Find the value of x pQ=QR the coordinates of p,q and r and (6,-1),(1,3)and(x,8) respectively.
If PQ = QR , (where P(6,-1), Q(1,3), R(x,8))
√( (3+1)^2 + (1-6)^2) = √( (8-3)^2 + (x-1)^2 )
√41 = √(x^2 - 2x + 26)
square both sides
41 = x^2 - 2x + 26
x^2 - 2x - 15 = 0
(x-5)(x+3) = 0
x = 5 or x = -3
what a exact answer no doubt
To find the value of x and the coordinates of point R, we can use the concept of slope.
1. Calculate the slope of line PQ using the formula: slope = (y2 - y1) / (x2 - x1), where (x1, y1) = (6, -1) and (x2, y2) = (1, 3).
slope(PQ) = (3 - (-1)) / (1 - 6)
= 4 / (-5)
= -4/5
2. Since line PQ and QR are collinear (lie on the same line), they have the same slope.
slope(PQ) = slope(QR)
-4/5 = (3 - 8) / (1 - x)
3. Cross-multiply to solve for x:
-4/5 = -5 / (1 - x)
-4(1 - x) = -5 * 5
-4 + 4x = -25
4x = -25 + 4
4x = -21
x = -21/4
So, the value of x is -21/4.
Now, let's find the coordinates of point R using the determined value of x:
Substitute x = -21/4 into the equation of the line containing points Q and R:
y - 3 = (8 - 3) / (x - 1)
y - 3 = 1 / (-21/4 - 1)
y - 3 = 1 / (-21/4 - 4/4)
y - 3 = 1 / (-25/4)
y - 3 = -4/25
y = -4/25 + 75/25
y = 71/25
Therefore, the coordinates of point R are (-21/4, 71/25).