Coordinate Geometry: questions with answers

The gaps are DIFFERENCES: $-4 - (1) = -5$ and $-8 - (4) = -12$, so $PQ = \sqrt{ -5^2 + -12^2 } = 13$ units. Adding the coordinates instead gives $5$ units. That route is tempting exactly here, because subtracting a negative looks like adding -- but $-4 - (1)$ is a gap between two positions and $-4 + 1$ is not a length at all. Walking the two gaps gives $17$ units.

The distance from the origin is the distance formula with the second point at $(0, 0)$: $OP = \sqrt{ -8^2 + -15^2 } = \sqrt{ 289 } = 17$ units. Squaring removes the signs, so a point and its reflections are all the same distance out. Adding the two coordinates gives $23$ units, the two-step path along the axes, and the larger coordinate alone, $15$ units, ignores the other one entirely.

The horizontal gap is $5$ and the vertical gap is $12$. These are the two legs of a right triangle whose hypotenuse is $AB$, so they combine through squares and not by addition: $AB = \sqrt{ 5^2 + 12^2 } = \sqrt{ 169 } = 13$ units. Adding the gaps instead gives $17$ units, which is the distance walked along the grid rather than the straight line, and it is always the larger of the two. Stopping at $169$ reports $AB^2$, not $AB$.

The three distances are $AB = 13$, $BC = 15$ and $AC = 14$ units. Three points lie in a line exactly when the two smaller distances add to the largest, because the middle point then uses up the whole of the journey. Here $13 + 14 = 27$ against a largest distance of $15$, so they are not. That the three distances are whole numbers says nothing about it -- plenty of genuine triangles have whole sides. This test is about the SUM, not the kind of number.

Four equal sides make a rhombus, and every square is one, so the sides can never settle it. What separates them is the diagonals. Here $AC^2 = 50$ and $BD^2 = 50$, so the diagonals are equal and it is a square.

The distance of a point from the $x$-axis is its $y$-coordinate. The distance from the $y$-axis is its $x$-coordinate. So for $P(3, 2)$ the distance from the $x$-axis is $2$ units. Reading it the other way round gives $3$ units. Adding the two coordinates gives $5$ units, the length of the two-step path from the origin.

Reflecting in the $y$-axis leaves the ordinate unchanged and reverses the sign of the abscissa. The abscissa measures distance from that mirror, so the point stays in place along the axis but flips to the same distance on the other side. So the image is $(-2, 6)$. Reversing the other coordinate instead gives $(2, -6)$, which is the reflection in the OTHER axis, and reversing both gives $(-2, -6)$, which is the reflection through the origin.

In a parallelogram the DIAGONALS bisect each other, and with the vertices named in order the diagonals are $AC$ and $BD$ -- they join opposite corners. Setting the mid-point of $AC$ equal to the mid-point of $BD$ gives $D = A - B + C = (1, 2)$. Treating $AB$ and $CD$ as the diagonals gives $(-5, -4)$, but $AB$ and $CD$ are opposite SIDES: they are parallel, never meet, and so cannot bisect each other at all.

Take the mid-point of $BC$ first -- the average, $(4, -7)$ -- and then measure from $A$ to it with the distance formula, which gives $13$ units. Starting from the halved DIFFERENCE $(-4, 1)$ instead gives $5$ units, but that point is not on $BC$ at all. And $17$ units comes from adding the horizontal and vertical gaps instead of using the square root, which measures a walk along the grid, not a straight line.

A squared quantity equal to $64$ has two solutions, $8$ and $-8$, and both must be carried through. The positive square root convention applies to the symbol $\sqrt{ }$, not to an unknown whose square is known. So $y + 1 = 8$ or $y + 1 = -8$, giving $y = 7$ and $y = -9$. These are TWO positions of $Q$, one above the fixed point's level and one below, both the same distance away. Stopping at $\pm8$ without undoing the shift leaves the answer as $\pm8$, which solves $y^2 = 64$ rather than the condition actually given.

Equal distances from two fixed points means the point lies on the perpendicular bisector of the segment joining them. That line passes through the mid-point of $AB$ and crosses $AB$ at right angles. The line $AB$ itself is wrong: on $AB$ only the mid-point is equidistant. Every other point on $AB$ is nearer to one end than the other. A line through the mid-point parallel to $AB$ contains the mid-point and nothing else that works.

Setting the ratio to $k : 1$ and solving gives a NEGATIVE $k$, of size $3 / 2$, and a negative $k$ means $P$ is not between $A$ and $B$ at all: it is on the line, beyond $B$. So $P$ does not divide the segment $AB$ -- there is no internal ratio to report. Dropping the minus sign and answering $3 : 2$ names a point that really does divide $AB$ in that ratio, but it is a different point, sitting between $A$ and $B$ where $P$ is not.

A point ON the $y$-axis has $x = 0$, so the crossing point is found by setting its $x$-coordinate to zero and solving for $k$. That gives $k = 2 / 7$, a ratio of $2 : 7$ counted from $A$. An axis is named after the direction it RUNS ALONG, not after the coordinate it kills -- lying on the $y$-axis says nothing about $y$ and everything about $x$. Zeroing the $y$-coordinate instead gives $2 : 1$, which is where the OTHER axis cuts the segment, and $7 : 2$ is this answer counted from $B$.

Dividing $AB$ in the ratio $1 : 4$ means $AP : PB = 1 : 4$, so $P$ is $1$ parts of the way from $A$ towards $B$ out of $5$ parts in all. The section formula pairs the FIRST ratio term with the FAR endpoint's coordinates, giving $P = (-1, 0)$. Attaching the terms to the nearer ends instead gives $(5, 6)$, which is the point for the ratio $4 : 1$ -- the same segment cut the other way round. And $(2, 3)$ is the mid-point, which is the answer only when the two ratio terms are equal.

A ratio compares the two PIECES, $AP : PB$, while the fraction compares one piece to the WHOLE. If $AP$ is $2$ parts out of $7$, then $PB$ is the remaining $7 - 2 = 5$ parts, so $AP : PB = 2 : 5$. Reading the fraction straight off as a ratio gives $2 : 7$. That describes a point $2$ parts along a segment of $2 + 7$ parts -- a different point, always nearer $A$ than the right one.