write an equation in slope intercept form of a line that passes through the point (14,8) and is (a) parallel (b) perpendicular to the line that passes through the points (4,9) and (-3,6)

Respuesta :

Slope-intercept form:

y = mx + b       "m" is the slope, "b" is the y-intercept (the y value when x = 0)


First find the slope of the line that passes through (4,9) and (-3,6).

Use the slope formula and plug in the two points:

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{6-9}{-3-4}[/tex]

[tex]m=\frac{-3}{-7} =\frac{3}{7}[/tex]



A.) For lines to be parallel, their slopes have to be the same.

Since the given line's slope is 3/7, the parallel line's slope is also 3/7


y = 3/7x + b    To find "b", plug in the point (14,8) into the equation

8 = 3/7(14) + b

8 = 42/7 + b

8 = 6 + b    Subtract 6 on both sides

2 = b


[tex]y=\frac{3}{7}x+2[/tex]



B.) For lines to be perpendicular, their slopes have to be the opposite/negative reciprocal (flipped sign and number)

For example:

slope is 3

perpendicular line's slope is -1/3

slope is -2/3

perpendicular line's slope is 3/2


Since the given line's slope is 3/7, the perpendicular line's slope is -7/3

y = -7/3x + b    Plug in (14, 8) to find "b"

8 = -7/3(14) + b

8 = -98/3 + b   Add 98/3 on both sides

8 + 98/3 = b       Make the denominators the same

24/3 + 98/3 = b

122/3 = b


[tex]y=-\frac{7}{3}x+\frac{122}{3}[/tex]

Equation in slope intercept form (parallel line)

[tex]y=\frac{3}{7}x+2\\[/tex]

Equation of perpendicular line in slope intercept form is

[tex]y=-\frac{7}{3}x+\frac{122}{3}[/tex]

Given :

an equation in slope intercept form of a line that passes through the point (14,8)

Two given points are (4,9) and (-3,6)

Lets find out the slope of the given two points

[tex]slope =\frac{y_2-y_1}{x_2-x_1} =\frac{6-9}{-3-4}=\frac{3}{7}[/tex]

Slope of parallel lines are same. So we use the above slope to find the equation of parallel line

The line passes through the point (14,8)

Equation of line is

y-y1=m(x-x1)

use the slope m= 3/7 and point (14,8)

[tex]y-8=\frac{3}{7} (x-14)\\y-8=\frac{3}{7}x-6\\y=\frac{3}{7}x-6+8\\y=\frac{3}{7}x+2\\[/tex]

Slope of perpendicular lines are negative reciprocal of one another

Slope of perpendicular line is [tex]\frac{-7}{3}[/tex]

Now we find equation of line that passes through the point (14,8) and slope -7/3

[tex]y-8=\frac{-7}{3} (x-14)\\y-8=-\frac{7}{3}x+\frac{98}{3}\\y=-\frac{7}{3}x+\frac{98}{3}+8\\y=\frac{122}{3}-\frac{7}{3}x\\[/tex]

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