Precalculus for Everybody
Lines and Linear Equations
- Using slopes, prove that the \(A = (2, 1)\), \(B = (-4, -2)\) and \(C = (1, 1/2)\) are collinear.
In the exercises, from 2 to 9, find an equation of the line that satisfies the given conditions. Write the equation in this form: \(\boldsymbol{y = mx + b}\).
- Pass through (1, 3); has slope 5.
- Pass through the origin; has slope 5.
- Pass through (1, 1) and (2, 3).
- x-Intercept 5; y-Intercept 2
- Pass through (1, 3), and is parallel to the line \(5y + 3x - 6 = 0\).
- Pass through (4, 3), and is perpendicular to the line \(5x + y - 2 = 0\).
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Is parallel to \(2y + 4x - 5 = 0\), and pass through the intersection of the lines:
\[ 5x + y = 4 \quad \text{ and } \quad 2x + 5y - 3 = 0 \]
- Pass through (8,-6), and intersects the axes at equal distances from the origin.
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Given the line \(L:\, 2y - 4x - 7 = 0\):
- Find the line passing through \(P = (1, 1)\), and perpendicular to \(L\).
- Find the distance from the point \(P = (1, 1)\) to \(L\).
- Using slopes, prove that the points \(A = (3, 1)\), \(B = (6, 0)\) and \(C = (4, 4)\) are the vertices of a right triangle. Find the area of the triangle.
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Determine which of the following lines are parallel and which are perpendicular:
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\[ L_1:\, 2x + 5y - 6 = 0 \]
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\[ L_2: \, 4x + 3y - 6 = 0 \]
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\[ L_3: \, -5x + 2y - 8 = 0 \]
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\[ L_4: \, 5x + y - 3 = 0 \]
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\[ L_5: \, 4x + 3y - 9 = 0 \]
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\[ L_6: \, -x + 5y - 20 = 0 \]
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Find the perpendicular bisector of the segment joining the given points:
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\[ (1, \, 0) \, \text{ and } \, (2, \, -3) \]
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\[ (-1, \, 2) \, \text{ and } \, (3, \, 10) \]
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\[ (-2, \, 3) \, \text{ and } \,(-2, \, -1) \]
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The endpoints of one of the diagonals of a rhombus are \((2, -1)\) and \((14, 3)\). Find an equation of the line that contains the other diagonal.
Hint: the diagonals of a rhombus are perpendicular
- Find the distance from the origin to the line \(4x + 3y -15 = 0\).
- Find the distance from the point (0,-3) to the line \(5x - 12y - 10 = 0\).
- Find the distance from the point (1,-2) to the line \(x - 3y = 5\).
- Find the distance between the parallel lines\(3x - 4y = 0\) and \(3x - 4y = 10\).
- Find the distance between the parallel lines \(3x - y + 1 = 0\) and \(3x - y + 9 = 0\).
- Find the distance from the point \(Q = (6, -3)\) to the line passing through \(P = (-4, 1)\) and parallel to the line \(4x + 3y = 0\).
- Determine the value of \(C\) in the equation of the line \(L\): \(4x +3y + C = 0\). It is known that the distance from the point \(Q = (5, 9)\) to the line \(L\) is 4 times the distance from the point \(P = (-3, 3)\) to the line \(L\).
- Find the lines parallel to the line \(5x + 12y - 12 = 0\) that are 4 units away from this line.
- Find the equation of the tangent line to the circle \(x^2 + y^2 - 4x + 6y - 12 = 0\) at the point (-1, 1).
- Find the equations of the two lines passing through the point \(P = (2, -8)\), and are also tangent to the circle \(x^2 + y^2 = 34\).
- In the above exercise, find the points where the tangent lines make contact with the circle.
- Find the equation of each of the two lines parallel to the line \(2x - 2y + 5 = 0\), which are also tangent to the circle \(x^2 + y^2 = 9\).
- Find the equation of the tangent line to the circle \(x^2 + y^2 + 2x + 4y - 20 = 0\) at the point (2, 2).
- Find the equation of the circle with center \(C = (1, -1)\), and is also tangent to the line \(5x - 12y + 22 = 0\).
- Find the equation of the circle passing through the point $Q = (4, 0)$, and is also tangent to the line \(3x - 4y + 20 = 0\) at the point \(P = (-12/5, \,16/5)\).
- Find the equation of the circle passing through the points (3, 1) and (-1, 3), with center in the line \(3x - y - 2 = 0\).
- Both parallel lines, \(2x + y -5 = 0\) and \(2x + y +15 = 0\), are tangent to a circle. One point of tangency is \(B = (2, 1)\). Find an equation of the circle.
- Find an equation of the line passing through the point \(P = (8, 6)\), which also forms a triangle of area 12 with the coordinate axes.
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Determine the values of \(k\) and \(n\) in the equations of the lines:
\[ L_1:kx - 2y - 3 = 0 \quad \text{and} \quad L_2:6x - 4y - n = 0, \]\[ \begin{aligned} &L_1: kx - 2y - 3 = 0 \text{ and} \\[1em] &L_2: 6x - 4y - n = 0, \end{aligned} \]
- if \(L_1\) intersects \(L_2\) in a single point.
- if \(L_1\) and \(L_2\) are perpendicular.
- if \(L_1\) and \(L_2\) are parallel and not coincident.
- if \(L_1\) and \(L_2\) are coincident.
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Determine for what values of \(k\) and \(n\) the lines:
\[ kx + 8y + n = 0 \quad \text{ and } \quad 2x + ky - 1 = 0, \]\[ \begin{aligned} &kx + 8y + n = 0 \text{ and} \\[1em] &2x + ky - 1 = 0, \end{aligned} \]
- are parallel and not coincident.
- are coincident.
- are perpendicular.
- The center of a square is \(C = (1, -1)\), and one of its sides is on the line \(x-2y = -12\). Find the equations of the lines that containing the other sides.
- Prove that the points \(A = (1, 4)\), \(B = (5, 1)\), \(C = (8, 5)\) and \(D = (4, 8)\) are the vertices of a rhombus (a quadrilateral whose sides have equal length). Verify that the diagonals are perpendicular.
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Let \(a\) and \(b\) be the x-intersection and the y-intersection of a line.
If \(a \neq 0\) and \(b \neq 0\), prove that an equation for this line is \(\frac{x}{a} + \frac{y}{b} = 1\).
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Roberto is playing pool in a championship.
He must hit, without spin, the eight ball with the white ball using two sides of the table(as the figure indicates).
If the white ball is on the point \(P = (2, 6)\), and the red ball on \(Q = (3, 2)\), find the points \(A\) and \(B\) of the sides of the table where the ball must make contact to be successful.
If the white ball is on the point \(P = (2, 6)\), and the red ball on \(Q = (3, 2)\), find the points \(A\) and \(B\) of the sides of the table where the ball must make contact to be successful.
Answers
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\[ y = 5x – 2 \]
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\[ y = -3x \]
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\[ y = 2x – 1 \]
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\[ y = – \frac{2}{5}x + 2 \]
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\[ y = – \frac{3}{5} x + \frac{18}{5} \]
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\[ y = \frac{x}{5}+ \frac{11}{5} \]
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\[ y = -2x + \frac{41}{23} \]
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\[ x + y = 2;\,x – y = 14 \]
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\[ y = – \frac{x}{2} + \frac{3}{2} \]
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\[ \frac{ 9 \sqrt{5}}{10} \]
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\[ 5 \]
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\(L_2\) is paralel to \(L_5\); \(L_3\) is perpendicular to \(L_1\); \(L_4\) is perpendicular to \(L_6\).
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\[ x – 3y – 6 = 0 \]
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\[ x + 2y – 13 = 0 \]
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\[ y = 1 \]
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\[ y + 3x – 25 = 0 \]
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\[ 3 \]
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\[ 2 \]
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\[ \frac{2}{\sqrt{10}} \]
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\[ 2 \]
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\[ \frac{ 4 \sqrt{10} }{5} \]
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\[ \frac{28}{5} \]
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\[ C = -7 \; \text{ or } \; C = \frac{59}{3} \]
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\(5x + 12y + 40 = 0\); \(5x + 12y – 64 = 0\)
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\[ 3x – 4y + 7 = 0 \]
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\(5x – 3y – 34 = 0\); \(3x + 5y + 34 = 0\)
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\[ (5, \, -3) \; \text{ y } \; (-3, \, -5) \]
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\[ x – y – 3 \sqrt{2} = 0; \; x – y + 3 \sqrt{2} = 0 \]\[ \begin{aligned} &x – y – 3 \sqrt{2} = 0; \\[.5em] &x – y + 3 \sqrt{2} = 0 \end{aligned} \]
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\[ 3x + 4y – 14 = 0 \]
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\[ (x – 1)^2 + (y + 1)^2 = 9 \]
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\[ x^2 + y^2 = 16 \]
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\[ (x – 2)^2 + (y – 4)^2 = 10 \]
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\[ (x + 2)^2 + (y + 1)^2 = 20 \]
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\[ 3x – 2y – 12 = 0; \, 3x – 8y + 24 = 0 \]
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\[ k \neq 3, \, \text{any } n \]
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\[ k = -\frac{4}{3}, \, \text{any } n \]
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\[ k = 3, \, n \neq 6 \]
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\[ k = 3, \, n = 6 \]
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\(k = -4\) and \(n \neq 2\) or \(k = 4\) and \(n \neq -2\)
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\(k = -4\) and \(n = 2\) or \(k = 4\) and \(n = -2\)
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\(k = 0\) and any \(n\)
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\(x – 2y – 18 = 0\); \(2x + y + 14 = 0\); \(2x + y – 16 = 0\)
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\(A = \left( 0, \, \frac{14}{5} \right)\), \(B = \left( \frac{7}{4}, \, 0\right)\)