Perhaps it's " the length of AC is 111", but the problem says "CV=√111".

Let's look at the text: "if AV=2BV and CV=√111, calculate the length of AC."

Perhaps in the diagram, CV is not 10, but let's calculate what it should be.

Maybe I have a calculation error in C.

V(-1,-1), T(3,2), midpoint, so C = 2*T - V = 2*(3,2) - (-1,-1) = (6,4) + (1,1) = (7,5), yes.

Distance from C to V: from (7,5) to (-1,-1): dx=8, dy=6, dist=10.

√111 = sqrt(111) ≈ 10.535, not 10.

Perhaps "CV" means something else, like from C to another point, but unlikely.

Another possibility: "CV" in part (f) is the length, but in the context, perhaps it's a different C, but that doesn't make sense.

Perhaps for part (f), the "if" includes that T is not midpoint, but the problem states it is.

I think the only logical thing is to assume that "CV=√111" is a typo, and it's supposed to be "AC= something" or omitted, or perhaps " the length of BV is √10" which we have, but it's given.

Perhaps "CV=√111" is for the length in general, but in this case it's 10, so for part (f), we proceed with the coordinates.

With A(-3,5), C(7,5), AC=10.

And since the problem asks to calculate, and with the given, it's 10.

Moreover, in the law of cosines, if we use CV=10, we get AC=10, as above.

So probably "CV=√111" is a mistake, and it's 10, or perhaps it's " the square of CV is 111", but 10^2=100, not 111.

111 is 3*37, not nice.

Perhaps it's " the length of AC is to be found, and CV is given as √111 for this part", but then we need to find C such that T is midpoint and CV=√111, but T is fixed, V is fixed, so C is fixed, so CV is fixed.

Unless V is not fixed, but it is given as (-1,-1).

I think we have to proceed with the calculation as per coordinates.

So for part (f), with AV=2BV, we have A(-3,5), C(7,5), so AC= |7- (-3)| = 10, since same y.

So AC = 10.

And ignore the "CV=√111" as likely a typo, or perhaps it's " and the length of CV is 10", but written as √111 by mistake.

Perhaps in some print, 100 is written as 111, but unlikely.

Another idea: perhaps "CV" means the vector or something, but no.

Or perhaps " the length from C to V is √111", but then we can solve for C, but T is midpoint, so if V(-1,-1), T(3,2), then C must be (7,5), so CV=10, so contradiction.

Unless T is not the midpoint for this part, but the problem says it is for the diagram.

I think for the sake of progress, we'll take AC=10 for part (f).

So ✔ AC = 10

(g) Calculate the area of ΔAVC.

Points A, V, C.

From above, with A(-3,5), V(-1,-1), C(7,5)

We can use shoelace formula.

Shoelace:

List the points: A(-3,5), V(-1,-1), C(7,5), back to A(-3,5)

Area = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) |

Or better:

Sum x_i y_{i+1} - x_{i+1} y_i

So:

A to V: x1= -3, y1=5; x2= -1, y2= -1

V to C: x2= -1, y2= -1; x3=7, y3=5

C to A: x3=7, y3=5; x1= -3, y1=5

So sum = [ (-3)*(-1) + (-1)*5 + 7*5 ] - [ 5*(-1) + (-1)*7 + 5*(-3) ]

Standard shoelace:

Area = 1/2 | sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i) | with (x_{n+1},y_{n+1}) = (x1,y1)

So:

i=1: x1= -3, y1=5; x2= -1, y2= -1 → (-3)*(-1) - (-1)*5 = 3 - (-5) = 3+5=8

i=2: x2= -1, y2= -1; x3=7, y3=5 → (-1)*5 - 7*(-1) = -5 - (-7) = -5+7=2

i=3: x3=7, y3=5; x1= -3, y1=5 → 7*5 - (-3)*5 = 35 - (-15) = 35+15=50

Sum = 8 + 2 + 50 = 60

Area = 1/2 * |60| = 30

Since all points, and we can see that A and C have same y=5, V at y=-1, so base AC=10, height is vertical distance from V to line AC.

Line AC is y=5, V is at y=-1, so height = |5 - (-1)| = 6

Area = 1/2 * base * height = 1/2 * 10 * 6 = 30

Yes.

So area = 30

(h) Determine the equation of the perpendicular bisector of line VC.

V(-1,-1), C(7,5)

First, midpoint of VC: which is T(3,2), given.

Gradient of VC: from V to C, (5 - (-1))/(7 - (-1)) = 6/8 = 3/4

So gradient of perpendicular is negative reciprocal: -4/3

So perpendicular bisector passes through T(3,2) with slope -4/3

Equation: y - 2 = (-4/3)(x - 3)

Multiply both sides by 3: 3(y - 2) = -4(x - 3)

3y - 6 = -4x + 12

4x + 3y - 6 - 12 = 0

4x + 3y - 18 = 0

Or y = (-4/3)x + 6

But usually standard form.

So equation: 4x + 3y = 18

Or 4x + 3y - 18 = 0

Now, back to part (d): size of α.

We have α = arctan(3)

Numerically, arctan(3) = approximately 71.565 degrees.

In many textbooks, they might expect the exact value or the degree measure.

Since the problem says "calculate", and other answers are numbers, perhaps compute.

Note that in part (g), area is 30, nice number, so perhaps α is to be left as is.

But let's see the answer format.

Perhaps they want the measure in degrees.

I recall that tanα=3, so α = \tan^{-1}(3)

But for boxing, perhaps write the value.

Another way: since in the triangle, with sides, but I think it's fine.

For part (d), we'll put α = \tan^{-1}(3) or 71.6^\circ

But to be precise, let's calculate.

Using calculator, arctan(3) = 71.56505117707799 degrees, so approximately 71.6 degrees.

Perhaps 72 degrees, but not accurate.

Since cosα = 1/√10, and 1/√10 = √10/10, etc.

I think for the answer, we can put the exact expression.

But let's see the options.

Perhaps in the context, since it's an angle in a triangle, and we have coordinates, but I think \tan^{-1}(3) is acceptable.

For the final answer, we'll list all.

Now, summarize:

(a) k = 4

(b) C = (7,5)

(c) gradient of CV = 3/4

(d) α = \tan^{-1}(3) or approximately 71.6^\circ — but let's decide.

Upon second thought, in many similar problems, they expect the measure, so perhaps 71.6 degrees.

But to be safe, since the problem might expect exact, and tanα=3, we can leave it.

Note that in part (g), area is 30, which is integer, so perhaps α is to be calculated as number.

Another way: the angle between the lines.

Line AV: slope -3, line CV: slope 3/4

The angle θ between them is given by tanθ = |(m2-m1)/(1+m1m2)| = |(3/4 - (-3))/(1 + (-3)(3/4))| = |(15/4)/(1 - 9/4)| = |(15/4)/(-5/4)| = | -3 | = 3, so θ = arctan(3)

So I think it's fine.

For the answer, we'll put α = \tan^{-1}(3)

But for boxing, perhaps write the value.

Let's look at part (e): BV = 10, which is exact.

So for (d), perhaps leave as \tan^{-1}(3)

But the problem says "calculate the size", so likely numerical.

Perhaps in degrees, and round to nearest tenth.

So 71.6 degrees.

I think that's reasonable.

So α ≈ 71.6°

Now for (f), we have AC=10, despite the CV=√111 issue.

Perhaps "CV=√111" is a distractor or typo, and we proceed.

So AC=10

(g) area=30

(h) equation: 4x + 3y = 18 or 4x + 3y - 18 = 0

Now, let's write all answers.

Final Answer:

(a) k = 4

(b) C = (7, 5)

(c) gradient = \frac{3}{4}

(d) α = \tan^{-1}(3) \approx 71.6^\circ (but perhaps specify)

To be consistent, since other answers are exact, for (d) we can put the exact value.

But "calculate" suggests numerical.

Perhaps they want the measure in degrees.

I recall that in some systems, they accept arctan, but for this, let's put 71.6 degrees.

Note that tan^{-1}(3) is exact, so perhaps box that.

But for the sake of the problem, I'll put the numerical value.

Another idea: perhaps α is the angle at V, and in the diagram, it's shown, and we can use the vectors.

But I think it's fine.

Let's finalize.

For (d): since tanα = 3, and it's acute, α = \arctan(3) radians, but usually degrees in such contexts.

So I'll put 71.6 degrees.

But