Geometry problem involving triangle ACV with given coordinates and equations, focusing on finding values related to angles, midpoints, gradients, lengths, and equations.

Let’s solve this step by step. We’re given:

- Point V = (-1, -1)
- T(3, 2) is the midpoint of CV → so we can find C using midpoint formula.
- Equation of line AB: y + k + 5x = 0 → we’ll use point A or B to find k.
- AV produced cuts y-axis at B → so B is on y-axis (x=0), and lies on line AV extended.
- Angles θ, β, α are defined but not needed for all parts — we’ll focus on what’s asked.

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(a) Determine the value of k.

We know that point A lies on line AB: y + k + 5x = 0 → rearrange:
→ y = -5x - k

Also, since AV is a line from A to V(-1,-1), and when extended it hits y-axis at B, then points A, V, B are colinear.

But we don’t yet have coordinates of A or B. However, note that point T(3,2) is midpoint of CV, and V is known → let’s first find C in part (b), then come back? Wait — maybe we can find B first?

Actually, let’s think differently.

Since AV produced meets y-axis at B, and V is (-1,-1), then line AV passes through V and goes to B on y-axis (so B has x=0).

So let’s assume line AV has some slope m, and passes through V(-1,-1). Then its equation is:

y - (-1) = m(x - (-1)) → y + 1 = m(x + 1)

When x = 0 (at point B), y = m(0 + 1) - 1 = m - 1 → so B = (0, m - 1)

Now, point A is also on this line, and also on line AB: y = -5x - k

Wait — actually, line AB *is* the same as line AVB! Because A, V, B are colinear, and AB is the line containing them.

So line AB is the same as line AVB → which has equation y = -5x - k (from given: y + k + 5x = 0 → y = -5x - k)

And this line passes through V(-1, -1). So plug in x=-1, y=-1 into the equation:

-1 = -5*(-1) - k
→ -1 = 5 - k
→ k = 5 + 1 = 6

✔ So k = 6

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(b) Determine the coordinates of C.

T(3,2) is midpoint of CV, and V = (-1, -1)

Midpoint formula: if M is midpoint of P(x₁,y₁) and Q(x₂,y₂), then
M = ((x₁+x₂)/2 , (y₁+y₂)/2)

Here, T(3,2) is midpoint of C(x,y) and V(-1,-1)

So:

( x + (-1) ) / 2 = 3 → x - 1 = 6 → x = 7
( y + (-1) ) / 2 = 2 → y - 1 = 4 → y = 5

✔ So C = (7, 5)

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(c) Determine the gradient of CV.

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

Gradient = (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 is angle AVC — that is, angle at V between points A, V, C.

So vectors VA and VC form angle α at V.

First, find coordinates of A.

We know line AB: y = -5x - k, and k=6 → y = -5x -6

Point A is intersection of... wait, do we have another condition? Actually, we don’t need A explicitly for angle α? Or do we?

Angle α is ∠AVC — so at vertex V, between points A, V, C.

So we need vectors from V to A and from V to C.

We already have vector VC: from V(-1,-1) to C(7,5) → (7 - (-1), 5 - (-1)) = (8,6)

Vector VA: from V(-1,-1) to A. But we don’t have A yet.

Wait — perhaps we can find A as the other end of the line from V along line AB.

Line AB: y = -5x -6, and passes through V(-1,-1) — which we already used.

But where is point A? The diagram shows triangle ACV, with A connected to C and V.

Actually, we might not need A for angle α if we consider triangle AVC and use gradients?

Alternative approach: angle between two lines meeting at V: line VA and line VC.

We know gradient of VC is 3/4 (from part c).

What is gradient of VA? Since VA is part of line AB, which has gradient -5 (because y = -5x -6 → slope = -5)

So at point V, we have two lines:

- Line VA: gradient m1 = -5
- Line VC: gradient m2 = 3/4

The angle α between them can be found by:

tan(α) = |(m2 - m1)/(1 + m1*m2)|

Plug in:

m1 = -5, m2 = 3/4

Numerator: (3/4 - (-5)) = 3/4 + 5 = 3/4 + 20/4 = 23/4

Denominator: 1 + (-5)(3/4) = 1 - 15/4 = (4/4 - 15/4) = -11/4

So tan(α) = | (23/4) / (-11/4) | = | -23/11 | = 23/11

Then α = arctan(23/11)

Calculate that:

23 ÷ 11 ≈ 2.0909

arctan(2.0909) ≈ ? Let's compute.

Using calculator: tan⁻¹(2.0909) ≈ 64.4 degrees? But let’s check.

Actually, better to leave exact or compute precisely.

But perhaps they want exact expression? Or decimal?

Wait — maybe I made a mistake. Is α the angle inside the triangle? And is it acute or obtuse?

Looking at diagram: point A is left-up, V is bottom-left, C is right-up. So angle at V between A and C should be acute? But our calculation gives tanα = 23/11 > 1, so α > 45°, which is fine.

But let me verify the formula.

Formula for angle φ between two lines with slopes m1 and m2:

tanφ = |(m2 - m1)/(1 + m1m2)|

Yes.

So tanα = 23/11 → α = tan⁻¹(23/11)

Compute numerically:

23/11 ≈ 2.090909

tan⁻¹(2.090909) ≈ 64.4 degrees? Let me calculate:

tan(60°) = √3 ≈ 1.732
tan(65°) ≈ 2.1445
tan(64°) ≈ 2.0503
tan(64.4°) ≈ ?

Interpolate:

At 64°: 2.0503
At 65°: 2.1445
Difference: 0.0942 per degree

We need 2.0909 - 2.0503 = 0.0406

So fraction: 0.0406 / 0.0942 ≈ 0.431

So α ≈ 64 + 0.431 ≈ 64.43°

But perhaps we should give exact value or round appropriately.

Wait — maybe there’s a better way? Or did I misidentify the lines?

Another thought: perhaps α is not the angle between VA and VC, but something else? The diagram says “∠AVC = α”, which is angle at V in triangle AVC, so yes, between vectors VA and VC.

But let me confirm with coordinates once we have A.

Let’s find point A.

We know line AB: y = -5x -6

And point A is on this line, and also, since T is midpoint of CV, and we have C and V, but how does A relate?

Actually, we don’t have direct info about A except that it’s on line AB and forms triangle with C and V.

But for angle at V, we only need directions from V to A and V to C.

We have direction to C: (8,6) as before.

Direction to A: since A is on line y = -5x -6, and V is (-1,-1), then any point on the line can be written as parameterized.

Let’s take a general point on line AB: let x = t, then y = -5t -6

Vector from V(-1,-1) to A(t, -5t-6) is (t - (-1), (-5t-6) - (-1)) = (t+1, -5t -5)

This vector is proportional to (1, -5) because:

(t+1, -5(t+1)) = (t+1)(1, -5)

So direction vector of VA is (1, -5) — which matches gradient -5.

Similarly, direction vector of VC is (8,6) or simplify to (4,3) — gradient 3/4.

So angle between vectors u = (1, -5) and v = (4, 3)

Use dot product formula:

cosα = (u • v) / (|u| |v|)

u • v = (1)(4) + (-5)(3) = 4 - 15 = -11

|u| = sqrt(1² + (-5)²) = sqrt(1 + 25) = sqrt(26)

|v| = sqrt(4² + 3²) = sqrt(16 + 9) = sqrt(25) = 5

So cosα = (-11) / (sqrt(26) * 5) = -11/(5√26)

But angle in triangle should be between 0 and 180, and cosine negative means obtuse angle? But in diagram, it looks acute.

Problem: the angle between the vectors as drawn might be the reflex or the smaller one.

In the context of triangle AVC, angle at V is the smaller angle between the two sides, so we take absolute value or consider the acute angle.

Actually, the formula for angle between two lines gives the acute angle, but here since we're dealing with vectors from V, and depending on direction, it could be obtuse.

But in the diagram, point A is above and left, C is above and right, V is below, so angle at V should be less than 180, and likely acute.

But our dot product gave negative, meaning the angle between the vectors as directed is obtuse, but in the triangle, it's the internal angle, which might be the supplement.

Let me sketch mentally:

V at (-1,-1)

A is on line y=-5x-6. For example, if x=0, y=-6 → B(0,-6)? Earlier I thought B is on y-axis, and from earlier, when x=0, y= -5*0 -6 = -6, so B(0,-6)

Then A is another point on the line. Since V is (-1,-1), and line goes through V and B(0,-6), then extending beyond V would go to A.

From V(-1,-1) to B(0,-6): delta x=1, delta y=-5, so slope -5.

To go to A, probably in the opposite direction from B relative to V.

So from V, moving in direction opposite to B: so delta x = -1, delta y = +5? Let's see.

Parametrize: from V, move s units in direction of the line.

Direction vector of line: since slope -5, direction vector can be (1, -5) or (-1, 5)

If we go from V in direction (-1,5), we get to A.

For example, let s=1: A = V + (-1,5) = (-1-1, -1+5) = (-2,4)

Check if on line: y = -5x -6 → x=-2, y= -5*(-2) -6 = 10-6=4 → yes!

So A = (-2,4)

Great! Now we have all points:

A(-2,4), V(-1,-1), C(7,5)

Now, angle at V: vectors VA and VC.

Vector VA = A - V = (-2 - (-1), 4 - (-1)) = (-1, 5)

Vector VC = C - V = (7 - (-1), 5 - (-1)) = (8,6)

Now, angle between vectors u = VA = (-1,5), v = VC = (8,6)

Dot product u•v = (-1)(8) + (5)(6) = -8 + 30 = 22

|u| = sqrt((-1)^2 + 5^2) = sqrt(1+25)=sqrt(26)

|v| = sqrt(8^2 + 6^2)=sqrt(64+36)=sqrt(100)=10

So cosα = (u•v) / (|u||v|) = 22 / (sqrt(26)*10) = 22/(10√26) = 11/(5√26)

Rationalize or compute numerical value.

First, √26 ≈ 5.099

So 5 * 5.099 ≈ 25.495

11 / 25.495 ≈ 0.4315

So α = arccos(0.4315) ≈ ?

cos(64°) ≈ 0.4384
cos(65°) ≈ 0.4226

So interpolate:

0.4384 - 0.4226 = 0.0158 difference for 1 degree

We have 0.4384 - 0.4315 = 0.0069

So fraction: 0.0069 / 0.0158 ≈ 0.4367

So α ≈ 64 + 0.4367 ≈ 64.44°

Same as before? Previously with tan we had similar.

Earlier with tan we got tanα=23/11≈2.0909, arctan(2.0909)≈64.43° — consistent.

But now with dot product, cosα=11/(5√26)≈11/(5*5.099)≈11/25.495≈0.4315, arccos(0.4315)≈64.44°

Good.

But perhaps they want exact value or rounded.

Since it's a geometry problem, maybe leave as arccos(11/(5√26)) but probably expect numerical.

Notice that in the tan method, we had tanα=23/11, which is exact.

How did we get that?

Earlier, when I used gradients, I took m1=-5 (for VA), m2=3/4 (for VC)

Then tanφ = |(m2 - m1)/(1 + m1m2)| = |(3/4 - (-5))/(1 + (-5)(3/4))| = |(3/4 + 20/4)/(1 - 15/4)| = |(23/4)/(-11/4)| = | -23/11 | = 23/11

This gives the tangent of the acute angle between the two lines.

In the triangle, since both vectors are going upwards from V, and A is left, C is right, the angle between them should be acute, and 23/11 >1, so about 64.4°, which matches.

In the dot product, we got positive dot product (22), so acute angle, good.

So α = arctan(23/11) or approximately 64.4 degrees.

But let's compute exactly.

Perhaps calculate in degrees.

I think for school level, they might expect the calculation.

Another way: use the formula.

But I think 64.4° is fine, but let's see if it's nice number.

Compute numerically more accurately.

23/11 = 2.090909...

arctan(2.090909) = ?

Using calculator: tan⁻¹(2.090909) = 64.439... degrees

So approximately 64.4° or 64.44°.

But perhaps keep as fraction or see.

Maybe they want in radians? Unlikely.

Or perhaps I can leave it, but I think for answer, we'll put the value.

But let's proceed and come back.

Actually, for now, I'll note α = \tan^{-1}(23/11) but probably compute.

Since in part (g) we need area, etc., but for now.

I think for (d), we can say α ≈ 64.4° but let's see the other parts.

Perhaps calculate exactly later.

Move on.

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(e) Calculate the length of BV.

B is on y-axis, and on line AB: y = -5x -6

When x=0, y= -6, so B(0, -6)

V is (-1, -1)

Distance BV = sqrt[ (0 - (-1))^2 + (-6 - (-1))^2 ] = sqrt[ (1)^2 + (-5)^2 ] = sqrt[1 + 25] = sqrt(26)

✔ Length of BV = √26

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

First, we already have coordinates, so we can compute directly, but the problem says "if", so perhaps it's hypothetical? But we already have actual values.

Let me check what we have.

From earlier:

A(-2,4), V(-1,-1), so AV = distance = sqrt[ (-2 - (-1))^2 + (4 - (-1))^2 ] = sqrt[ (-1)^2 + 5^2 ] = sqrt[1+25]=√26

BV = √26, as above.

But AV = √26, BV = √26, so AV = BV, but the problem says "if AV = 2BV" — contradiction?

Oh! So this part is conditional: "if AV = 2BV and CV = √111", but in reality from previous calculations, AV = BV = √26, and CV = distance from C(7,5) to V(-1,-1) = sqrt[(8)^2 + (6)^2]=sqrt[64+36]=sqrt[100]=10, not √111.

So this part is a separate scenario? Or did I make a mistake?

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

But in the diagram and previous parts, we have specific points, but this "if" suggests that for this part, we ignore previous coordinates and use these conditions.

That makes sense, because otherwise it's inconsistent.

So for part (f), we are to assume that in triangle AVC, with V fixed? Or not?

The problem doesn't specify, but likely, we are to use the given conditions without relying on previous coordinates.

But V is given as (-1,-1), and probably the configuration is similar, but with different lengths.

Moreover, T is midpoint of CV, but if CV changes, T might change, but the problem doesn't mention T for this part.

Perhaps we need to reinterpret.

Another thought: perhaps the "if" is based on the diagram, but with scaled lengths, but that might be complicated.

Let me read carefully: "If AV=2BV and CV=√111, calculate the length of AC."

But BV is part of the line, and B is defined as where AV produced meets y-axis.

So, in general, for any position, but V is fixed at (-1,-1), and line AV is still the same line? Or not?

The line AB is given as y + k + 5x = 0, and we found k=6, so the line is fixed: y = -5x -6.

So the line is fixed, V is fixed on it, B is fixed at (0,-6), as intersection with y-axis.

Then AV is distance from A to V, BV is from B to V.

In our calculation, with A(-2,4), V(-1,-1), B(0,-6), we have AV = √[(-1)^2 + 5^2] = √26, BV = √[1^2 + (-5)^2] = √26, so AV = BV.

But the problem says "if AV=2BV", which is not true in our case, so for this part, we must be considering a different point A on the same line, such that AV = 2 BV.

That makes sense.

So, the line is fixed: y = -5x -6, V(-1,-1), B(0,-6) are fixed.

Point A is on the line, and we need AV = 2 BV.

First, BV is fixed: distance from B to V is √[(0-(-1))^2 + (-6-(-1))^2] = √[1 + 25] = √26, as before.

So AV = 2 * √26

Now, A is on the line y = -5x -6, and distance from V(-1,-1) is 2√26.

Let A have coordinates (x, -5x -6)

Distance to V(-1,-1):

√[ (x - (-1))^2 + ((-5x -6) - (-1))^2 ] = 2√26

Simplify:

√[ (x+1)^2 + (-5x -5)^2 ] = 2√26

Note that -5x -5 = -5(x+1)

So:

√[ (x+1)^2 + [ -5(x+1) ]^2 ] = 2√26

= √[ (x+1)^2 + 25(x+1)^2 ] = √[26(x+1)^2] = √26 * |x+1|

Set equal to 2√26:

√26 * |x+1| = 2√26

Divide both sides by √26 (assuming ≠0):

|x+1| = 2

So x+1 = 2 or x+1 = -2

Thus x = 1 or x = -3

Now, corresponding y:

If x=1, y= -5(1) -6 = -11 → A(1,-11)

If x=-3, y= -5(-3) -6 = 15-6=9 → A(-3,9)

Now, which one is it? In the diagram, A is above and left of V, while B is below and right.

V(-1,-1), B(0,-6) is down-right.

A should be in the opposite direction, so up-left, so x < -1, y > -1.

So x=-3, y=9 → A(-3,9)

Check distance: from V(-1,-1) to A(-3,9): dx=-2, dy=10, dist=√[4+100]=√104=2√26, yes.

While if x=1, y=-11, dx=2, dy=-10, dist=√[4+100]=√104=2√26, but that would be in the direction of B, beyond B, but the problem says "AV produced cuts y-axis at B", implying that B is beyond V from A, so A and B are on opposite sides of V.

In this case, if A is at (-3,9), V(-1,-1), B(0,-6), then from A to V to B, so yes, AV produced meets y-axis at B.

If A were at (1,-11), then from A(1,-11) to V(-1,-1) to B(0,-6)? From V to B is to (0,-6), which is not on the line from A to V extended? Let's see the line.

The line is y=-5x-6, all points are on it.

From A(1,-11) to V(-1,-1): as x decreases from 1 to -1, y increases from -11 to -1.

Then continuing, to x=0, y=-6, which is B.

But in this case, from A to V to B, so B is on AV produced, yes.

But in the diagram, typically A is not on the same side as B relative to V; usually A is on one side, B on the other.

In the original diagram description, it says "AV produced cuts the y-axis at B", which implies that starting from A, going through V, then hitting B on y-axis.

So if A is at (-3,9), V(-1,-1), then extending beyond V, when x=0, y=-6, which is B, so yes.

If A is at (1,-11), then from A(1,-11) to V(-1,-1), then to B(0,-6)? From V(-1,-1) to B(0,-6) is increasing x, decreasing y, but from A to V is decreasing x, increasing y, so the direction from A to V is towards decreasing x, but B is at x=0 > x_V=-1, so it's not on the ray from A through V; it's on the extension beyond A or something.

Let's parametrize.

The line: from V(-1,-1), direction vector.

The line has direction: since slope -5, direction vector (1,-5) or (-1,5)

From V, moving in direction (-1,5) gets to A(-3,9): from (-1,-1) + t*(-1,5), set t=2: (-1-2, -1+10)=(-3,9)

Moving in direction (1,-5) gets to B(0,-6): from (-1,-1) + s*(1,-5), set s=1: (0,-6)

So if we start at A(-3,9), go to V(-1,-1) (which is adding (2,-10) or t from 2 to 0 in the parameter), then continuing in the same direction, we add more (1,-5) per unit, but from V, to go to B, we add (1,-5), which is the same direction as from A to V? From A to V: from (-3,9) to (-1,-1): delta (2,-10) = 2*(1,-5)

From V to B: (1,-5), so same direction vector (1,-5)

So if we go from A to V to B, it's continuous in the direction of (1,-5)

But in the diagram, typically, "AV produced" means extending beyond V away from A, so if A is at one end, V in middle, B further.

In this case, with A(-3,9), V(-1,-1), B(0,-6), then from A to V is vector (2,-10), from V to B is (1,-5), which is half, so not collinear in the sense of same line but different scale, but it is collinear.

Points are collinear, and B is on the line AV extended beyond V.

Similarly, if A were at (1,-11), then from A(1,-11) to V(-1,-1): vector (-2,10) = -2*(1,-5)

Then from V to B(0,-6): (1,-5), which is opposite direction? Vector from V to B is (1,-5), while from A to V is (-2,10) = -2*(1,-5), so same line, but from A to V is in direction -(1,-5), then from V to B is +(1,-5), so it's extending in the opposite direction, which would mean that B is not on the ray from A through V, but on the other side.

The phrase "AV produced" usually means the ray starting at A, passing through V, and extending beyond V.

So for B to be on that ray, it should be that V is between A and B.

In the first case, with A(-3,9), V(-1,-1), B(0,-6): let's see the parameter.

Let P(t) = V + t * D, where D is direction.

Set D = (1,-5), then V(-1,-1)

P(t) = (-1 + t, -1 -5t)

When t=0, V

When t=1, (0,-6) = B

When t=-2, (-1-2, -1+10)=(-3,9)=A

So A at t=-2, V at t=0, B at t=1

So on the line, A--V--B, with V between A and B? t=-2,0,1, so from A to V to B, yes, V is between A and B only if t_A < t_V < t_B, here -2<0<1, yes.

t_A=-2, t_V=0, t_B=1, so order on line: A, then V, then B.

So AV produced beyond V hits B.

In the other choice, if A is at (1,-11), which is P(t) = (-1+t, -1-5t) = (1,-11) → -1+t=1 => t=2, -1-5*2=-11, yes.

So A at t=2, V at t=0, B at t=1

So order: V(t=0), B(t=1), A(t=2), so from A to V is from t=2 to t=0, then produced beyond V would be to t<0, but B is at t=1>0, so not on the ray from A through V beyond V; it's between A and V or something.

From A(t=2) to V(t=0), then beyond V is t<0, but B is at t=1, which is not in that direction.

So only A at t=-2, i.e., (-3,9) satisfies that B is on AV produced beyond V.

Moreover, in the diagram, A is likely above, so y=9 is reasonable.

So for part (f), we take A(-3,9)

Now, also given CV = √111

C is such that T is midpoint of CV, but T is not specified for this part? The problem doesn't mention T for this part, so perhaps C is not constrained by T here.

The condition is only AV=2BV and CV=√111, and we have V fixed, and the line fixed, but C is another point.

In the triangle, C is connected, but no other constraint, so we need to find AC, but we have two points A and V, and C is at distance √111 from V, but we don't know where.

The problem is to find length of AC, but with only CV=√111, and A and V fixed, then C can be anywhere on circle center V radius √111, so AC is not determined.

That can't be.

Perhaps in this context, the point C is still such that T is midpoint, but T is not given.

Another possibility: perhaps "CV = √111" is given, and we have to use the fact that in the diagram, but it's messy.

Perhaps for this part, we are to assume that the configuration is the same, but with the given lengths, but that doesn't work.

Let's read the problem again: "If AV=2BV and CV=√111, calculate the length of AC."

But in the initial setup, we have points, but this "if" suggests a different scenario.

Perhaps BV is not the distance, but something else, but unlikely.

Another thought: perhaps "BV" refers to the length from B to V, which is fixed, as we have B and V fixed from the line and y-axis.

In our case, BV = √26, as calculated.

Then AV = 2 * BV = 2√26

Then we found A at (-3,9) or (1,-11), and we chose (-3,9)

Now, CV = √111, and C is a point, but in the triangle, and T is midpoint of CV, but T is not specified, so perhaps C is not constrained, but then AC is not unique.

Unless we assume that C is in the plane, but we need another condition.

Perhaps in this context, the angle or something is the same, but not specified.

Another idea: perhaps "CV = √111" is a typo or something, but let's see the number.

In our original calculation, CV was 10, and √111 ≈ 10.535, close but not same.

Perhaps for this part, we are to use the coordinates we have, but with the given lengths, but it's inconsistent.

Let's look back at the problem: " (f) If AV=2BV and CV=√111, calculate the length of AC."

But in the diagram, there is point T, midpoint of CV, and T(3,2) is given, but if CV changes, T may change.

Perhaps for this part, we ignore T and just use the distances.

But then with A and V fixed, and C at distance √111 from V, then AC depends on the angle at V.

But the angle at V is not given for this part.

Unless we assume that the angle α is the same, but not specified.

Perhaps in the context, the line CV is the same as before, but that doesn't make sense.

Another interpretation: perhaps "BV" is not the distance, but the length along the line or something, but unlikely.

Let's calculate what CV was in original: from C(7,5) to V(-1,-1) = sqrt(8^2 +6^2)=10, and √111≈10.535, not match.

Perhaps for this part, we are to use the values from previous parts, but the "if" suggests otherwise.

Let's see the next part (g) : "Calculate the area of ΔAVC" — which in original we can do, but for (f) it's conditional.

Perhaps the "if" is for a different configuration, but we need to find AC given those lengths and the geometry.

Recall that in triangle AVC, we have points A, V, C, with V fixed, A on the line, with AV = 2 BV, and BV is fixed, so AV = 2*√26, and we have A at (-3,9)

Then CV = √111, so C is at distance √111 from V(-1,-1)

But to find AC, we need the distance between A and C, which requires knowing where C is.

Unless there is additional information. Perhaps from the diagram, C is such that the line CV has a certain property, but not specified.

Another thought: in the original problem, T is midpoint of CV, and T(3,2) is given, but for this part, if CV = √111, and T is midpoint, then VT = CV/2 = √111 /2, but T is not specified, so perhaps T is still (3,2)? But then CV would be distance from C to V, with T midpoint, so if T(3,2), V(-1,-1), then C = 2T - V = 2*(3,2) - (-1,-1) = (6,