Free math worksheets

- If the midpoint of the base is the origin, how far left and right must the endpoints be when the total base length is \(2a\)? - A point on the y-axis has x-coordinate \(0\). - Use the altitude length \(h\) to determine the y-coordinate of \(C\).

1. A horizontal base of length \(2a\) centered at the origin has endpoints \(A(-a,0)\) and \(B(a,0)\). 2. The altitude lies on the y-axis and has length \(h\), so the opposite vertex is \(C(0,h)\). 3. This placement builds the symmetry into the coordinates: the base endpoints have opposite x-coordinates, and the y-axis is simultaneously the altitude, median, and symmetry axis.

\(A(-a,0)\), \(B(a,0)\), \(C(0,h)\). The y-axis is the triangle's symmetry axis, so corresponding coordinate differences on the two sides differ only by sign.

- A midpoint is halfway in both coordinate directions. - Average the two x-coordinates and the two y-coordinates separately.

1. Average the x-coordinates: \(\frac{-2+6}{2}=2\). 2. Average the y-coordinates: \(\frac{4+0}{2}=2\). 3. The midpoint is \((2, 2)\).

\((2, 2)\)

- Translate “bisect each other” into a statement about the intersection point. - Ask which coordinate formula locates the halfway point of a segment.

1. To bisect each other, the two diagonals must cross at the midpoint of each diagonal. 2. Therefore, compute the midpoint of each diagonal and check whether the two midpoint coordinates are equal.

B

- Compare the change in y with the change in x. - Equal y-coordinates produce no vertical change.

1. The slope is \(\frac{2-2}{4-(-3)}=\frac{0}{7}=0\). 2. A segment with slope \(0\) is horizontal.

The slope is \(0\), so \(\overline{AB}\) is horizontal.

- Compare all four side lengths rather than only the diagonals. - Group equal sides by whether they share a vertex. - Check the stronger side-length requirement for a rhombus.

1. The squared side lengths are \(AB^2=25\), \(BC^2=13\), \(CD^2=13\), and \(DA^2=25\). 2. Thus \(AB=AD\) and \(BC=CD\), giving two distinct pairs of congruent adjacent sides. 3. Therefore, \(ABCD\) is a kite. 4. Since \(AB^2=25\ne13=BC^2\), not all four sides are congruent. 5. Therefore, \(ABCD\) is not a rhombus.

\(ABCD\) is a kite because \(AB=AD=5\) and \(BC=CD=\sqrt{13}\). It is not a rhombus because the two pairs of side lengths are different.

- Treat the two diagonal properties as separate coordinate checks. - Compare the midpoint of each diagonal first. - Then compare their length expressions without substituting particular values.

1. The midpoint of \(\overline{AC}\) is \(\left(\frac{a}{2}, \frac{b}{2}\right)\). 2. The midpoint of \(\overline{BD}\) is also \(\left(\frac{a}{2}, \frac{b}{2}\right)\). 3. Therefore, the diagonals bisect each other. 4. The diagonal lengths are \(AC=\sqrt{a^2+b^2}\) and \(BD=\sqrt{(-a)^2+b^2}=\sqrt{a^2+b^2}\). 5. Therefore, the diagonals are congruent.

Both diagonals have midpoint \(\left(\frac{a}{2}, \frac{b}{2}\right)\), so they bisect each other. Both have length \(\sqrt{a^2+b^2}\), so they are congruent.

- Test the two sides meeting at each possible right-angle vertex. - Use one coordinate relationship for perpendicularity and another for congruent sides. - Keep the shared vertex of the equal sides in view when naming the triangle.

1. The slope of \(\overline{AB}\) is \(\frac{3-1}{5-1}=\frac{1}{2}\). 2. The slope of \(\overline{BC}\) is \(\frac{7-3}{3-5}=-2\). 3. The slopes are negative reciprocals, so \(AB\perp BC\). Thus \(\angle ABC=90^\circ\). 4. The squared lengths are \(AB^2=(5-1)^2+(3-1)^2=20\) and \(BC^2=(3-5)^2+(7-3)^2=20\). 5. Therefore, \(AB=BC\), so the triangle is isosceles with vertex \(B\). 6. Hence \(\triangle ABC\) is an isosceles right triangle with right angle at \(B\).

\(AB\perp BC\) because their slopes are \(\frac{1}{2}\) and \(-2\). Also, \(AB^2=BC^2=20\). Therefore, \(\triangle ABC\) is an isosceles right triangle with the right angle at \(B\).

- Use slopes to determine which pairs of opposite sides are parallel. - Use the distance formula on the two nonparallel sides. - Match the resulting properties to the definition of an isosceles trapezoid.

1. The slopes of \(\overline{AB}\) and \(\overline{CD}\) are both \(0\), so \(\overline{AB}\parallel\overline{CD}\). 2. The slopes of \(\overline{AD}\) and \(\overline{BC}\) are \(\frac{3}{2}\) and \(-\frac{3}{2}\), so those sides are not parallel. 3. The leg lengths are \(AD=\sqrt{(2)^2+(3)^2}=\sqrt{13}\) and \(BC=\sqrt{(-2)^2+(3)^2}=\sqrt{13}\). 4. The quadrilateral has exactly one pair of parallel opposite sides and congruent legs, so it is an isosceles trapezoid.

Since \(\overline{AB}\parallel\overline{CD}\), \(\overline{AD}\not\parallel\overline{BC}\), and \(AD=BC=\sqrt{13}\), quadrilateral \(ABCD\) is an isosceles trapezoid.

- Read the center and radius from the circle equation. - Find the perpendicular distance from the center to each coordinate axis. - Verify the perpendicular feet in the circle equation.

1. The circle has center \(O(a, a)\) and radius \(a\). 2. The perpendicular distance from \(O\) to the x-axis is \(a\), equal to the radius. The perpendicular foot is \(T_x(a, 0)\). 3. Substitution gives \((a-a)^2+(0-a)^2=a^2\), so \(T_x\) lies on the circle. Therefore, the x-axis is tangent at \((a, 0)\). 4. The perpendicular distance from \(O\) to the y-axis is also \(a\), with foot \(T_y(0, a)\). 5. Substitution gives \((0-a)^2+(a-a)^2=a^2\), so \(T_y\) lies on the circle. Therefore, the y-axis is tangent at \((0, a)\).

The circle is tangent to the x-axis at \((a, 0)\) and to the y-axis at \((0, a)\), because the center is exactly one radius from each axis.

- Write the squared distance between the original points. - Write the squared distance after the coordinates are exchanged. - Compare the two sums without expanding them.

1. The squared length of \(\overline{AB}\) is \(AB^2=(x_2-x_1)^2+(y_2-y_1)^2\). 2. The squared length of \(\overline{A''B''}\) is \(A''B''^2=(y_2-y_1)^2+(x_2-x_1)^2\). 3. Addition is commutative, so \(A''B''^2=AB^2\). 4. Segment lengths are nonnegative, so taking square roots gives \(A''B''=AB\). Therefore, reflection across \(y=x\) preserves distance.

\(A''B''^2=(y_2-y_1)^2+(x_2-x_1)^2=AB^2\), so \(A''B''=AB\). The reflection preserves segment length.

- Rigid motions let you choose a convenient position without changing the rectangle's geometry. - Put one side on an axis and one vertex at the origin. - Prefer coordinates that avoid unnecessary parameters.

1. Place one vertex at the origin and align the sides with the coordinate axes. 2. A convenient placement is \(A(0, 0)\), \(B(w, 0)\), \(C(w, h)\), and \(D(0, h)\). 3. Horizontal and vertical sides then have immediate slopes, side lengths are direct coordinate differences, and both diagonal midpoints are \(\left(\frac{w}{2},\frac{h}{2}\right)\).

One efficient placement is \(A(0, 0)\), \(B(w, 0)\), \(C(w, h)\), and \(D(0, h)\). It aligns the sides with the axes and makes distance, slope, and midpoint calculations simple.

- Recompute each slope with the same order of endpoints in numerator and denominator. - Pay special attention to the sign of the change in \(y\) for \(CD\). - Use the slope product only after both slopes are correct.

1. From the graph, \(A=(-2,1)\), \(B=(2,3)\), \(C=(0,4)\), and \(D=(2,0)\). 2. The slope of \(AB\) is \(\frac{3-1}{2-(-2)}=\frac{1}{2}\). 3. The slope of \(CD\) is \(\frac{0-4}{2-0}=-2\), not \(2\). 4. The correct slopes have product \(\frac{1}{2}\cdot(-2)=-1\). 5. Therefore, lines \(AB\) and \(CD\) are perpendicular. Lorenzo’s error was dropping the negative sign in the vertical change for \(CD\).

The slope of \(CD\) is \(-2\), not \(2\). Since \(\frac{1}{2}\cdot(-2)=-1\), the lines are perpendicular.

- From vertex \(A\), the two known adjacent sides give two displacement vectors. - The opposite vertex is reached by applying both displacements. - Verify the result with the midpoint property of parallelogram diagonals.

1. In a parallelogram, \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\). 2. \(\overrightarrow{AB}=(5,1)\) and \(\overrightarrow{AD}=(1,4)\), so \(\overrightarrow{AC}=(6,5)\). 3. Starting from \(A(-1,1)\), this gives \(C=(5,6)\). 4. The midpoint of \(\overline{AC}\) is \(\left(\frac{-1+5}{2},\frac{1+6}{2}\right)=\left(2,\frac{7}{2}\right)\). 5. The midpoint of \(\overline{BD}\) is \(\left(\frac{4+0}{2},\frac{2+5}{2}\right)=\left(2,\frac{7}{2}\right)\). 6. The diagonals have the same midpoint, verifying the parallelogram.

\(C=(5, 6)\). Both diagonals have midpoint \(\left(2,\frac{7}{2}\right)\).

- Check enough coordinate relationships to distinguish a rhombus from a square. - Compare the side lengths and the diagonal lengths separately. - Ask which verified properties do not, by themselves, force right angles.

1. From the graph, the vertices are \(A(-5,0)\), \(B(0,4)\), \(C(5,0)\), and \(D(0,-4)\). 2. The squared side lengths are \(AB^2=41\), \(BC^2=41\), \(CD^2=41\), and \(DA^2=41\). Thus all four sides are congruent. 3. The midpoint of \(\overline{AC}\) is \((0,0)\), and the midpoint of \(\overline{BD}\) is also \((0,0)\), so the diagonals bisect each other. 4. Diagonal \(\overline{AC}\) is horizontal and diagonal \(\overline{BD}\) is vertical, so they are perpendicular. 5. The diagonal lengths are \(AC=10\) and \(BD=8\), so the diagonals are not congruent. 6. Therefore, \(ABCD\) is a rhombus, but it is not a square. Nadira failed to verify a property that would force right angles, such as congruent diagonals.

\(ABCD\) is a rhombus, not a square. Its four sides are congruent, and its diagonals are perpendicular and bisect each other, but the diagonals have lengths \(10\) and \(8\). The missing square property is congruent diagonals, equivalently four right angles.

- Begin by expressing both midpoint coordinates in terms of \(a\) and \(b\). - Use one coordinate comparison to establish direction and another to compare size. - Simplify the two segment-length expressions before comparing them.

1. The midpoint coordinates are \(M(a, 0)\) and \(N(0, b)\). 2. The slope of \(\overline{MN}\) is \(\frac{b-0}{0-a}=-\frac{b}{a}\). 3. The slope of \(\overline{BC}\) is \(\frac{2b-0}{0-2a}=-\frac{b}{a}\). Equal slopes show that \(\overline{MN}\parallel\overline{BC}\). 4. The lengths satisfy \(MN=\sqrt{a^2+b^2}\) and \(BC=\sqrt{(2a)^2+(2b)^2}=2\sqrt{a^2+b^2}\). 5. Therefore, \(MN=\frac{1}{2}BC\).

The midpoints are \(M(a, 0)\) and \(N(0, b)\). Both \(\overline{MN}\) and \(\overline{BC}\) have slope \(-\frac{b}{a}\), so they are parallel. Their lengths are \(\sqrt{a^2+b^2}\) and \(2\sqrt{a^2+b^2}\), so \(MN=\frac{1}{2}BC\).

- Compare opposite sides using coordinate changes in both directions. - After establishing a broad classification, test the extra properties of its special cases. - Use separate evidence for right angles and for congruent adjacent sides.

1. The displacement from \(A\) to \(B\) is \((5, 1)\), and the displacement from \(D\) to \(C\) is also \((5, 1)\). Thus \(\overline{AB}\parallel\overline{DC}\) and \(AB=DC\). 2. The displacement from \(B\) to \(C\) is \((2, 4)\), and the displacement from \(A\) to \(D\) is also \((2, 4)\). Thus \(\overline{BC}\parallel\overline{AD}\) and \(BC=AD\). 3. Both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram. 4. The slopes of adjacent sides \(\overline{AB}\) and \(\overline{BC}\) are \(\frac{1}{5}\) and \(2\). Their product is \(\frac{2}{5}\), not \(-1\), so the sides are not perpendicular and the parallelogram is not a rectangle. 5. The squared adjacent side lengths are \(AB^2=26\) and \(BC^2=20\). Since adjacent sides are not congruent, the parallelogram is not a rhombus.

\(ABCD\) is a parallelogram. Its opposite sides have matching displacement vectors. It is not a rectangle because adjacent side slopes are not negative reciprocals, and it is not a rhombus because \(AB^2=26\ne20=BC^2\).

- Locate the midpoint of the segment before testing the proposed bisector. - Compare the directions of the segment and the line from its midpoint to \(P\). - Use a second coordinate check to connect the bisector result to the triangle's side lengths.

1. The midpoint of \(\overline{AB}\) is \(M(-1, 3)\). 2. The slope of \(\overline{AB}\) is \(\frac{5-1}{2-(-4)}=\frac{2}{3}\). 3. The slope of \(\overline{MP}\) is \(\frac{0-3}{1-(-1)}=-\frac{3}{2}\). The slope product is \(-1\), so \(MP\perp AB\). 4. Since line \(MP\) passes through midpoint \(M\) and is perpendicular to \(\overline{AB}\), point \(P\) lies on the perpendicular bisector of \(\overline{AB}\). 5. Also, \(PA^2=(1+4)^2+(0-1)^2=26\), \(PB^2=(1-2)^2+(0-5)^2=26\), and \(AB^2=(2+4)^2+(5-1)^2=52\). 6. Therefore, exactly two sides are congruent, so \(\triangle APB\) is isosceles with base \(\overline{AB}\).

The midpoint is \(M(-1, 3)\). The slopes of \(\overline{AB}\) and \(\overline{MP}\) are \(\frac{2}{3}\) and \(-\frac{3}{2}\), so \(MP\perp AB\). Thus \(P\) lies on the perpendicular bisector of \(\overline{AB}\). Since \(PA^2=PB^2=26\) while \(AB^2=52\), \(\triangle APB\) is isosceles with base \(\overline{AB}\).

- Verify first that the claimed point belongs to both geometric objects. - Compare the direction of the radius with the direction of the line. - Connect that direction relationship at a point on the circle to tangency.

1. Substituting \(T(3, 4)\) into the circle equation gives \(3^2+4^2=9+16=25\), so \(T\) lies on the circle. 2. Substituting \(T(3, 4)\) into the line equation gives \(3(3)+4(4)=9+16=25\), so \(T\) lies on \(m\). 3. The slope of radius \(\overline{OT}\) is \(\frac{4}{3}\). 4. Rewriting the line equation gives \(y=-\frac{3}{4}x+\frac{25}{4}\), so the slope of \(m\) is \(-\frac{3}{4}\). 5. The slopes are negative reciprocals, so \(OT\perp m\). 6. A line perpendicular to a radius at its endpoint on the circle is tangent there. Therefore, \(m\) is tangent to the circle at \(T\).

Point \(T\) lies on both the circle and line \(m\). The radius \(OT\) has slope \(\frac{4}{3}\), while \(m\) has slope \(-\frac{3}{4}\), so they are perpendicular. Hence \(m\) is tangent at \(T\).

- Compare all four side-length expressions before using diagonal information. - Express each diagonal length in terms of one parameter. - Identify the extra rhombus property that distinguishes a square.

1. The squared side lengths are \(AB^2=a^2+b^2\), \(BC^2=a^2+b^2\), \(CD^2=a^2+b^2\), and \(DA^2=a^2+b^2\). 2. All four sides are congruent, so \(ABCD\) is a rhombus. 3. Diagonal \(\overline{AC}\) has length \(2a\), and diagonal \(\overline{BD}\) has length \(2b\). 4. A rhombus is a square exactly when its diagonals are congruent, so \(2a=2b\). 5. Therefore, the rhombus is a square exactly when \(a=b\).

a) Each side has length \(\sqrt{a^2+b^2}\), so \(ABCD\) is a rhombus. b) It is a square exactly when \(a=b\), because then diagonals \(AC=2a\) and \(BD=2b\) are congruent.

- Use special center locations in a right triangle. - Write the centroid as the average of the vertex coordinates. - Compare the vectors from the orthocenter to the other two centers.

1. Since \(\angle A=90^\circ\), the orthocenter is \(H=A=(0, 0)\). 2. The centroid is \(G\left(\frac{0+2a+0}{3}, \frac{0+0+2b}{3}\right)=\left(\frac{2a}{3}, \frac{2b}{3}\right)\). 3. The circumcenter is the midpoint of hypotenuse \(\overline{BC}\), so \(O=(a, b)\). 4. Vector \(\overrightarrow{HG}=\left(\frac{2a}{3}, \frac{2b}{3}\right)=\frac{2}{3}(a, b)=\frac{2}{3}\overrightarrow{HO}\). 5. Therefore, \(H\), \(G\), and \(O\) are collinear. Also, \(HG=\frac{2}{3}HO\) and \(GO=\frac{1}{3}HO\), so \(HG:GO=2:1\).

The points are \(H(0, 0)\), \(G\left(\frac{2a}{3}, \frac{2b}{3}\right)\), and \(O(a, b)\). They are collinear, with \(HG:GO=2:1\).

- Apply the distance formula to each diagonal separately. - Simplify the horizontal changes carefully, including their signs. - Compare the squared lengths before taking square roots.

1. For diagonal \(\overline{AC}\), \(AC^2=(b-(-a))^2+(h-0)^2=(a+b)^2+h^2\). 2. For diagonal \(\overline{BD}\), \(BD^2=(-b-a)^2+(h-0)^2=(a+b)^2+h^2\). 3. Since \(AC^2=BD^2\) and both lengths are positive, \(AC=BD\). 4. Therefore, the diagonals of the isosceles trapezoid are congruent.

The squared diagonal lengths are \(AC^2=(a+b)^2+h^2\) and \(BD^2=(-a-b)^2+h^2=(a+b)^2+h^2\). Therefore, \(AC=BD=\sqrt{(a+b)^2+h^2}\), so the diagonals are congruent.

- Find the midpoint coordinates of the hypotenuse first. - Compare squared distances from that midpoint to each vertex. - Look for sign changes that disappear when squared.

1. The midpoint of \(\overline{BC}\) is \(M(a, b)\). 2. \(MA^2=(a-0)^2+(b-0)^2=a^2+b^2\). 3. \(MB^2=(a-2a)^2+(b-0)^2=a^2+b^2\). 4. \(MC^2=(a-0)^2+(b-2b)^2=a^2+b^2\). 5. Since the three squared distances are equal, \(MA=MB=MC\). Thus \(M\) is equidistant from all three vertices.

\(M=(a, b)\) and \(MA=MB=MC=\sqrt{a^2+b^2}\). Therefore, the midpoint of the hypotenuse is equidistant from the three vertices.

- Find the slope of each side before writing the corresponding altitude. - Start with the altitude to the horizontal side. - Verify that the intersection of two altitudes also lies on the third.

1. Since \(\overline{AB}\) is horizontal, the altitude from \(C\) is the vertical line \(x=2\). 2. The slope of \(\overline{BC}\) is \(-1\), so the altitude from \(A\) has slope \(1\) and equation \(y=x\). 3. These two altitudes intersect at \((2, 2)\). 4. The slope of \(\overline{AC}\) is \(2\), so the altitude from \(B\) has slope \(-\frac{1}{2}\) and equation \(y=-\frac{1}{2}(x-6)\). 5. Substituting \((2, 2)\) gives \(2=-\frac{1}{2}(2-6)\), so the point lies on the third altitude. 6. Therefore, all three altitudes are concurrent at \((2, 2)\).

The three altitudes intersect at \((2, 2)\), the orthocenter of the triangle.

- Write a general point on the proposed ray. - Express its perpendicular distances to the coordinate axes. - Prove the reverse direction using a general point in the angle.

1. A point on the ray has coordinates \(P(t, t)\) with \(t\ge0\). 2. Its perpendicular distance to the x-axis is \(t\), and its perpendicular distance to the y-axis is also \(t\). Thus every point on the ray is equidistant from the two axes. 3. Conversely, let \(Q(a, b)\) be a point in the closed first quadrant that is equidistant from the axes. 4. Its distances to the x-axis and y-axis are \(b\) and \(a\), respectively. Equal distances give \(a=b\). 5. Therefore, \(Q\) lies on \(y=x\). Hence the ray is exactly the angle-bisector locus.

Points \((t, t)\) are distance \(t\) from both axes, and any point \((a, b)\) in the closed first quadrant that is equidistant from the axes must satisfy \(a=b\). Therefore, the ray \(y=x\) is exactly the internal angle bisector.

- Compare all four side lengths using squared distances. - Use slopes of one pair of adjacent sides to test for a right angle. - Match the verified properties to a square characterization.

1. The squared side lengths are \(AB^2=3^2+2^2=13\), \(BC^2=2^2+(-3)^2=13\), \(CD^2=(-3)^2+(-2)^2=13\), and \(DA^2=(-2)^2+3^2=13\). Thus all four sides are congruent. 2. The slope of \(\overline{AB}\) is \(\frac{2}{3}\), and the slope of \(\overline{BC}\) is \(-\frac{3}{2}\). 3. Their product is \(-1\), so \(\overline{AB}\perp\overline{BC}\). 4. A quadrilateral with four congruent sides and a right angle is a square.

All four sides have length \(\sqrt{13}\), and adjacent sides \(\overline{AB}\) and \(\overline{BC}\) are perpendicular. Therefore, \(ABCD\) is a square.

- Use a diagonal property of parallelograms rather than side slopes. - Find the midpoint of the diagonal whose endpoints are known. - Set the midpoint of the other diagonal equal to that point.

1. In a parallelogram, the diagonals bisect each other. The midpoint of \(\overline{AC}\) is \(\left(\frac{-2+5}{2}, \frac{1+(-1)}{2}\right)=\left(\frac{3}{2}, 0\right)\). 2. Let \(D=(x, y)\). The midpoint of \(\overline{BD}\) must also be \(\left(\frac{3}{2}, 0\right)\). 3. Therefore, \(\frac{3+x}{2}=\frac{3}{2}\), so \(x=0\), and \(\frac{4+y}{2}=0\), so \(y=-4\). 4. Thus \(D=(0, -4)\). The midpoint of \(\overline{BD}\) is \(\left(\frac{3+0}{2}, \frac{4+(-4)}{2}\right)=\left(\frac{3}{2}, 0\right)\), matching the midpoint of \(\overline{AC}\). 5. Since the diagonals bisect each other, the four points form a parallelogram.

\(D=(0, -4)\). Both diagonals have midpoint \(\left(\frac{3}{2}, 0\right)\), verifying the parallelogram.

- Find the midpoint and slope of the original segment. - Use the negative reciprocal slope for the perpendicular line. - Verify the point both in the line equation and with squared distances.

1. The midpoint of \(\overline{AB}\) is \(M\left(\frac{-2+4}{2}, \frac{3+(-1)}{2}\right)=(1, 1)\). 2. The slope of \(\overline{AB}\) is \(\frac{-1-3}{4-(-2)}=-\frac{2}{3}\), so the perpendicular slope is \(\frac{3}{2}\). 3. An equation of the perpendicular bisector is \(y-1=\frac{3}{2}(x-1)\). 4. For \(P(3, 4)\), the equation gives \(4-1=\frac{3}{2}(3-1)=3\), so \(P\) lies on the line. 5. Also, \(PA^2=(3-(-2))^2+(4-3)^2=26\) and \(PB^2=(3-4)^2+(4-(-1))^2=26\). Thus \(PA=PB=\sqrt{26}\).

a) \(y-1=\frac{3}{2}(x-1)\). b) Point \(P(3, 4)\) satisfies the equation and has \(PA=PB=\sqrt{26}\), so it lies on the perpendicular bisector.

- Write the slopes from \(P\) to both diameter endpoints. - Use the circle equation to replace \(y^2\). - Simplify the product of the slopes.

1. Because \(P\) is distinct from \(A\) and \(B\), \(x\ne-3\) and \(x\ne3\). The slope of \(\overline{PA}\) is \(\frac{y}{x+3}\), and the slope of \(\overline{PB}\) is \(\frac{y}{x-3}\). 2. Their product is \(\frac{y^2}{(x+3)(x-3)}=\frac{y^2}{x^2-9}\). 3. Since \(x^2+y^2=9\), \(y^2=9-x^2\). 4. Therefore, the slope product is \(\frac{9-x^2}{x^2-9}=-1\). 5. The two segments are perpendicular, so \(\angle APB=90^\circ\).

The slopes of \(\overline{PA}\) and \(\overline{PB}\) have product \(-1\), so \(PA\perp PB\) and \(\angle APB=90^\circ\).

- Find two side midpoints and perpendicular slopes. - Intersect two perpendicular bisectors to locate the center. - Use the distance from the center to one vertex for the radius.

1. The perpendicular bisector of \(\overline{AB}\) is \(x=4\). 2. Segment \(\overline{AC}\) has midpoint \((1, 3)\) and slope \(3\), so its perpendicular bisector has slope \(-\frac{1}{3}\) and equation \(y-3=-\frac{1}{3}(x-1)\). 3. Substituting \(x=4\) gives \(y-3=-1\), so \(y=2\). The circumcenter is \(O(4, 2)\). 4. The squared radius is \(OA^2=4^2+2^2=20\). 5. The circle equation is \((x-4)^2+(y-2)^2=20\). 6. Also, \(OB^2=(-4)^2+2^2=20\) and \(OC^2=(-2)^2+4^2=20\), confirming that all three vertices lie on the circle.

The circumcenter is \((4, 2)\), and the circumscribed circle is \((x-4)^2+(y-2)^2=20\).

- Write the image coordinates of two general points. - Compare the two squared-distance expressions. - Notice that the coordinate differences are exchanged and one sign changes.

1. Let \(P(x_1, y_1)\) and \(Q(x_2, y_2)\). Their images are \(P''(-y_1, x_1)\) and \(Q''(-y_2, x_2)\). 2. The squared original distance is \(PQ^2=(x_2-x_1)^2+(y_2-y_1)^2\). 3. The squared image distance is \(P''Q''^2=((-y_2)-(-y_1))^2+(x_2-x_1)^2\). 4. This simplifies to \((y_2-y_1)^2+(x_2-x_1)^2=PQ^2\). 5. Since distances are nonnegative, \(P''Q''=PQ\). Therefore, the rotation preserves distance.

\(P''Q''^2=(y_2-y_1)^2+(x_2-x_1)^2=PQ^2\), so \(P''Q''=PQ\). The rotation preserves every distance.

- Use the given ratio to find the coordinates of \(D\). - Determine the equation of the line through \(A\) and \(D\). - Compare that line's direction with the two coordinate-axis sides of the triangle.

1. By the section formula, \(D=\left(\frac{4(6)+3(0)}{7}, \frac{4(0)+3(8)}{7}\right)=\left(\frac{24}{7}, \frac{24}{7}\right)\). 2. Therefore, line \(AD\) has slope \(1\) and equation \(y=x\). 3. Rays \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) lie on the positive x-axis and positive y-axis, so \(\angle A=90^\circ\). 4. Line \(y=x\) makes a \(45^\circ\) angle with each positive coordinate axis. 5. Thus \(\angle BAD\cong\angle DAC\), so \(\overline{AD}\) bisects \(\angle A\). 6. Also, \(AB:AC=6:8=3:4=BD:DC\), consistent with the Angle Bisector Theorem.

\(D=\left(\frac{24}{7}, \frac{24}{7}\right)\), so \(\overline{AD}\) lies on \(y=x\), the \(45^\circ\) bisector of the coordinate axes. Therefore, \(\overline{AD}\) bisects \(\angle A\).

- Verify each proposed point in the circle equation. - Compare the slope of each radius with the slope of the segment from \(P\). - Use the distance formula for one tangent segment and symmetry for the other.

1. For either point, \(x^2+y^2=\frac{625+3600}{169}=25\), so both points lie on the circle. 2. The slope of \(\overline{OT_1}\) is \(\frac{60}{25}=\frac{12}{5}\). 3. The slope of \(\overline{PT_1}\) is \(\frac{60}{25-169}=-\frac{5}{12}\). Their product is \(-1\), so \(OT_1\perp PT_1\), making \(\overline{PT_1}\) tangent. 4. The reflected coordinates of \(T_2\) produce slopes \(-\frac{12}{5}\) and \(\frac{5}{12}\), so \(OT_2\perp PT_2\) as well. 5. \(PT_1^2=\left(13-\frac{25}{13}\right)^2+\left(0-\frac{60}{13}\right)^2=144\), so \(PT_1=12\). 6. By the same calculation or reflection symmetry, \(PT_2=12\).

Both segments are tangent because each is perpendicular to the radius at its circle point. Their lengths are \(PT_1=PT_2=12\).

- Compare the slopes from the three equations. - Use the distance formula for parallel lines in standard form. - Compare each distance from \(n\) with the full distance between the outer lines.

1. Rewriting each equation in slope-intercept form gives slope \(\frac{3}{4}\), so the three distinct lines are parallel. 2. The distance between \(\ell\) and \(m\) is \(\frac{|12-(-8)|}{\sqrt{3^2+(-4)^2}}=\frac{20}{5}=4\). 3. The distance between \(n\) and \(\ell\) is \(\frac{|2-12|}{5}=2\). 4. The distance between \(n\) and \(m\) is \(\frac{|2-(-8)|}{5}=2\). 5. Therefore, \(n\) is parallel to both lines and is equidistant from them, so it lies halfway between \(\ell\) and \(m\).

All three lines have slope \(\frac{3}{4}\). Line \(n\) is \(2\) units from each of \(\ell\) and \(m\), while \(\ell\) and \(m\) are \(4\) units apart. Thus \(n\) lies halfway between them.

- Write all four squared lengths from the coordinates. - Add the two side-length expressions before expanding fully. - Compare the result with twice the sum involving the median and half the base.

1. The squared side lengths are \(CA^2=(u+a)^2+v^2\) and \(CB^2=(u-a)^2+v^2\). 2. Adding gives \(CA^2+CB^2=(u+a)^2+(u-a)^2+2v^2=2u^2+2a^2+2v^2\). 3. The squared median length is \(CM^2=u^2+v^2\), and \(AM^2=a^2\). 4. Therefore, \(2(CM^2+AM^2)=2(u^2+v^2+a^2)=2u^2+2a^2+2v^2\). 5. The two expressions are equal, proving \(CA^2+CB^2=2(CM^2+AM^2)\).

Both sides simplify to \(2u^2+2a^2+2v^2\), so \(CA^2+CB^2=2(CM^2+AM^2)\).

- Apply the dilation rule to the original center. - Express the image coordinates in terms of the original coordinates. - Compare the squared distance from an image point to the image center with the original radius equation.

1. The center maps as \(C(3, -2)\mapsto C''(-6, 4)\). 2. Let \(P(x, y)\) be any point on the original circle, so \((x-3)^2+(y+2)^2=16\). 3. Its image is \(P''(X, Y)=(-2x, -2y)\). Then \(X+6=-2(x-3)\) and \(Y-4=-2(y+2)\). 4. Therefore, \((X+6)^2+(Y-4)^2=4[(x-3)^2+(y+2)^2]=4\cdot16=64\). 5. Every image point lies on the circle centered at \((-6, 4)\) with radius \(8\). Conversely, reversing the dilation maps every point of that circle back to the original circle.

The image is the circle centered at \((-6, 4)\) with radius \(8\). Its equation is \((x+6)^2+(y-4)^2=64\).

- Write vectors for both sides from \(A\) and both diagonals. - Add the two squared diagonal lengths before simplifying. - Watch how the mixed terms cancel.

1. The squared side lengths are \(AB^2=u^2+v^2\) and \(AD^2=r^2+s^2\). 2. Diagonal \(\overline{AC}\) has vector \((u+r, v+s)\), so \(AC^2=(u+r)^2+(v+s)^2\). 3. Diagonal \(\overline{BD}\) has vector \((r-u, s-v)\), so \(BD^2=(r-u)^2+(s-v)^2\). 4. Adding the diagonal expressions cancels the cross terms: \(AC^2+BD^2=2u^2+2v^2+2r^2+2s^2\). 5. This equals \(2[(u^2+v^2)+(r^2+s^2)]=2(AB^2+AD^2)\).

Expanding the two diagonal-length expressions cancels their cross terms and gives \(AC^2+BD^2=2u^2+2v^2+2r^2+2s^2=2(AB^2+AD^2)\).

- Write both squared distances before taking their difference. - Expand only enough to see which terms cancel. - Check the converse by substituting the resulting coordinate condition back into the equation.

1. The squared distances are \(PA^2=(x+3)^2+y^2\) and \(PB^2=(x-3)^2+y^2\). 2. Subtracting gives \(PA^2-PB^2=(x+3)^2-(x-3)^2=12x\). 3. The condition \(PA^2-PB^2=24\) becomes \(12x=24\), so \(x=2\). 4. Every point with \(x=2\) satisfies the original equation because substituting \(x=2\) makes the difference of squared distances equal to \(24\), regardless of \(y\).

The locus is the vertical line \(x=2\).

- Find the slope of the given line before writing the perpendicular line. - The desired point satisfies two line equations at once. - Verify both incidence and perpendicularity.

1. Rewrite \(\ell\) as \(y=2x-3\), so its slope is \(2\). 2. A perpendicular line has slope \(-\frac{1}{2}\). The perpendicular through \(P\) is \(y-1=-\frac{1}{2}(x-5)\). 3. Solving this equation with \(y=2x-3\) gives \(x=\frac{13}{5}\) and \(y=\frac{11}{5}\). 4. Substitution verifies \(2\left(\frac{13}{5}\right)-\frac{11}{5}=3\), so \(H\) lies on \(\ell\). 5. The slope of \(PH\) is \(-\frac{1}{2}\), whose product with the slope \(2\) of \(\ell\) is \(-1\). Therefore, \(PH\perp\ell\).

The perpendicular foot is \(H\left(\frac{13}{5}, \frac{11}{5}\right)\).

- Write both squared distances before adding them. - Look for cancellation of the linear terms. - Identify the standard geometric figure represented by the simplified equation.

1. The squared distances are \(PA^2=(x+2)^2+y^2\) and \(PB^2=(x-2)^2+y^2\). 2. Adding gives \(PA^2+PB^2=2x^2+2y^2+8\). 3. Setting this equal to \(20\) gives \(2x^2+2y^2=12\), so \(x^2+y^2=6\). 4. Every point on the circle \(x^2+y^2=6\) makes the original sum equal to \(20\), so the implication works in both directions.

The locus is the circle centered at the origin with radius \(\sqrt{6}\), whose equation is \(x^2+y^2=6\).

- Write each centroid as the average of its three vertices. - Subtract the centroid coordinates before simplifying. - Interpret the resulting vector multiple geometrically.

1. Let \(A(x_1, y_1)\), \(B(x_2, y_2)\), \(C(x_3, y_3)\), and \(D(x_4, y_4)\). 2. The centroids are \(G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\) and \(H\left(\frac{x_1+x_2+x_4}{3}, \frac{y_1+y_2+y_4}{3}\right)\). 3. Therefore, \(\overrightarrow{GH}=\left(\frac{x_4-x_3}{3}, \frac{y_4-y_3}{3}\right)=\frac{1}{3}\overrightarrow{CD}\). 4. Since one vector is a positive scalar multiple of the other, the segments are parallel and \(GH=\frac{1}{3}CD\).

The centroid difference vector is \(\overrightarrow{GH}=\frac{1}{3}\overrightarrow{CD}\). Hence \(GH\parallel CD\) and \(GH=\frac{1}{3}CD\).

- Replace the distance ratio with an equation involving squared distances. - Expand and gather the \(x\)- and \(y\)-terms. - Complete the square to identify the locus.

1. Square the distance condition to obtain \((x+3)^2+y^2=4[(x-3)^2+y^2]\). 2. Expanding and simplifying gives \(x^2-10x+y^2+9=0\). 3. Completing the square gives \((x-5)^2+y^2=16\). 4. Every point on this circle satisfies the squared distance equation. Since distances are nonnegative, the squared equation is equivalent to \(PA=2PB\).

The locus is the circle centered at \((5, 0)\) with radius \(4\), whose equation is \((x-5)^2+y^2=16\).

- Use \((A, B)\) as a direction perpendicular to the line. - Move from \(P\) by an unknown multiple of that normal vector. - Choose the multiple so the new point satisfies the line equation.

1. A normal vector to \(\ell\) is \((A, B)\). 2. Let \(k=\frac{Ax_0+By_0+C}{A^2+B^2}\), and define \(H=(x_0-kA, y_0-kB)\). 3. Substitution gives \(A(x_0-kA)+B(y_0-kB)+C=Ax_0+By_0+C-k(A^2+B^2)=0\), so \(H\) lies on \(\ell\). 4. Vector \(\overrightarrow{PH}=(-kA, -kB)\) is parallel to the normal vector, so \(PH\perp\ell\). 5. Its length is \(PH=|k|\sqrt{A^2+B^2}=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\).

Let \(k=\frac{Ax_0+By_0+C}{A^2+B^2}\) and \(H=(x_0-kA, y_0-kB)\). Then \(H\) lies on \(\ell\), and \(\overrightarrow{PH}=(-kA, -kB)\) is perpendicular to \(\ell\). Therefore, \(PH=|k|\sqrt{A^2+B^2}=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\).

- Write the coordinates of all four midpoints first. - Compare the horizontal and vertical changes along opposite sides of the midpoint quadrilateral. - Use a parallelogram test based on opposite sides.

1. The midpoint coordinates are \(E\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\), \(F\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\), \(G\left(\frac{x_3+x_4}{2}, \frac{y_3+y_4}{2}\right)\), and \(H\left(\frac{x_4+x_1}{2}, \frac{y_4+y_1}{2}\right)\). 2. The displacement vector from \(E\) to \(F\) is \(\overrightarrow{EF}=\left(\frac{x_3-x_1}{2}, \frac{y_3-y_1}{2}\right)\). 3. The displacement vector from \(H\) to \(G\) is also \(\overrightarrow{HG}=\left(\frac{x_3-x_1}{2}, \frac{y_3-y_1}{2}\right)\). Thus \(\overline{EF}\) and \(\overline{HG}\) are parallel and congruent. 4. Similarly, \(\overrightarrow{FG}=\left(\frac{x_4-x_2}{2}, \frac{y_4-y_2}{2}\right)=\overrightarrow{EH}\). 5. Therefore, both pairs of opposite sides are parallel and congruent, so \(EFGH\) is a parallelogram.

The opposite sides of \(EFGH\) have equal displacement vectors: \(\overrightarrow{EF}=\overrightarrow{HG}=\left(\frac{x_3-x_1}{2}, \frac{y_3-y_1}{2}\right)\) and \(\overrightarrow{FG}=\overrightarrow{EH}=\left(\frac{x_4-x_2}{2}, \frac{y_4-y_2}{2}\right)\). Therefore, \(EFGH\) is a parallelogram.

- Write the midpoint coordinates of \(M\). - Compare the displacement vectors from \(A\) to \(M\) and from \(A\) to \(G\). - Interpret a scalar multiple of one directed segment by another.

1. The midpoint is \(M\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\). 2. The displacement vector from \(A\) to \(M\) is \(\overrightarrow{AM}=\left(\frac{x_2+x_3-2x_1}{2}, \frac{y_2+y_3-2y_1}{2}\right)\). 3. The displacement vector from \(A\) to \(G\) is \(\overrightarrow{AG}=\left(\frac{x_2+x_3-2x_1}{3}, \frac{y_2+y_3-2y_1}{3}\right)\). 4. Therefore, \(\overrightarrow{AG}=\frac{2}{3}\overrightarrow{AM}\). This shows that \(A\), \(G\), and \(M\) are collinear and that \(AG=\frac{2}{3}AM\). 5. Then \(GM=AM-AG=\frac{1}{3}AM\), so \(AG:GM=2:1\).

The midpoint is \(M\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\). The displacement vectors satisfy \(\overrightarrow{AG}=\left(\frac{x_2+x_3-2x_1}{3}, \frac{y_2+y_3-2y_1}{3}\right)=\frac{2}{3}\overrightarrow{AM}\). Thus \(A\), \(G\), and \(M\) are collinear, \(AG=\frac{2}{3}AM\), and \(GM=\frac{1}{3}AM\). Therefore, \(AG:GM=2:1\).

- Find the three side midpoints first. - Intersect two median equations, then test the point on the third median. - Use convenient bases on the coordinate axes and a determinant for the remaining pieces.

1. The midpoints are \(D(3, 3)\), \(E(0, 3)\), and \(F(3, 0)\). 2. Median \(AD\) lies on \(y=x\). Median \(BE\) has equation \(y=-\frac{1}{2}x+3\). Their intersection is \(G(2, 2)\). 3. The line through \(C(0, 6)\) and \(F(3, 0)\) has equation \(y=-2x+6\), and \(G(2, 2)\) satisfies it. Thus all three medians are concurrent at \(G\). 4. Using \(\frac{1}{2}(\text{base})(\text{height})\), \(\operatorname{Area}(\triangle AGF)=\operatorname{Area}(\triangle BGF)=3\) because \(AF=BF=3\) and the height from \(G\) to \(\overline{AB}\) is \(2\). 5. Similarly, \(\operatorname{Area}(\triangle AGE)=\operatorname{Area}(\triangle CGE)=3\) because \(AE=CE=3\) and the distance from \(G\) to line \(AC\) is \(2\). 6. The determinant area formula gives \(\operatorname{Area}(\triangle BGD)=\frac{1}{2}|(-4)(3)-2(-3)|=3\) and \(\operatorname{Area}(\triangle CGD)=\frac{1}{2}|2(-3)-(-4)(3)|=3\). 7. Therefore, the medians meet at \(G(2, 2)\) and form six triangles, each with area \(3\) square units.

The three medians are concurrent at \(G(2, 2)\). Each of the six smaller triangles has area \(3\) square units.

- Write squared lengths for two consecutive sides. - Subtract the expressions instead of expanding each completely on its own. - Connect the resulting dot-product term to the perpendicular diagonals.

1. The squared length of \(\overline{AB}\) is \(AB^2=(p-r)^2+(q-s)^2\). 2. The squared length of \(\overline{BC}\) is \(BC^2=(-p-r)^2+(-q-s)^2=(p+r)^2+(q+s)^2\). 3. Subtracting gives \(BC^2-AB^2=4pr+4qs=4(pr+qs)\). 4. Since the diagonals are perpendicular, \(pr+qs=0\). Therefore, \(BC^2-AB^2=0\), so \(AB=BC\). 5. A parallelogram with congruent consecutive sides is a rhombus. Hence \(ABCD\) is a rhombus.

The perpendicular-diagonal condition gives \(pr+qs=0\). Since \(BC^2-AB^2=4(pr+qs)=0\), consecutive sides are congruent. Therefore, the parallelogram is a rhombus.

- Use the line's normal vector to locate the perpendicular foot from the origin. - Parameterize the chord line using a direction vector perpendicular to that normal vector. - Substitute the parameterization into the circle equation and compare the two parameter values.

1. The circle's center is \(O(0, 0)\). A normal vector to the line is \((3, 4)\), so the perpendicular from \(O\) follows the direction \((3, 4)\). 2. A point on this perpendicular has the form \((3t, 4t)\). Substituting into \(3x+4y=10\) gives \(25t=10\), so the perpendicular foot is \(M\left(\frac{6}{5}, \frac{8}{5}\right)\). 3. A direction vector for the chord line is \((-4, 3)\). Since \(\left(\frac{6}{5}, \frac{8}{5}\right)\cdot(-4, 3)=0\), \(\overline{OM}\) is perpendicular to the chord. 4. Parameterize the chord line by \((x, y)=\left(\frac{6}{5}, \frac{8}{5}\right)+t(-4, 3)\). 5. Substitution into the circle equation gives \(4+25t^2=25\), so the two intersection parameters are \(t=\pm\frac{\sqrt{21}}{5}\). 6. The two parameters are opposites, so their points have midpoint \(M\left(\frac{6}{5}, \frac{8}{5}\right)\). Their parameter difference is \(\frac{2\sqrt{21}}{5}\), and \(\|(-4, 3)\|=5\), so the chord length is \(2\sqrt{21}\).

The chord's midpoint is \(\left(\frac{6}{5}, \frac{8}{5}\right)\), and its length is \(2\sqrt{21}\). The center-to-midpoint direction is parallel to \((3, 4)\), which is perpendicular to the chord direction \((-4, 3)\).

- Use slopes to find two altitudes and their intersection. - Reflect the orthocenter by changing the sign of its y-coordinate. - Find the circle equation through the three triangle vertices and test the reflected point.

1. Since \(AB\) is horizontal, the altitude from \(C\) is \(x=2\). 2. Line \(AC\) has slope \(2\), so the altitude from \(B\) has slope \(-\frac{1}{2}\) and equation \(y=-\frac{1}{2}x+3\). 3. Intersecting this altitude with \(x=2\) gives \(H(2, 2)\). Reflecting across \(AB\), which lies on the x-axis, gives \(H''(2, -2)\). 4. A circle through \(A(0, 0)\) has equation \(x^2+y^2+Dx+Ey=0\). 5. Substituting \(B(6, 0)\) gives \(D=-6\), and substituting \(C(2, 4)\) gives \(E=-2\). Thus the circumcircle is \(x^2+y^2-6x-2y=0\). 6. Substituting \(H''(2, -2)\) gives \(4+4-12+4=0\). Therefore, \(H''\) lies on the circumcircle.

The orthocenter is \(H(2, 2)\), so its reflection across \(AB\) is \(H''(2, -2)\). This point satisfies the circumcircle equation \(x^2+y^2-6x-2y=0\), so \(H''\) lies on the circumcircle.

- Use a rigid motion to put one base on the x-axis; the other base must then lie on a horizontal line. - Keep the horizontal offset of the second base as a variable rather than assuming an isosceles trapezoid. - Parameterize both diagonals and use the equal y-coordinates at their intersection before forming segment ratios.

1. A translation and rotation preserve lengths, parallelism, intersection ratios, and the shape of the trapezoid. Therefore, place \(A(0, 0)\), \(B(b, 0)\), \(D(d, h)\), and \(C(d+c, h)\), where \(h>0\). This puts the parallel bases on horizontal lines without loss of generality. 2. A point on \(\overline{AC}\) has coordinates \((t(d+c), th)\), where \(0<t<1\). 3. A point on \(\overline{BD}\) has coordinates \((b+s(d-b), sh)\), where \(0<s<1\). 4. At the intersection \(P\), the y-coordinates give \(t=s\). 5. Equating x-coordinates gives \(t(d+c)=b+t(d-b)\), so \(t(b+c)=b\) and \(t=\frac{b}{b+c}\). 6. Therefore, \(\frac{AP}{PC}=\frac{t}{1-t}=\frac{b}{c}\). 7. The same parameter \(s=t\) divides \(\overline{BD}\), so \(\frac{BP}{PD}=\frac{s}{1-s}=\frac{b}{c}\).

Place \(A(0, 0)\), \(B(b, 0)\), \(D(d, h)\), and \(C(d+c, h)\); translations and rotations preserve the required ratios, so this loses no generality. Parameterizing the diagonals gives \(t=s=\frac{b}{b+c}\). Hence \(\frac{AP}{PC}=\frac{BP}{PD}=\frac{t}{1-t}=\frac{b}{c}\).