Free math worksheets

- Compare the x-coordinates of the two points. - Think about the direction of a line whose x-coordinate never changes. - Recall how vertical lines relate to the y-axis.

1. Points \(C\) and \(D\) have the same x-coordinate, \(120\). 2. A line through two distinct points with the same x-coordinate is vertical and has equation \(x=120\). 3. Every vertical line is parallel to the y-axis.

Line \(g\) is parallel to the y-axis because both points have x-coordinate \(120\), so \(g\) is the vertical line \(x=120\).

- Compare the y-coordinates of each pair of points. - A horizontal line has a constant y-coordinate. - Determine which two points lie at the same height.

1. Points \(P\) and \(Q\) have the same y-coordinate, \(30\). 2. A line through two distinct points with the same y-coordinate is horizontal. 3. Therefore, \(\overleftrightarrow{PQ}\) is parallel to the x-axis.

The pair is \(P\) and \(Q\). Both have y-coordinate \(30\), so \(\overleftrightarrow{PQ}\) is horizontal and parallel to the x-axis.

- Compare the slopes of the four boundary lines. - Equal slopes tell you that two distinct lines are parallel, and horizontal lines have slope \(0\). - Identify the quadrilateral property determined by two pairs of parallel opposite sides.

1. Lines \(g_1\) and \(g_2\) both have slope \(0.5\), so they are parallel. 2. Lines \(g_3\) and \(g_4\) are horizontal, so both have slope \(0\) and are parallel. 3. The enclosed quadrilateral therefore has two pairs of opposite sides that are parallel, so it is a parallelogram.

A parallelogram. The lines form two pairs of parallel sides: \(g_1\parallel g_2\) and \(g_3\parallel g_4\).

- How do you calculate the slope of a line from two known points? - What must be true about the slopes of two parallel lines? - Can you rewrite the equation of \(h\) so its slope is visible directly? - What coordinates does the origin have?

1. The slope of \(g\) is \(m_g = \frac{5 - 0}{2 - 0} = \frac{5}{2} = 2.5\). 2. Rewrite \(3x - y = 4\) as \(y = 3x - 4\), so \(m_h = 3\). 3. Since \(2.5 \neq 3\), the slopes are different, so the lines are not parallel.

The lines are not parallel because their slopes are \(2.5\) and \(3\).

- Find the slope of each line. - Lines with equal slopes are parallel. - Horizontal and vertical lines are perpendicular.

1. Lines \(\overleftrightarrow{AB}\) and \(\overleftrightarrow{CD}\) are horizontal, so both have slope \(0\). Therefore, \(\overleftrightarrow{AB} \parallel \overleftrightarrow{CD}\). 2. Lines \(\overleftrightarrow{BC}\) and \(\overleftrightarrow{DA}\) are vertical, so both have undefined slope. Therefore, \(\overleftrightarrow{BC} \parallel \overleftrightarrow{DA}\). 3. Every horizontal line is perpendicular to every vertical line. Thus, \(\overleftrightarrow{AB} \perp \overleftrightarrow{BC}\), \(\overleftrightarrow{BC} \perp \overleftrightarrow{CD}\), \(\overleftrightarrow{CD} \perp \overleftrightarrow{DA}\), and \(\overleftrightarrow{DA} \perp \overleftrightarrow{AB}\).

Parallel: \(\overleftrightarrow{AB} \parallel \overleftrightarrow{CD}\) and \(\overleftrightarrow{BC} \parallel \overleftrightarrow{DA}\). Perpendicular: \(\overleftrightarrow{AB} \perp \overleftrightarrow{BC}\), \(\overleftrightarrow{BC} \perp \overleftrightarrow{CD}\), \(\overleftrightarrow{CD} \perp \overleftrightarrow{DA}\), and \(\overleftrightarrow{DA} \perp \overleftrightarrow{AB}\).

- Find the slope of each line. - Lines with equal slopes are parallel. - Perpendicular nonvertical lines have slopes that are negative reciprocals.

1. The slope of \(\overleftrightarrow{PQ}\) is \(\frac{1-1}{8-1}=0\), and the slope of \(\overleftrightarrow{RS}\) is \(\frac{5-5}{3-10}=0\). Therefore, \(\overleftrightarrow{PQ} \parallel \overleftrightarrow{RS}\). 2. The slope of \(\overleftrightarrow{QR}\) is \(\frac{5-1}{10-8}=2\), and the slope of \(\overleftrightarrow{SP}\) is \(\frac{1-5}{1-3}=2\). Therefore, \(\overleftrightarrow{QR} \parallel \overleftrightarrow{SP}\). 3. Perpendicular nonvertical lines have slopes whose product is \(-1\). The only slopes here are \(0\) and \(2\), and there are no vertical lines. Therefore, no pair is perpendicular.

Parallel: \(\overleftrightarrow{PQ} \parallel \overleftrightarrow{RS}\) and \(\overleftrightarrow{QR} \parallel \overleftrightarrow{SP}\). No pair of lines is perpendicular.

- Find the slope of \(g\) from its two given points. - What must be true about the slopes of parallel lines? - Use the point on \(h\) to determine its y-intercept.

1. The slope of \(g\) is \(m=\frac{4-2}{5-1}=\frac{1}{2}\). 2. Parallel lines have the same slope, so line \(h\) has slope \(\frac{1}{2}\). 3. Substitute \(A(6, -1)\) into \(y=\frac{1}{2}x+b\): \(-1=\frac{1}{2}\cdot 6+b\), so \(b=-4\). 4. Therefore, \(f(x)=\frac{1}{2}x-4\).

\(f(x)=\frac{1}{2}x-4\)

- First use the two points to find the slope of \(g\). - Perpendicular nonvertical lines have slopes whose product is \(-1\). - Take the negative reciprocal of the original slope.

1. Find the slope of \(g\): \(m_g=\frac{6-(-3)}{5-2}=\frac{9}{3}=3\). 2. Slopes of nonvertical perpendicular lines are negative reciprocals, so \(m_h=-\frac{1}{m_g}=-\frac{1}{3}\).

\(m_h=-\frac{1}{3}\)

- Read each slope from slope-intercept form. - The product of the slopes of perpendicular nonvertical lines is \(-1\). - Solve the resulting equation for \(a\).

1. The slope of \(g\) is \(\frac{3}{4}\), and the slope of \(h\) is \(a\). 2. Perpendicular slopes satisfy \(\frac{3}{4}a=-1\). 3. Solving gives \(a=-\frac{4}{3}\).

\(a=-\frac{4}{3}\)

- Perpendicular nonvertical lines have negative-reciprocal slopes. - Substitute the given point into \(y=mx+b\) to find the intercept. - Write the final equation in slope-intercept form.

1. The slope of \(f\) is \(\frac{3}{7}\). A perpendicular line has slope \(-\frac{7}{3}\). 2. Substitute \(P(6, -2)\) into \(y=-\frac{7}{3}x+b\): \(-2=-\frac{7}{3}\cdot 6+b=-14+b\). Thus, \(b=12\). 3. Therefore, \(g(x)=-\frac{7}{3}x+12\).

\(g(x)=-\frac{7}{3}x+12\)

- Use the two points to find the slope of \(\ell\). - Parallel lines have the same slope. - Substitute the given point on \(k\) to find its y-intercept.

1. Find the slope of \(\ell\): \(m_\ell=\frac{0-3}{2-0.5}=\frac{-3}{1.5}=-2\). 2. Parallel lines have equal slopes, so \(m_k=-2\). 3. Substitute \(S(-4, 5)\) into \(y=-2x+b\): \(5=-2\cdot(-4)+b\), so \(b=-3\). 4. Therefore, \(k\) has equation \(y=-2x-3\).

\(y=-2x-3\)

- Parallel lines have equal slopes. - Use the given point to find the y-intercept. - Set \(y=0\) to find the x-intercept.

1. Because \(h\) is parallel to \(k\), its slope is \(\frac{2}{3}\). 2. Substitute \(P(6, 1)\) into \(y=\frac{2}{3}x+b\): \(1=4+b\), so \(b=-3\). Thus, \(h(x)=\frac{2}{3}x-3\). 3. Set \(h(x)=0\): \(0=\frac{2}{3}x-3\), so \(x=\frac{9}{2}=4.5\). The x-intercept is \((4.5, 0)\).

Equation: \(h(x)=\frac{2}{3}x-3\) x-intercept: \((4.5, 0)\)

- Classify \(v\) from the form of its equation before thinking about slopes. - Ask what must stay constant on a line parallel to a vertical line. - Recall which direction is perpendicular to vertical.

1. Line \(v\) is vertical. A parallel line through \(P(-2,3)\) is therefore the vertical line \(x=-2\). 2. A line perpendicular to a vertical line is horizontal. The horizontal line through \(P\) is \(y=3\). 3. A vertical line has undefined slope, so there is no numerical slope for which a negative reciprocal can be taken.

a) \(x=-2\) b) \(y=3\) c) Line \(v\) has undefined slope, so the negative-reciprocal rule for two finite nonzero slopes does not apply directly.

- Classify \(v\) from the form of its equation before thinking about slopes. - Ask what must stay constant on a line parallel to a vertical line. - Recall which direction is perpendicular to vertical.

1. Line \(v\) is vertical. A parallel line through \(P(-2,3)\) is therefore the vertical line \(x=-2\). 2. A line perpendicular to a vertical line is horizontal. The horizontal line through \(P\) is \(y=3\). 3. A vertical line has undefined slope, so there is no numerical slope for which a negative reciprocal can be taken.

a) \(x=-2\) b) \(y=3\) c) Line \(v\) has undefined slope, so the negative-reciprocal rule for two finite nonzero slopes does not apply directly.

- Find the slope of each nonvertical line. - Equal slopes indicate parallel lines. - Slopes that are negative reciprocals indicate perpendicular lines.

1. The slope of \(\overleftrightarrow{KL}\) is \(\frac{3-1}{5-1}=\frac{1}{2}\). 2. The slope of \(\overleftrightarrow{LM}\) is \(\frac{7-3}{3-5}=-2\). 3. The slope of \(\overleftrightarrow{MN}\) is \(\frac{6-7}{1-3}=\frac{1}{2}\). Since \(\overleftrightarrow{KL}\) and \(\overleftrightarrow{MN}\) have equal slopes, \(\overleftrightarrow{KL} \parallel \overleftrightarrow{MN}\). 4. Because \(\frac{1}{2} \cdot (-2)=-1\), \(\overleftrightarrow{LM}\) is perpendicular to both \(\overleftrightarrow{KL}\) and \(\overleftrightarrow{MN}\). 5. Line \(\overleftrightarrow{NK}\) is vertical. None of the other lines is vertical or horizontal, so it is neither parallel nor perpendicular to any of them.

Parallel: \(\overleftrightarrow{KL} \parallel \overleftrightarrow{MN}\). Perpendicular: \(\overleftrightarrow{KL} \perp \overleftrightarrow{LM}\) and \(\overleftrightarrow{LM} \perp \overleftrightarrow{MN}\).

- Find the slope of each line. - Parallel distinct lines have equal slopes. - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. The slope of \(f\) is \(m_f=1.5=\frac{3}{2}\). 2. The slope of \(g\) is \(m_g=\frac{5-2}{3-1}=\frac{3}{2}\). Since \(g\) passes through \((1, 2)\), its y-intercept is \(0.5\), so \(g\) is distinct from \(f\). 3. The slope of \(h\) is \(m_h=\frac{4-6}{2-(-1)}=-\frac{2}{3}\). 4. Lines \(f\) and \(g\) have equal slopes and different intercepts, so they are parallel. Also, \(\frac{3}{2}\cdot\left(-\frac{2}{3}\right)=-1\), so \(h\) is perpendicular to both \(f\) and \(g\).

Lines \(f\) and \(g\) are parallel. Line \(h\) is perpendicular to both \(f\) and \(g\).

- Parallel lines have equal slopes. - Perpendicular nonvertical lines have negative-reciprocal slopes. - At an intersection, the two function values are equal.

a) For parallel lines, set the slopes equal: \(2k-1=-\frac{1}{3}\). Then \(2k=\frac{2}{3}\), so \(k=\frac{1}{3}\). b) For perpendicular lines, the slope of \(f_k\) must be the negative reciprocal of \(-\frac{1}{3}\), which is \(3\). Thus, \(2k-1=3\), so \(k=2\). c) When \(k=2\), \(f_2(x)=3x+3\). Solve \(3x+3=-\frac{1}{3}x-2\). Multiplying by \(3\) gives \(9x+9=-x-6\), so \(x=-1.5\). Then \(y=3\cdot(-1.5)+3=-1.5\). The intersection is \((-1.5, -1.5)\).

a) \(k=\frac{1}{3}\) b) \(k=2\) c) \((-1.5, -1.5)\)

- Use the two points to find the slope of \(f\). - Parallel lines have equal slopes. - Perpendicular lines have negative-reciprocal slopes. - Use a known point to determine an unknown intercept.

1. The slope of \(f\) is \(m_f=\frac{2-5}{4-(-2)}=-\frac{1}{2}\). Substitute \(B(4, 2)\): \(2=-\frac{1}{2}\cdot 4+b\), so \(b=4\). Thus, \(f(x)=-\frac{1}{2}x+4\). 2. A parallel line has the same slope, and the given intercept is \(-1\). Therefore, \(g(x)=-\frac{1}{2}x-1\). 3. The negative reciprocal of \(-\frac{1}{2}\) is \(2\), so \(h\) has slope \(2\). Substitute \(B(4, 2)\): \(2=2\cdot 4+b\), so \(b=-6\). Therefore, \(h(x)=2x-6\).

1. \(f(x)=-\frac{1}{2}x+4\) 2. \(g(x)=-\frac{1}{2}x-1\) 3. \(h(x)=2x-6\)

- An x-intercept has y-coordinate \(0\). - A line through the origin has y-intercept \(0\). - Set two function equations equal to find their intersection.

1. Line \(p\) has slope \(0.5\). Substitute \((4, 0)\) into \(y=0.5x+b\): \(0=2+b\), so \(b=-2\). Thus, \(p(x)=0.5x-2\). 2. A perpendicular line has slope \(-2\). Since \(q\) passes through the origin, \(q(x)=-2x\). 3. Set the equations equal: \(0.5x+3=-2x\). Then \(2.5x=-3\), so \(x=-1.2\). Substituting into \(q\) gives \(y=2.4\). The intersection is \((-1.2, 2.4)\).

1. \(p(x)=0.5x-2\) 2. \(q(x)=-2x\) 3. \((-1.2, 2.4)\)

- Read all four vertex coordinates from the graph before comparing sides. - Compare the slopes of each pair of opposite sides. - After establishing a parallelogram, check whether one pair of adjacent sides is perpendicular.

1. From the graph, \(A=(1, 1)\), \(B=(7, 3)\), \(C=(6, 6)\), and \(D=(0, 4)\). 2. \(m_{AB}=\frac{3-1}{7-1}=\frac{1}{3}\) and \(m_{CD}=\frac{4-6}{0-6}=\frac{1}{3}\), so \(AB\parallel CD\). 3. \(m_{BC}=\frac{6-3}{6-7}=-3\) and \(m_{DA}=\frac{1-4}{1-0}=-3\), so \(BC\parallel AD\). Thus, \(ABCD\) is a parallelogram. 4. The slopes \(\frac{1}{3}\) and \(-3\) are negative reciprocals, so adjacent sides are perpendicular. A parallelogram with a right angle is a rectangle.

\(ABCD\) is a rectangle because both pairs of opposite sides have equal slopes, and adjacent sides have negative-reciprocal slopes.

- Check the condition that the product of two perpendicular nonvertical slopes must satisfy. - Compare the student's proposed slope with the reciprocal as well as the sign change. - After fixing the slope, use the given point to determine the intercept.

1. Perpendicular nonvertical lines have slopes whose product is \(-1\), so the perpendicular slope must be the negative reciprocal of \(\frac{2}{3}\), which is \(-\frac{3}{2}\), not \(-\frac{2}{3}\). 2. Use \(P=(6,1)\) in \(y=-\frac{3}{2}x+b\): \(1=-\frac{3}{2}\cdot6+b=-9+b\), so \(b=10\). 3. The correct perpendicular line is \(y=-\frac{3}{2}x+10\).

The student changed the sign but did not take the reciprocal. The correct line is \(y=-\frac{3}{2}x+10\).

- Check the condition that the product of two perpendicular nonvertical slopes must satisfy. - Compare the student's proposed slope with the reciprocal as well as the sign change. - After fixing the slope, use the given point to determine the intercept.

1. Perpendicular nonvertical lines have slopes whose product is \(-1\), so the perpendicular slope must be the negative reciprocal of \(\frac{2}{3}\), which is \(-\frac{3}{2}\), not \(-\frac{2}{3}\). 2. Use \(P=(6,1)\) in \(y=-\frac{3}{2}x+b\): \(1=-\frac{3}{2}\cdot6+b=-9+b\), so \(b=10\). 3. The correct perpendicular line is \(y=-\frac{3}{2}x+10\).

The student changed the sign but did not take the reciprocal. The correct line is \(y=-\frac{3}{2}x+10\).

- Compare the slope of an adjacent side with the slope of either given side. - Use the fact that the square is centered at the origin to constrain where the missing parallel pair can lie. - Once all four side equations are known, think about a convenient length that determines the area of this rotated square.

1. The given sides have slope \(1\). Adjacent sides of a square are perpendicular, so the missing sides have slope \(-1\). 2. Because the square is centered at the origin, the missing parallel pair is symmetric about the origin. The two lines are \(y=-x+4\) and \(y=-x-4\). 3. The four vertices are \((0, 4)\), \((4, 0)\), \((0, -4)\), and \((-4, 0)\), so each diagonal has length \(8\). 4. Using \(A=\frac{1}{2}d_1d_2\), \(A=\frac{1}{2}\cdot 8\cdot 8=32\) square units.

a) \(y=-x+4\) and \(y=-x-4\) b) \(32\) square units

- The distance from a point to a line is measured along a perpendicular segment. - Find the slope of \(h\), then use the negative reciprocal for the perpendicular slope. - Find the intersection of the two lines before using the distance formula.

a) Line \(g\) is horizontal with equation \(y=2\). The distance from \(R(5, 5)\) to \(g\) is \(5-2=3\,\text{cm}\). b) The slope of \(h\) is \(\frac{5-2}{5-9}=-\frac{3}{4}\), so a perpendicular line has slope \(\frac{4}{3}\). The perpendicular line through \(P(1, 2)\) is \(y-2=\frac{4}{3}(x-1)\), or \(4x-3y+2=0\). Line \(h\) has equation \(3x+4y-35=0\). Solving the two line equations gives \(F\left(\frac{97}{25},\frac{146}{25}\right)\), and \(PF=\sqrt{\left(\frac{97}{25}-1\right)^2+\left(\frac{146}{25}-2\right)^2}=\frac{24}{5}\,\text{cm}=4.8\,\text{cm}\).

a) \(3\,\text{cm}\) b) The perpendicular line is \(y-2=\frac{4}{3}(x-1)\), and the distance from \(P\) to \(h\) is \(4.8\,\text{cm}\).