Connecting Algebra and Geometry Through Coordinates
Unit 4 of Geometry. 15 questions below, each with the working. Every answer was checked by a second pass before it was published.
Coordinate proofs, slope criteria for parallel and perpendicular lines, distance and partitioning segments, the equation of a circle and a parabola.
How this unit is tested
Start by identifying which tool the problem is really asking for. If two lines or a right angle are involved, you need slope. If lengths, midpoints, or a point along a segment are involved, you need the distance, midpoint, or partition formula. If the shape is a circle, you are really just applying the distance formula to a fixed center. If it is a parabola, you are applying the distance formula twice — once to the focus, once to the directrix — and setting the two distances equal.
For coordinate proofs specifically, the setup matters as much as the algebra. Place the figure so that a vertex sits at the origin and a side lies along an axis whenever possible; this turns messy numbers into variables like $(0,0)$, $(a,0)$, $(0,b)$ and keeps the proof general instead of proving one specific example. Never plug in numbers for a proof that's supposed to hold for every such figure — use letters for the unknown coordinates.
Once the figure is set up, translate the geometric claim into an algebraic one: 'opposite sides parallel' becomes 'equal slopes,' 'right angle' becomes 'slopes multiply to −1,' 'congruent sides' becomes 'equal distances,' and 'bisects' becomes 'shares a midpoint.' Compute each quantity carefully, then state the conclusion in words, tying the algebra back to the geometric claim.
For circles and parabolas, memorize the derivation, not just the final formula. A circle is the set of points a fixed distance from a center — set the distance formula equal to $r$ and square both sides. A parabola is the set of points equidistant from a focus and a directrix — set those two distances equal and simplify. If you forget a formula on the exam, you can rebuild it from this definition in under a minute.
What you have to know
Slope formula
The slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\dfrac{y_2-y_1}{x_2-x_1}$.
Parallel lines criterion
Two distinct non-vertical lines are parallel if and only if they have equal slopes.
Perpendicular lines criterion
Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ (their slopes are opposite reciprocals).
Distance formula
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, from the Pythagorean theorem.
Midpoint and partition formula
The midpoint of a segment from $A(x_1,y_1)$ to $B(x_2,y_2)$ is $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$. More generally, the point that divides $\overline{AB}$ in the ratio $a:b$ from $A$ is $\left(x_1+\dfrac{a}{a+b}(x_2-x_1),\; y_1+\dfrac{a}{a+b}(y_2-y_1)\right)$.
Equation of a circle and a parabola
A circle with center $(h,k)$ and radius $r$ satisfies $(x-h)^2+(y-k)^2=r^2$. A parabola with vertex $(h,k)$ opening vertically satisfies $(x-h)^2=4p(y-k)$, with focus $(h,k+p)$ and directrix $y=k-p$; opening horizontally, $(y-k)^2=4p(x-h)$, with focus $(h+p,k)$ and directrix $x=h-p$.
15 practice questions
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Find the slope of the line through (2, −3) and (7, 7).
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Answer. 2
Apply the slope formula: $m=\dfrac{7-(-3)}{7-2}=\dfrac{10}{5}=2$. The rise is 10 and the run is 5, giving a slope of 2. -
Write the equation of the line through (1, 4) that is parallel to y = 3x − 2.
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Answer. y = 3x + 1
Parallel lines share a slope, so the new line has slope 3. Using point-slope form with (1,4): $y-4=3(x-1)$, which simplifies to $y=3x+1$. -
Are the lines y = 2x + 5 and y = −(1/2)x + 3 parallel, perpendicular, or neither?
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Answer. Perpendicular
Their slopes are 2 and −1/2. Multiplying gives $2\cdot(-\tfrac12)=-1$, which is exactly the perpendicular criterion, so the lines meet at a right angle. -
Find the equation of the line through (4, −1) that is perpendicular to 2x + 3y = 6.
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Answer. y = (3/2)x − 7
Rewriting 2x+3y=6 as $y=-\tfrac23x+2$ gives slope −2/3, so a perpendicular line has slope 3/2. Point-slope form: $y+1=\tfrac32(x-4)$, which simplifies to $y=\tfrac32x-7$. -
Find the distance between (−3, 4) and (5, −2).
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Answer. 10
Using the distance formula: $d=\sqrt{(5-(-3))^2+(-2-4)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10$. -
Find the midpoint of the segment with endpoints (−6, 3) and (2, −9).
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Answer. (−2, −3)
Average the x-coordinates and the y-coordinates: $\left(\dfrac{-6+2}{2},\dfrac{3+(-9)}{2}\right)=(-2,-3)$. -
Point P divides the segment from A(−4, −2) to B(8, 10) in the ratio 1:3, measured from A. Find P.
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Answer. (−1, 1)
The fraction of the way from A to B is $\tfrac{1}{1+3}=\tfrac14$. Then $x=-4+\tfrac14(8-(-4))=-4+3=-1$ and $y=-2+\tfrac14(10-(-2))=-2+3=1$, giving P(−1, 1). -
Show that the triangle with vertices A(0,0), B(4,3), and C(−3,4) is a right triangle, and state where the right angle is.
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Answer. Right angle at A, since slope AB × slope AC = −1
Slope AB = 3/4 and slope AC = 4/(−3) = −4/3. Their product is $\tfrac34\cdot(-\tfrac43)=-1$, so AB ⟂ AC, meaning the right angle is at vertex A. -
Which equation represents a circle with center (−2, 5) and radius 6?
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Answer. (x + 2)^2 + (y − 5)^2 = 36
Standard form is $(x-h)^2+(y-k)^2=r^2$ with (h,k)=(−2,5) and r=6, so $x-(-2)=x+2$, $y-5$, and $r^2=36$. -
Convert x^2 + y^2 − 6x + 4y − 3 = 0 to standard form and state the center and radius.
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Answer. (x − 3)^2 + (y + 2)^2 = 16, center (3, −2), radius 4
Group and complete the square: $(x^2-6x+9)+(y^2+4y+4)=3+9+4$, giving $(x-3)^2+(y+2)^2=16$. So the center is (3,−2) and $r=\sqrt{16}=4$. -
Find the equation of the parabola with vertex (0,0) and focus (0,3).
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Answer. x^2 = 12y
The focus lies above the vertex on the y-axis, so the parabola opens upward with $p=3$. Using $x^2=4py$: $x^2=4(3)y=12y$. -
Find the equation of the parabola with focus (2,0) and directrix x = −2.
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Answer. y^2 = 8x
The vertex is midway between focus and directrix, at (0,0), and it opens rightward since the focus is to the right, with $p=2$. Using $y^2=4px$: $y^2=4(2)x=8x$. -
Triangle ABC has vertices A(0,0), B(8,0), C(4,6). Prove that the segment joining the midpoints of AC and BC is parallel to AB and half its length.
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Answer. Midsegment goes from (2,3) to (6,3): it's horizontal like AB (slope 0), with length 4, which is half of AB's length 8
Midpoint of AC is (2,3) and midpoint of BC is (6,3). This segment has slope 0, matching AB's slope of 0, so they're parallel. Its length is $6-2=4$, while AB has length 8, so the midsegment is exactly half as long. -
Use the distance formula to show that quadrilateral ABCD with A(0,0), B(3,4), C(8,4), D(5,0) is a rhombus.
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Answer. All four sides have length 5
AB = $\sqrt{3^2+4^2}=5$, BC = $\sqrt{5^2+0^2}=5$, CD = $\sqrt{3^2+4^2}=5$, DA = $\sqrt{5^2+0^2}=5$. Since all four sides are congruent, ABCD is a rhombus. -
Point P divides the segment from A(1,1) to B(11,11) in the ratio AP:PB = 3:2. Find P.
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Answer. (7, 7)
The fraction of the way from A to B is $\tfrac{3}{3+2}=\tfrac35$. Then $x=1+\tfrac35(11-1)=1+6=7$ and $y=1+\tfrac35(11-1)=7$, giving P(7,7).
What people get wrong
- Proving a claim with one specific numerical figure instead of general coordinates. If the problem says 'prove for any rectangle,' place the vertices as $(0,0)$, $(a,0)$, $(a,b)$, $(0,b)$ with letters, not a single numeric example — a numeric case only proves that one case.
- Taking the reciprocal of a slope without also flipping its sign when finding a perpendicular slope. Slope $-\tfrac{2}{3}$ becomes $\tfrac{3}{2}$, not $-\tfrac{3}{2}$ or $\tfrac{2}{3}$; always negate AND invert.
- Concluding two segments are parallel just because their slopes match, without checking they aren't the same line. Two identical lines are not 'parallel sides' of a figure — confirm the points are actually distinct segments.
- Applying the partition formula in the wrong direction. The formula $x_1+\tfrac{a}{a+b}(x_2-x_1)$ finds the point from $A$ toward $B$ in ratio $a:b$; swapping which endpoint is $A$ gives a different point on the segment.
- Completing the square incorrectly when converting a circle's general equation to standard form, especially forgetting to add the same constant to both sides or mishandling a leading coefficient other than 1.
- Confusing $p$ with $4p$ in a parabola equation. $4p$ is the coefficient in $(x-h)^2=4p(y-k)$, but $p$ itself is the distance from vertex to focus — solve for $p$ before locating the focus or directrix.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Geometry. Free, and no account needed to start.