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Coordinate Geometry: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two measurements

A segment joins $A(-13,4)$ to $B(2,-9)$ . Find its exact length and midpoint. Explain why the length calculation uses perpendicular gaps and why your midpoint is on the segment.

The drawing records only the two given endpoints.

Text description of this figure

A at (-13,4) and B at (2,-9) are joined on equal-scale axes. No midpoint, perpendicular corner, or computed length is drawn.
1. Hint 1
  A horizontal trip followed by a vertical trip forms a right triangle with the direct segment as hypotenuse.
2. Hint 2
  A halfway point makes half of each signed coordinate change from the same endpoint.
Answer

Length $\sqrt{394}$ units; midpoint $(-11/2,-5/2)$ .

Full solution

The horizontal gap has length 15 and the vertical gap has length 13.
The coordinate axes are perpendicular, so these are right-triangle legs.
$\begin{gathered}AB^2=15^2+13^2\\AB^2=394\\AB=\sqrt{394}.\end{gathered}$

Averaging the matching coordinates gives
$\begin{gathered}x_{M}=\frac{-13+2}{2}\\x_{M}=-\frac{11}2\\y_{M}=\frac{4-9}{2}\\y_{M}=-\frac52.\end{gathered}$

The signed changes from A to M are $15/2$ and negative $13/2$ .
These are half of the changes from A to B.
A dilation of B about A with positive factor one half therefore places M on AB and gives two equal half-segments.

Answer

Length $\sqrt{394}$ units; midpoint $(-11/2,-5/2)$ .

Key idea

A midpoint halves both signed changes, while straight-line length comes from their perpendicular square sum.
Problem 2 An endpoint to recover

The endpoint is $B(-24,13)$ and the internal point is $P(1,3)$ . The ordered ratio is $AP:PB=2:5$ . Find A, verify the same internal fraction in both coordinates, and state $BP:PA$ .

Explain geometrically why the same fraction governs both coordinates, how the reasoning handles horizontal or vertical segments, and why the endpoint weights reverse the ordered ratio parts.

The written ratio fixes the continuation beyond P.

Text description of this figure

P at (1,3) and B at (-24,13) are the only marked points, joined by the known subsegment PB. The coordinate scales are equal. The unknown endpoint A is not plotted.
1. Hint 1
  The known change from P to B represents the second ratio part.
2. Hint 2
  Find the change for one part, then move two such changes backward from P.
3. Hint 3
  Reversing the endpoint order exchanges the two positive subsegment lengths.
4. Hint 4
  For the explanation, compare the smaller and full coordinate right triangles, then expand the fraction-of-change expression.
Answer

A is $(11,-1)$ . Both fractions are $2/7$ . Reversed ratio:
$BP:PA=5:2.$
Weights are 5 on A and 2 on B; a fixed coordinate is unchanged.

Full solution

The signed changes from P to B are negative 25 and 10.
These are five ratio parts, so one part changes the coordinates by negative 5 and 2.

Move backward two parts from P.
$\begin{gathered}x_{A}=1-2(-5)\\x_{A}=11\\y_{A}=3-2(2)\\y_{A}=-1.\end{gathered}$

Verify the complete segment rather than one coordinate alone.
$\begin{gathered}\frac{1-11}{-24-11}=\frac27\\\frac{3-(-1)}{13-(-1)}=\frac27.\end{gathered}$

The common fraction is strictly between zero and one, so P is internal.
Two sevenths lie before P from A and five sevenths after it.
Starting from B therefore gives
$BP:PA=5:2.$

When both coordinate changes are nonzero, the smaller and whole coordinate right triangles share the angle at A and each has a right angle.
AA similarity gives the same leg-length fraction as AP to AB.
The small and full signed changes have matching directions, so the same positive fraction applies to both signed coordinate changes.

For distinct endpoints and positive ordered ratio parts u and v, that fraction is u divided by their sum.
The horizontal coordinate therefore satisfies
$x_{P}=x_{A}+\frac{u}{u+v}(x_{B}-x_{A}).$
Rearranging gives
$(u+v)x_{P}=vx_{A}+ux_{B}.$
The height equation has the same form.
Thus the first ratio number weights B, while the second weights A.
The first part measures how far the point moves toward B.

On a horizontal or vertical segment, the fixed coordinate has zero change and stays fixed under this same fraction.
Apply the fraction only to the changing coordinate; there is no division by a zero coordinate change.

Answer

A is $(11,-1)$ . Both fractions are $2/7$ . Reversed ratio:
$BP:PA=5:2.$
Weights are 5 on A and 2 on B; a fixed coordinate is unchanged.

Key idea

A recovered endpoint must satisfy both coordinate fractions and the named endpoint order.
Problem 3 A four-sided boundary

A simple quadrilateral has boundary vertices $A(-4,-2)$ , $B(10,-2)$ , $C(t,4)$ , and $D(-1,4)$ in that order, where $-1<t<10$ . Its area is 69 square units. Find t and the exact perimeter. Check that the recovered value preserves the stated boundary.
1. Hint 1
  Compare the region with the rectangle spanning the extreme coordinate values.
2. Hint 2
  The two omitted corner triangles have the same perpendicular height as the quadrilateral.
3. Hint 3
  After finding the missing coordinate, add all four outside lengths and retain exact radicals.
Answer

$t=8$ ; perimeter $23+2\sqrt{10}+3\sqrt5$ units. The recovered t satisfies $-1<t<10$ .

Full solution

The containing rectangle has width 14 and height 6, so its area is 84.
The upper-left missing triangle has base 3 and height 6, giving area 9.
The upper-right missing triangle has base $10-t$ and height 6, giving area $3(10-t)$ .
These two triangles have disjoint interiors and fill the missing region.

The required area therefore gives
$\begin{gathered}84-9-3(10-t)=69\\45+3t=69\\t=8.\end{gathered}$

Now the horizontal sides have lengths 14 and 9.
The two slanted side lengths follow from their perpendicular gaps.
$\begin{gathered}BC^2=2^2+6^2\\BC=2\sqrt{10}\\DA^2=3^2+6^2\\DA=3\sqrt5.\end{gathered}$

Only the outside edges contribute to perimeter.
$P=23+2\sqrt{10}+3\sqrt5.$

The value 8 lies strictly between negative 1 and 10.
Thus the upper edge has positive length, both corner cuts remain valid, and the four vertices form the stated simple boundary.

The missing corner triangles justify the recovered area.

Answer

$t=8$ ; perimeter $23+2\sqrt{10}+3\sqrt5$ units. The recovered t satisfies $-1<t<10$ .

Key idea

Recovering a coordinate from area still requires checking the inequalities that define the polygon.
Problem 4 An altitude constraint

Triangle ABC has $A(-5,-4)$ and $B(7,2)$ . Vertex C lies on $x=-1$ . The altitude from C to the full line of AB has length $4\sqrt5$ units. Find every possible C and the median line from C to AB for each case. Check the altitude length and nondegeneracy. Explain whether a nondegenerate triangle would be possible if the required altitude length were zero.

The unknown vertex lies somewhere on the dashed line.

Text description of this figure

A(-5,-4) and B(7,2) determine the solid side AB. The dashed vertical line x equals negative 1 is the given constraint on C. Equal coordinate scales are used. Neither possible C nor an altitude or foot is plotted.
1. Hint 1
  An altitude must meet the full opposite side line at a right angle; its foot may be an endpoint or lie beyond a side.
2. Hint 2
  The slope of AB fixes the perpendicular direction. Represent the signed horizontal change from C to its foot by one variable.
3. Hint 3
  The distance condition allows two signs. Recover each vertex, then use the fixed midpoint of AB for its median.
Answer

$C(-1,8)$ :
$9x+2y=7.$
Or $C(-1,-12)$ :
$11x-2y=13.$
Both triangles are nondegenerate. No nondegenerate triangle is possible for zero altitude.

Full solution

The line AB has slope one half and equation
$y=\frac{x-3}{2}.$
Every perpendicular altitude therefore has slope negative 2.

Let u be the signed horizontal change from C to its foot H on AB.
The signed vertical change is negative $2u$ .
The altitude length gives
$\begin{gathered}u^2+(-2u)^2=80\\5u^2=80\\u=\pm4.\end{gathered}$

Write C at $(-1,c)$ .
The foot has first coordinate $-1+u$ , so its height on AB is $(u-4)/2$ .
Its signed height relation is
$y_{H}=c-2u.$
Substituting the foot height gives
$c=\frac{5u-4}{2}.$

Case $u=4$ .
The vertex is $C(-1,8)$ and its foot is $H(3,0)$ .
The altitude gaps are 4 and negative 8, whose square sum is 80.
Its positive length is the required $4\sqrt5$ .

Case $u=-4$ .
The vertex is $C(-1,-12)$ and its foot is $H(-5,-4)$ , which is A.
The altitude gaps are negative 4 and 8, again with square sum 80.

The midpoint of AB is $M(1,-1)$ .
In the first case the line CM has slope negative $9/2$ , giving
$9x+2y=7.$
In the second case its slope is $11/2$ , giving
$11x-2y=13.$

Each median equation contains its recovered C and M.
At first coordinate negative 1, the line AB has height negative 2; neither recovered C has that height.
Both triangles are nondegenerate.
The only two real signs for u have been retained, so the list is complete.

For zero altitude length the vertex would be on the full line AB.
Its only point with first coordinate negative 1 is (-1,-2), so A, B, and C would be collinear.
No nondegenerate triangle would satisfy that altered condition.

The full side line admits both checked altitude feet.

Answer

$C(-1,8)$ :
$9x+2y=7.$
Or $C(-1,-12)$ :
$11x-2y=13.$
Both triangles are nondegenerate. No nondegenerate triangle is possible for zero altitude.

Key idea

A prescribed altitude length can give two vertices, and each must be checked against its own geometric lines.
Problem 5 Two interior segments

In any nondegenerate triangle ABC, M and N are the midpoints of AB and AC, respectively. D is the midpoint of BC, G is the centroid, and S is the intersection of AD with MN.

Determine the direction and length of MN relative to BC, and the ordered ratio SG:GD. Derive the centroid coordinate rule as part of your argument. Justify your conclusions for every allowed triangle, including triangles with an obtuse angle.

State the most specific quadrilateral family guaranteed for BCNM. Justify this classification and any excluded stronger families.

The diagram records the named midpoints; the argument keeps the triangle general.

Text description of this figure

A nondegenerate triangle ABC has M on AB, N on AC, and D on BC, each at its given midpoint. Segment MN is solid blue and median AD is dashed red. No parallel, length-ratio, centroid, or intersection label is supplied. This is a schematic of the given relationships, not a restriction on the triangle shape.
1. Hint 1
  Choose axes that simplify one side while leaving the third vertex’s horizontal position free.
2. Hint 2
  Find the midpoint coordinates of M, N, and D. A dilation about A can compare MN with BC.
3. Hint 3
  Locate S on AD and compare its fraction of that median with the centroid’s fraction from A.
4. Hint 4
  Use the already established relations between opposite sides to compare the quadrilateral families.
Answer

$MN\parallel BC.$ $MN=\frac12BC.$ $SG:GD=1:2.$
The centroid averages the three first coordinates and the three heights separately. BCNM has exactly one parallel pair: a trapezoid, not a parallelogram.

Full solution

Choose
$\begin{gathered}A=(0,0)\\B=(2a,0)\\C=(2b,2c),\end{gathered}$
where a and c are positive and b is any real number.
Every nondegenerate triangle can be placed this way by directing the vertical axis toward C.
No restriction on b adds a right angle or excludes an obtuse triangle.

The given midpoints have coordinates
$\begin{gathered}M=(a,0)\\N=(b,c)\\D=(a+b,c).\end{gathered}$

A dilation centered at A with factor one half sends B to M and C to N.
The line BC does not pass through A.
Its image MN is therefore a distinct parallel line and its segment length is half of BC.
This covers horizontal or vertical directions without dividing by a zero coordinate change.

Quadrilateral BCNM has parallel opposite sides BC and MN, so it is a trapezoid under the inclusive definition.
The length of MN is half the positive length BC, so these opposite sides are unequal.
Every parallelogram has congruent opposite sides; BCNM therefore cannot be a parallelogram.
Consequently it has exactly one parallel pair.
Rectangles, rhombi, and squares are excluded because they are parallelograms.

The midpoint of MN is
$S=\left(\frac{a+b}{2},\frac c2\right).$
These are also the midpoint coordinates of AD.
Thus this point is the stated intersection, and it lies halfway from A to D.

The centroid is two thirds from A to D.
Comparing the two fractions gives
$\begin{gathered}\frac{SG}{AD}=\frac23-\frac12\\\frac{SG}{AD}=\frac16\\\frac{GD}{AD}=\frac13\\SG:GD=1:2.\end{gathered}$

For arbitrary vertex coordinates, D is the midpoint of BC and G is two thirds along AD from A.
The first-coordinate relations therefore give
$\begin{gathered}2x_{D}=x_{B}+x_{C}\\3x_{G}=x_{A}+2x_{D}\\3x_{G}=x_{A}+x_{B}+x_{C}.\end{gathered}$
The same argument for heights gives
$3y_{G}=y_{A}+y_{B}+y_{C}.$
Dividing each sum by 3 derives the centroid coordinate rule from the internal median fraction.

Since one half is less than two thirds and both lie between zero and one, the order on the median is A, S, G, D.
Every segment in the requested ratio is positive.
The free parameters were never specialized, so the conclusions hold for every allowed triangle.

Answer

$MN\parallel BC.$ $MN=\frac12BC.$ $SG:GD=1:2.$
The centroid averages the three first coordinates and the three heights separately. BCNM has exactly one parallel pair: a trapezoid, not a parallelogram.

Key idea

Comparing two internal fractions on the same median connects midpoint geometry with the centroid ratio.
Problem 6 A circle with two constraints

A circle passes through $T(2,-5)$ and $U(-11,-4)$ . At T its tangent line is
$6x-7y=47.$
Find the circle’s center and equation. Verify that both given points lie on it and that the line is tangent at T. Explain why the constraints determine only one circle.

The center and circle must satisfy both given constraints.

Text description of this figure

The given points are T(2,-5) and U(-11,-4). The dashed line is the given tangent candidate, six x minus seven y equals 47, passing through T. Equal coordinate scales are used. No circle, center, or perpendicular construction is shown.
1. Hint 1
  A circle center is equally far from its two boundary points and lies on the perpendicular to a tangent at contact.
2. Hint 2
  Build the perpendicular bisector of TU and the perpendicular to the tangent through T.
3. Hint 3
  Their intersection supplies a center. Check the radius, both incidences, and the perpendicular contact direction.
Answer

Center $(-4,2)$ .
$(x+4)^2+(y-2)^2=85.$
The circle is unique and the line is tangent at T.

Full solution

The tangent has slope $6/7$ .
Its perpendicular through T has slope negative $7/6$ , so the center lies on
$7x+6y=-16.$

The midpoint of TU is $(-9/2,-9/2)$ and TU has slope negative $1/13$ .
Its perpendicular bisector has slope 13 and equation
$y=13x+54.$

Intersect the two required center lines.
$\begin{gathered}7x+6(13x+54)=-16\\85x=-340\\x=-4\\y=2.\end{gathered}$

The squared radius is the squared distance from C at $(-4,2)$ to T.
$\begin{gathered}r^2=6^2+(-7)^2\\r^2=85\\(x+4)^2+(y-2)^2=85.\end{gathered}$

T has squared center distance 85.
U has gaps negative 7 and negative 6, giving squared distance 85 as well.
Both points therefore lie on the positive-radius circle.

The given contact lies on the tangent because
$6(2)-7(-5)=47.$
The radius slope is negative $7/6$ , whose product with the line slope $6/7$ is negative 1.
Incidence and perpendicularity prove tangency at T.

Any circle satisfying the data must have its center on both constructed lines.
Their slopes, negative $7/6$ and 13, differ, so their intersection is unique.
The distance to T then fixes one positive radius.

The revealed radius and incidence checks establish the contact.

Answer

Center $(-4,2)$ .
$(x+4)^2+(y-2)^2=85.$
The circle is unique and the line is tangent at T.

Key idea

A boundary pair and a contact direction locate a center by two independent geometric lines.
Problem 7 All values of the constant

For every real k, describe the real graph below as a positive-radius circle, a single point, or empty. Find all of its common points with the horizontal line. State exactly when the line is tangent to a positive-radius circle. Whenever the graph is a circle, classify H at $(8,-4)$ as inside, on, or outside.
$x^2+y^2-16x+18y+k=0.$ $y=-4.$
1. Hint 1
  Recover the completed-square right side before classifying the graph.
2. Hint 2
  After substituting the line height, distinguish a positive, zero, and negative remaining square.
3. Hint 3
  The graph can stop being a circle even after the line has stopped meeting it.
Answer

Circle below 145; point $(8,-9)$ at 145; empty above 145. In circle cases, H is inside below 120, on at 120, outside above 120. Intersections: $(8\pm\sqrt{120-k},-4)$ below 120; H alone at 120 (tangent); none above 120.

Full solution

Completing both squares gives
$(x-8)^2+(y+9)^2=145-k.$
Thus the center is $(8,-9)$ whenever the right side is positive.

On the required line, the vertical square is 25, leaving
$(x-8)^2=120-k.$

Case $k<120$ .
The graph is a positive-radius circle and the remaining square is positive.
There are exactly two common points,
$(8\pm\sqrt{120-k},-4).$
Each has the required height and circle square sum $145-k$ .
H has squared center distance 25, less than the squared radius, so H is inside.

Case $k=120$ .
The circle has radius 5 and the only common point is $(8,-4)$ .
Its contact radius from $(8,-9)$ is vertical, while the line is horizontal.
Hence the line is tangent.
Thus H is on the circle.

Case $120<k<145$ .
A positive-radius circle still exists, but the remaining square is negative.
The line has no common points.
Its squared radius is positive but less than 25, so H is outside.

Case $k=145$ .
Both completed squares must vanish, so the graph is only $(8,-9)$ .
That point has the wrong height for the line; there are no common points and no positive-radius circle tangency.

Case $k>145$ .
The completed-square right side is negative, so there are no real graph points or common points.
These intervals and endpoint cases exhaust all real k.

Answer

Circle below 145; point $(8,-9)$ at 145; empty above 145. In circle cases, H is inside below 120, on at 120, outside above 120. Intersections: $(8\pm\sqrt{120-k},-4)$ below 120; H alone at 120 (tangent); none above 120.

Key idea

Graph classification and line-intersection classification can change at different parameter values.
Problem 8 A different boundary

Triangle ABC has $A(-9,-4)$ , $B(9,-4)$ , and $C(-27,11)$ . Its centroid is G. Find G and the exact perimeter and area of the simple quadrilateral whose boundary order is A, G, C, B.

The centroid determines the new boundary.

Text description of this figure

The original triangle has vertices A(-9,-4), B(9,-4), and C(-27,11), joined in boundary order. The coordinate scales are equal. The requested centroid and inward boundary are not plotted.
1. Hint 1
  Locate G from the three vertex coordinate averages before tracing the new boundary.
2. Hint 2
  Compare the new region with the whole triangle ABC; its inward edge removes one triangle.
3. Hint 3
  A vertical base can make the removed triangle’s perpendicular height a horizontal distance.
Answer

$G=(-9,1)$ . Perimeter $62+2\sqrt{106}$ units; area $90$ square units.

Full solution

The centroid averages corresponding vertex coordinates.
$\begin{gathered}x_{G}=\frac{-9+9-27}{3}\\x_{G}=-9\\y_{G}=\frac{-4-4+11}{3}\\y_{G}=1.\end{gathered}$

The ordered boundary has vertical edge AG of length 5 and horizontal edge BA of length 18.
The other two edge lengths are
$\begin{gathered}GC^2=18^2+10^2\\GC=2\sqrt{106}\\CB^2=36^2+15^2\\CB=39.\end{gathered}$

Add those four outside edges.
The original side AC is absent from the new boundary.
$P=62+2\sqrt{106}.$

The original triangle has horizontal base AB of length 18 and perpendicular height 15.
$A_{ABC}=135.$

The removed triangle ACG has vertical base AG of length 5.
C is 18 units horizontally from the full line of AG, so its perpendicular height is 18.
$\begin{gathered}A_{ACG}=\frac12(5)(18)\\A_{ACG}=45\\A_{AGCB}=135-45\\A_{AGCB}=90.\end{gathered}$

G is inside the nondegenerate original triangle.
Removing triangle ACG leaves exactly the simple region with boundary A, G, C, B, without overlapping interiors or a missing extra region.

The dashed original side bounds the removed triangle.

Answer

$G=(-9,1)$ . Perimeter $62+2\sqrt{106}$ units; area $90$ square units.

Key idea

An inward boundary changes which edges count and which region must be removed.
Problem 9 A graph and its translation

A parabola has equation
$x^2-12x+20y+116=0.$
Recover its vertex, axis, focus, directrix, and opening direction. Starting from the defining distances for your recovered geometry, justify that they describe exactly the original graph.

Translate the parabola 5 units left and 2 units up. Give its new equation, and verify whether $Q(11,-7)$ is on it using both the equation and the two defining distances.
1. Hint 1
  Expose a squared coordinate gap before reading the vertex and signed focus step.
2. Hint 2
  The multiplier is four times the signed parameter; line distance uses an absolute value when reversing the squared equality.
3. Hint 3
  Translate every defining object together, leaving the signed parameter unchanged.
4. Hint 4
  For Q, check both distance measurements as well as substitution into the translated equation.
Answer

Vertex $(6,-4)$ , focus $(6,-9)$ ; opens down. Axis
$x=6.$
Directrix
$y=1.$
Translated:
$(x-1)^2=-20(y+2).$
Q is on it; both distances are 10.

Full solution

Completing the x-square gives
$\begin{gathered}(x-6)^2=-20(y+4)\\p=-5.\end{gathered}$

The vertex is $(6,-4)$ and the axis is vertical through first coordinate 6.
The signed parameter is negative 5, placing the focus at $(6,-9)$ and the directrix at height 1.
The curve opens down.

For any real point $(x,y)$ , the recovered focus and directrix give the distance equality
$\begin{gathered}\sqrt{(x-6)^2+(y+9)^2}\\=|y-1|.\end{gathered}$

Both distances are nonnegative, so squaring is reversible.
Their squared equality is
$\begin{gathered}(x-6)^2+(y+9)^2\\=(y-1)^2.\end{gathered}$
Expanding the two y-squares and canceling matching terms gives
$(x-6)^2=-20(y+4).$

Conversely, this last equation restores the squared equality when substituted into the focus-distance square.
Taking nonnegative square roots restores the absolute line distance.
Thus no extra points have been introduced.

The translation moves the vertex to $(1,-2)$ , the focus to $(1,-7)$ , and the directrix to height 3.
The signed parameter stays negative 5, so the new equation is
$(x-1)^2=-20(y+2).$

At Q, the equation has left side 100 and right side 100.
The focus distance is the horizontal gap 10.
Its perpendicular foot on the translated directrix is $(11,3)$ , giving line distance 10.
Hence Q belongs to the translated locus in both representations.

The completed configuration supplies the two equal distances.

Answer

Vertex $(6,-4)$ , focus $(6,-9)$ ; opens down. Axis
$x=6.$
Directrix
$y=1.$
Translated:
$(x-1)^2=-20(y+2).$
Q is on it; both distances are 10.

Key idea

Recovering and translating a locus remains valid only when the defining distances agree in both directions.
Problem 10 A centroid as a focus

For real t, triangle ABC has $A(-10,3)$ , $B(2,-9)$ , and $C(t,12)$ . Its centroid F is the focus of a parabola. The directrix is the vertical line through the midpoint of AB, and the parabola passes through $P(-8,2)$ .

Find every t satisfying the data. Give the focus, vertex, axis, directrix, opening direction, and equation. Check the triangle’s nondegeneracy, the nonzero focus-directrix spacing, and P’s defining distances.
1. Hint 1
  The two fixed vertices determine the midpoint line, and all three heights determine the centroid height.
2. Hint 2
  Keep the focus’s first coordinate unknown and apply the distance condition at P.
3. Hint 3
  Retain both algebraic candidates before checking whether either puts the focus on the directrix.
4. Hint 4
  Recover the vertex halfway across the valid spacing, then verify both the triangle and locus conditions.
Answer

$t=-28$ ; focus $(-12,2)$ ; vertex $(-8,2)$ ; opens left. Axis
$y=2.$
Directrix
$x=-4.$
Equation
$(y-2)^2=-16(x+8).$
P has both distances 4; the triangle is nondegenerate.

Full solution

The midpoint of AB is $(-4,-3)$ , so the directrix is
$x=-4.$
The centroid height is
$y_{F}=\frac{3-9+12}{3}.$
It equals 2.
Write the focus as $(f,2)$ , where
$f=\frac{t-8}{3}.$

P is 4 units horizontally from the directrix.
It shares the focus height, so its focus distance must satisfy
$\begin{gathered}|f+8|=4\\f=-12\text{ or }-4.\end{gathered}$

The candidate negative 4 places the focus on the directrix and is excluded.
The valid focus is $(-12,2)$ , giving
$\begin{gathered}\frac{t-8}{3}=-12\\t=-28.\end{gathered}$

The directrix foot from the focus is $(-4,2)$ .
Its midpoint with the focus is the vertex $(-8,2)$ , and the axis is horizontal at height 2.
The focus lies 4 units left of the vertex, so the signed parameter is negative 4 and the curve opens left.
$(y-2)^2=-16(x+8).$

At the recovered t, AB has slope negative 1 and AC has slope negative one half.
Their distinct slopes prove the triangle is nondegenerate.
The focus-directrix spacing is 8, which is nonzero.

At P, the squared-coordinate equation has both sides zero.
Its focus distance is 4 and its perpendicular distance to the directrix is also 4.
Thus every original condition holds.
The two real distance candidates were exhausted and the invalid one rejected, so there is no other t.

The rejected focus-on-directrix candidate is not drawn.

Answer

$t=-28$ ; focus $(-12,2)$ ; vertex $(-8,2)$ ; opens left. Axis
$y=2.$
Directrix
$x=-4.$
Equation
$(y-2)^2=-16(x+8).$
P has both distances 4; the triangle is nondegenerate.

Key idea

Connected constraints require checking the recovered triangle as well as the nondegenerate distance locus.

Multiple choice

Core practice