Question

line segment segment ab has a slope of start fraction five over four end fraction and contains point a(8, −7). what is the y-coordinate of point q(3, y) if segment qa is perpendicular to line segment segment ab question mark y=-2 y=-3 y=1 y=3

Answer 1

The y-coordinate of point Q on line segment QA is -4.

Step-by-step explanation:

The slope of a line is given by the formula m = change in y / change in x. From the given information, we know that the slope of line segment AB is 5/4. We also know that point A has coordinates (8, -7).

To find the equation of line segment AB, we can use the slope-intercept form of a line, y = mx + b. Since we have the slope and point A, we can substitute the values into the equation to find the y-intercept:

-7 = (5/4)(8) + b

-7 = 10 + b

b = -17

So the equation of line segment AB is y = (5/4)x - 17.

A line perpendicular to another line has a negative reciprocal slope. The negative reciprocal of 5/4 is -4/5.

To find the y-coordinate of point Q on line segment QA, we can use the equation of line segment AB and the slope of line segment QA:

y = (slope of QA)(x - x-coordinate of Q) + y-coordinate of Q

Substituting the known values, we get:

y = (-4/5)(3 - 8) + y-coordinate of Q

y = (-4/5)(-5) + y-coordinate of Q

y = 4 + y-coordinate of Q

To find the y-coordinate of point Q, we set y = 0 and solve for the y-coordinate:

0 = 4 + y-coordinate of Q

y-coordinate of Q = -4

Therefore, the y-coordinate of point Q is -4.

Answer 2

For perpendicular lines, the product of their slopes is -1.
Let the slope of the second line be m, then 5/4 m = -1
m = -4/5
The slope of a line passing through two given points is given by m = (y2 - y1)/(x2 - x1)
-4/5 = (y - (-7))/(3 - 8) = (y + 7)/-5
4 = y + 7
y = 4 - 7 = -3
y = -3