Graphing Lines: Star problems

The slope is

An equation through A is $2x+3y=13$.

If $x,y$ are integers, then $2x$ has remainder 1 on division by 3, so $x$ has remainder 2.

Thus write $x=-4+3j$ for an integer $j$; substituting gives $y=7-2j$.

A point on the segment has

Therefore $0\leq j\leq4$, and the five integer choices give exactly the listed points.

This proves completeness rather than merely spotting points on a drawing.

For five equal divisions, the first new point would be one-fifth of the way from A to B.

Its coordinates would be $(-4,7)+(12/5,-8/5)=(-8/5,27/5)$, which are not integers.

Hence the requested division is impossible.

Equivalently, the segment contains only four equal smallest lattice steps, each with coordinate change $(3,-2)$; four such steps cannot be split into five equal whole numbers of steps.

Rearrange the equation as

A point satisfying $2x+y=7$ and $-x+3y=2$ therefore lies on every member of the family.

Solving these two linear equations gives $x=19/7$ and $y=11/7$.

Substitution into the grouped equation proves that this point works for every real $t$.

It is the only common point: any point on both the $t=0$ and $t=1$ lines must satisfy the two equations obtained by subtracting their line equations, and those equations have the unique solution just found.

A vertical line requires the coefficient of $y$ to vanish, so $t=-3$.

Its equation becomes $-7x=-19$, or $x=19/7$.

A horizontal line requires the coefficient of $x$ to vanish, so $t=1/2$, giving $y=11/7$.

The coefficients $2t-1$ and $t+3$ cannot vanish together, because their zeros are different.

Thus the equation always has at least one nonzero coordinate coefficient and always defines a genuine line, including the two exceptional orientations.

The original line AB has slope

Write $D=(d,0)$ and $E=(0,e)$, where $d,e>0$.

Parallelism gives $-e/d=-2/3$, so $d=12s$ and $e=8s$ for a common positive scale factor $s$.

Because D and E lie on the original sides, $s\leq1$.

The original triangle has area

The smaller triangle has area

Requiring half the original area gives $s^2=1/2$.

Positivity selects $s=1/\sqrt2=\sqrt2/2$, so the new intercepts are $6\sqrt2$ and $4\sqrt2$.

A line with these intercepts and slope $-2/3$ is

The scale factor lies between 0 and 1, verifying that both intersection points lie on the required segments.

If both intercepts were halved, the base and height would each acquire a factor of $1/2$, so the area would acquire a factor of $1/4$.

Halving a length scale and halving an area are different conditions.

The slope of AB is

Since BC is perpendicular to AB, its slope is $-2$.

The line through B with that slope is $y-5=-2(x-7)$, or $y=19-2x$.

Combining it with $x+2y=23$ gives $x+2(19-2x)=23$, so $x=5$ and $y=9$.

Thus $C=(5,9)$.

The coordinate displacement from B to C is $(-2,4)$.

Apply the same displacement to A to obtain $D=(-1,6)$.

Then AD is parallel and equal in length to BC, while DC has displacement $(6,3)$ and is parallel and equal to AB.

The four points form a parallelogram, and the slopes $1/2$ and $-2$ give a right angle.

Hence it is a rectangle.

The side lengths, by the Pythagorean theorem, are $AB=\sqrt{6^2+3^2}=3\sqrt5$ and

Their product is $6(5)=30$ square units.

The two lines used to find C have different slopes, so they have exactly one intersection.

Once A, B, and C are fixed consecutive vertices, the parallel-side condition determines D uniquely.

The rectangle has area 96, so each region must have area 48.

A line coinciding with the bottom edge does not divide the interior.

Every other line through P that divides the interior meets the boundary again on the left side, right side, or top side.

If it exits on the left at height $h\leq8$, it cuts off a triangle with area

If it exits on the right at height $h\leq8$, it cuts off a triangle with area

Neither triangle can have area 48, so neither exit is possible.

This also rules out the top corners when treated as endpoints of vertical sides.

Therefore the line exits on the top at $(q,8)$, with $0<q<12$.

The left region is a trapezoid whose horizontal parallel sides have lengths 3 and $q$ and whose height is 8.

Its area is

Equality to 48 forces $q=9$.

The line through $(3,0)$ and $(9,8)$ has slope $8/6=4/3$, giving the stated equation.

Its two regions both have area 48, and every possible boundary exit has been considered, proving uniqueness.

The intercepts give a line equation $x/a+y/b=1$.

Substituting $(2,3)$ yields $2/a+3/b=1$.

Both summands are positive and less than 1, so $a>2$ and $b>3$.

Multiplying by $ab$ gives

Add 6 and group the terms:

This identity can be checked by expanding its left side.

Both factors are positive integers, so their ordered values must be $(1,6),(2,3),(3,2)$, or $(6,1)$.

Restoring the subtracted constants gives exactly $(3,9),(4,6),(5,5),(8,4)$.

Each pair satisfies the original equation, so none is extraneous.

The triangle areas $ab/2$, in the same order, are $27/2,12,25/2$, and 16.

Their smallest value is 12, attained uniquely by intercepts 4 and 6.

The line is $x/4+y/6=1$, equivalently $3x+2y=12$.

The finite comparison proves a global minimum because the factor argument first proved that the four candidates exhaust all possible positive integer intercepts.

Reflect A across the line $y=x+1$.

Its image is $A^{\prime}=(-1,1)$: the midpoint of $AA^{\prime}$ is $(-1/2,1/2)$ on the line, and $AA^{\prime}$ has slope $-1$, perpendicular to the line's slope 1.

Consequently every point P on the line is equally far from A and $A^{\prime}$.

One can verify this directly: for $P=(x,x+1)$, both squared distances are $x^2+(x+1)^2$.

Thus

Every such broken route from $A^{\prime}$ to B is at least the straight distance

Equality occurs exactly when P lies on the straight segment $A^{\prime} B$.

The line through $A^{\prime}$ and B has equation $y=(6-x)/7$.

Intersecting it with $y=x+1$ gives $7x+7=6-x$, hence $x=-1/8$ and $y=7/8$.

This point lies between $A^{\prime}$ and B, so it actually attains the bound.

The two lines have distinct slopes and meet once.

Therefore the minimizing point is unique; extending the allowed location of P beyond the segment does not create another equality case.

For distinct indices $i,j$, equating the line equations gives

Expanding $(j-i)(j+i)$ shows that it equals $j^2-i^2$, so division by $i-j\ne0$ yields $x=-(i+j)$.

Substitution gives $y=-ij$.

The slopes are distinct, so every pair meets exactly once.

No third line $L_k$ can pass through the same point.

If $k\ne i$, its intersection with $L_i$ would have horizontal coordinate $-(i+k)$.

Equality with $-(i+j)$ forces $k=j$.

Hence no three lines are concurrent.

If two different pairs had the same intersection, at least three lines would pass through it, which is impossible.

The number of distinct points is therefore the number of pairs: $6(5)/2=15$.

A vertical line groups pairs having the same sum $i+j$.

For sums 3 through 11, the respective numbers of pairs with $1\leq i<j\leq6$ are $1,1,2,2,3,2,2,1,1$.

This short list follows by choosing the smaller index and requiring the larger to remain at most 6.

The unique maximum is three pairs at sum 7: $(1,6),(2,5),(3,4)$.

They give points $(-7,-6),(-7,-10),(-7,-12)$, all on $x=-7$.

Thus the maximum is 3 and no other vertical line attains it.

The fixed intersection is $P=(4,4)$.

Solving each fixed line with $x+y=k$ gives $U=(2k-12,12-k)$ and

Their midpoint is

The line UV has slope $-1$, while PM lies on $y=x$ with slope 1, so PM is perpendicular to UV and is the triangle's altitude.

Let $d$ be the nonnegative distance between the real numbers $k$ and 8.

The two coordinate differences between U and V each have magnitude $3d$, so

The two coordinate differences between P and M each have magnitude $d/2$, so

The area is consequently

An area of 6 requires $d^2=4$, hence $d=2$.

Thus $k$ is 2 units from 8: $k=6$ or $k=10$.

Substituting into U and V gives the stated vertices; each case has positive area 6.

The derivation allowed k on either side of 8.

The excluded value $k=8$ would make all three lines meet at P and give zero area.

Since a nonnegative distance with square 4 must be 2, no further cases are possible.