Hints for solving problems in Topic 4.2

In these exercises there are several formulas that you must remember and know how to work with them. They are:

Slope formula:     

Standard form for an equation:    Ax + By = C

Point-Slope form:      

Slope-intercept form:     y = mx + b

In most of the problems, after the slope is determined, you start with the point-slope form and solve the problem from there according to the directions.

Finding the Equation I

Problems 1-10

Problem No. 9 - Find the equation of the line that passes through the point (-5. 3) and has slope m = 3/8. Write the answer in point-slope form.

Since we have a point and the slope, just substitute into the point-slope from for an equation.

                           (y - 3) = (3/8)[x -(-5)]   =   (y - 3) = (3/8)( x + 5)    ans.

Problems 11-19.

Problem No. 17 – Rewrite the equation  y + 5 = (-4/7)(x +7) in standard form.

Get rid of the denominator by multiplying both sides of the equation by 7.

       7(y) + 7(5) = 7(-4/7)(x + 7)       7y + 35 = -4(x + 7)

                   7y + 35 = -4x -28         4x + 7y = -28 -35

            4x + 7y = -63 ans. The book gives some of the answers with a fraction. Usually the instructor, including me, will require no fractions in the answer and A, numerical coefficient of x, must be positive.

Problems 20-28

Problem No. 25 – Find the equation of the line that passes through the points (6, 7) and (3, -2). Write your answer in point-slope form and standard form.

First find the slope:         Use one of the points (either one,

but I like to use the one that doesn’t have a negative no.), and the slope to substitute into the point-slope formula

     (6, 7)  m = 3      y - 7 = 3(x - 6)    ans.  Now put the equation into standard form.                  y - 7 = 3x - 18       -3x + y = -18 + 7

                            -3x + y = -11      3x - y = 11   ans.

Finding the Equations II

Problems 29-38

Problem No. 29 – Find the equation of the line in slope-intercept form that passes through the point (3, 1) and has slope m = 4.

Substitute the given numbers into the point-slope form and then solve for y.

     (y - 1) = 4(x - 3)       y - 1 = 4x - 12       y = 4x -12 + 1

                             y = 4x - 11    ans.

Problems 39-44

Problem No. 43 – Find the equation of the line in slope-intercept form that passes through the point (-2, 7) and is perpendicular to the line y = (1/5)x - 16.

The slope if the given equation is 1/5. To find an equation of a line parallel to the given line, you use the same slope (1/5). To find an equation of a line perpendicular to the given line, you use the negative reciprocal (-5).

Also, to start the problem, substitute the point and the correct slope into the point-slope formula.

       (-2, 7) m = -5       (y - 7) = -5[x - (-2)]      y - 7 = -5x - 10

                          y = -5x - 10 + 7       y = -5x - 3     ans.

Problems 45-50

Problem No. 45 – Find the slope and y-intercept of the line    -3x + y = 8

Solve for y by putting the equation in the slope-intercept form.

       y = mx + b      m = slope     b = y-intercept

       y = 3x  + 8    m = 3  slope      b = 8    y-intercept = (0, 8)    ans.

Problems 51-56

Problem No. 51 – Find the equation of the vertical line that passes through the point (7,3).

     If it asks for the equation of the vertical line, put    x = 7.

     If it asks for the equation of the horizontal line, put    y = 3.

If you don’t see this, plot the point and draw the line vertical or horizontal through the point and notice where it crosses the x or y axis. It can’t cross both, only one. Also a horizontal line has only a y in it and a vertical line has only an x in it. All other line are oblique, meaning they have an x and a y in them in the form of  y = mx + b or           Ax + By = C.