What is the equation of a line that passes through the point (-7,-8) and parallel to y=5/7x-5.

The equation of the line that passes through the point (-7, -8) and parallel to the given line is y = 5/7·x - 3

Step-by-step explanation:

The coordinates of the point the line passes = (-7, -8)

The equation of the line the required line is parallel to is y = 5/7·x - 5

The slope of of the given line is therefore in slope and intercept form with the slope = 5/7 and the y-intercept = -5

The slope of a line parallel to another line are equal, therefore the slope of the required line = The slope of the given line = 5/7

The equation of the required line in point slope form is therefore;

y - (-8) = 5/7 × (x - (-7))

y + 8 = 5/7 × x + 5/7 × 7

y + 8 = 5/7 × x + 5

∴ y = 5/7 × x + 5 - 8 = 5/7 × x - 3

y = 5/7 × x - 3

The equation of the line that passes through the point (-7, -8) and parallel to y = 5/7x - 5 is y = 5/7·x - 3.

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

~dasgupta/mathbook/append17dec2k8.pdf