Math Problem Statement

Solve the following problems: (1) Find the slope and angle of inclination of the line joining (8/3, 1/3) and (5/4, 13/4). (2) Determine x so that the three points (x, -3), (9, 1), and (13, 2) will lie on a straight line. (3) A line with slope 1/5 passes through (6, 3). If the abscissa of a point on the line is 11, what is its ordinate?

Solution

Here are the solutions to the problems in your image:


Problem 1: Find the slope and angle of inclination of the line joining (83,13)\left( \frac{8}{3}, \frac{1}{3} \right) and (54,134)\left( \frac{5}{4}, \frac{13}{4} \right).

Solution:

The slope formula is: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

Substitute the given points: m=134135483m = \frac{\frac{13}{4} - \frac{1}{3}}{\frac{5}{4} - \frac{8}{3}}

  1. Simplify the numerator: 13413=3912412=3512\frac{13}{4} - \frac{1}{3} = \frac{39}{12} - \frac{4}{12} = \frac{35}{12}

  2. Simplify the denominator: 5483=15123212=1712\frac{5}{4} - \frac{8}{3} = \frac{15}{12} - \frac{32}{12} = -\frac{17}{12}

Thus, the slope is: m=35121712=3517=3517m = \frac{\frac{35}{12}}{-\frac{17}{12}} = \frac{35}{-17} = -\frac{35}{17}

Angle of inclination (θ\theta):

The angle is given by: θ=tan1(m)=tan1(3517)\theta = \tan^{-1}(m) = \tan^{-1}(-\frac{35}{17})


Problem 2: Determine xx so that the three points (x,3),(9,1),(13,2)(x, -3), (9, 1), (13, 2) lie on a straight line.

Solution:

If three points lie on the same line, the slopes between consecutive points must be equal. Let the points be A(x,3)A(x, -3), B(9,1)B(9, 1), and C(13,2)C(13, 2).

  1. Slope of ABAB: mAB=1(3)9x=49xm_{AB} = \frac{1 - (-3)}{9 - x} = \frac{4}{9 - x}

  2. Slope of BCBC: mBC=21139=14m_{BC} = \frac{2 - 1}{13 - 9} = \frac{1}{4}

Set mAB=mBCm_{AB} = m_{BC}: 49x=14\frac{4}{9 - x} = \frac{1}{4}

Cross-multiply: 4(9x)=4    364x=4    4x=32    x=84(9 - x) = 4 \implies 36 - 4x = 4 \implies 4x = 32 \implies x = 8

Thus, x=8x = 8.


Problem 3: A line with slope 15\frac{1}{5} passes through (6,3)(6, 3). If the abscissa of a point on the line is 1111, what is its ordinate?

Solution:

The equation of the line in point-slope form is: yy1=m(xx1)y - y_1 = m(x - x_1)

Substitute m=15m = \frac{1}{5}, (x1,y1)=(6,3)(x_1, y_1) = (6, 3): y3=15(x6)y - 3 = \frac{1}{5}(x - 6)

Simplify: y3=15x65y - 3 = \frac{1}{5}x - \frac{6}{5} y=15x+95y = \frac{1}{5}x + \frac{9}{5}

Substitute x=11x = 11 (abscissa): y=15(11)+95=115+95=205=4y = \frac{1}{5}(11) + \frac{9}{5} = \frac{11}{5} + \frac{9}{5} = \frac{20}{5} = 4

Thus, the ordinate is 44.


Any questions, or should I provide further details on any solution?

Here are five related questions to expand your understanding:

  1. What does the slope of a line represent geometrically?
  2. How can you find the equation of a line given two points?
  3. How do you verify if three points are collinear using determinants?
  4. What is the significance of the angle of inclination of a line?
  5. How can point-slope and slope-intercept forms of a line be converted into each other?

Tip: Always double-check arithmetic in slope calculations, as small errors propagate through!

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Linear Equations
Slope of a Line
Angle of Inclination
Collinearity of Points

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Point-slope equation of a line: y - y1 = m(x - x1)
Angle of inclination: θ = tan^(-1)(m)

Theorems

Collinearity criterion (equality of slopes)
Properties of straight lines

Suitable Grade Level

Grades 9-12