Why don't you follow either of the two methods I just showed you in
http://www.jiskha.com/display.cgi?id=1364642553
Did you even look at it ?
http://www.jiskha.com/display.cgi?id=1364642553
Did you even look at it ?
Given that the line segment joins points P(-6,7) and Q(5,-2), we can first find the equation of the line passing through these two points using the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
Step 1: Find the slope (m) of the line.
The slope (m) of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1)
Taking the coordinates of P(-6,7) as (x1, y1) and Q(5,-2) as (x2, y2), we have:
m = (-2 - 7) / (5 - (-6))
= (-2 - 7) / (5 + 6)
= -9 / 11
Step 2: Find the y-intercept (b) of the line.
To find the y-intercept (b), we can substitute the coordinates of one of the given points (P or Q) into the slope-intercept form (y = mx + b) and solve for b.
Let's use point P(-6,7):
7 = (-9/11)(-6) + b
7 = 54/11 + b
To simplify, we can convert 7 to an equivalent fraction with a denominator of 11:
77/11 = 54/11 + b
Now, subtract 54/11 from both sides:
77/11 - 54/11 = b
23/11 = b
Therefore, the y-intercept (b) is 23/11.
Step 3: Write the equation of the line.
With the slope (m = -9/11) and the y-intercept (b = 23/11), we can write the equation of the line passing through points P and Q:
y = (-9/11)x + 23/11
Step 4: Find the x-coordinate where the line intersects the y-axis.
Since the y-coordinate on the y-axis is always 0, we can substitute y = 0 into the equation of the line and solve for x:
0 = (-9/11)x + 23/11
Rearranging the equation:
(9/11)x = 23/11
Multiplying both sides by 11/9:
x = (23/11) * (11/9)
= 23/9
Step 5: Find the coordinates of point R.
The coordinates of point R are (x, y), where x = 23/9 and y = 0.
Therefore, the coordinates of point R are (23/9, 0).