# The coordinates of P, Q & R are (6,-11), (k,-9) & (2k,-3) respectively. If gradient of PQ=gradient of PR, find the value of k.

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The gradient of a line joining two points A and B with co-ordinates (x1,y1) and (x2,y2) is given by (y1-y2)/(x1-x2)

Let's take the points P and Q, the gradient of PQ would be (-9-(-11))/(k-6)=(-9+11)/(k-6)=2/(k-6).

For PR the gradient is (-3-(-11))/(2k-6)=(-3+11)/(2k-6)=8/(2k-6)=4/(k-3)

As the gradients of PQ and PR as the same 2/(k-6)=4/(k-3) or 2*(k-3)=4*(k-6) or 2k-6=4k-24 or 4k-2k=-6+24 or 2k=18 or k=9

Therefore the value of k is 9.

The gradient of the segment is the slope of that segment.

We know that the slope of a line that passes through 2 given points is:

mPQ = (yQ-yP)/(xQ-xP)

mPQ = (-9+11)/(k-6)

mPQ = 2/(k-6)

mPR = (yR-yP)/(xR-xP)

mPR = (-3+11)/(2k-6)

mPR = 8/(2k-6)

We'll divide by 2:

mPR = 4/(k-3)

We know, from enunciation , that mPQ = mPR, so we'll substitute them by the found expressions.

2/(k-6) = 4/(k-3)

We'll cross multiply:

2(k-3) = 4(k-6)

We'll divide by 2:

(k-3) = 2(k-6)

We'll remove the brackets:

k-3 = 2k-12

We'll combine like terms:

2k-k = -3+12

**k = 9**

The gradient of the line PQ = Gradient of PR.

So, (yQ-yP)/(xQ-xP) = (yR-yP)/(xR-xP).....(1)

P(6,-11) and Q(k,-9) and R(2k,-3) . Substitute the coordinates in (1)

(-9- -11)/(k-6) = (-3- -9)/(2k-k)

2/(k-6) = 6/k. Multiply by k(k-6)

2k = 6(k-6)

2k = 6k -36

36 = 6k-2k

36 =4k

k = 36/4 = 9.