# The line joining the points A(2,t) & B(7,2t^2+7) has a gradient of 2. Find possible values of "t".The line joining the points A(2,t) & B(7,2t^2+7) has a gradient of 2. Find possible values of "t".

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### 5 Answers

A(2,t) B(7, 2t^2 + 7)

We knwo that the gradient of AB = 2

==> mAB = 2

==> (2t^2 +7 - t)/(7-2) = 2

==? (2t^2 -t +7)/5 = 2

==> 2t^2 -t + 7 = 10

==> 2t^2 - t - 3 = 0

==> t1 = [1 + sqrt(25)]/4 = (6/4) = 3/2

==> t2= (1-5)/4 = -1

Gradient of a line A(x1, y1) and B(x2, y2) is given by the formula:

Gradient = (y2 - y1)/(x2 - x1)

Substituting given values of the gradient, x1, y1, x2, and y2 in the formula for gradient:

2 = [2t^2 + 7) - t]/(7 - 2)

==> 2 = (2t^2 - t +7)/5

Multiplying both sides by 5 and transferring all terms to left hand side:

==> 2t^2 - t - 3 = 0

==> 2t^2 + 2t - 3t - 3 = 0

==> 2t(t + 1) - 3(t + 1) = 0

==> (t + 1)(2t - 3) = 0

Therefore there are following two roots or possible values of t.

t = -1 , and t = 3/2

The general equation of a line is y=mx+b, where m is the gradient which is given here as 2 here and the coordinates of any point on the line are (x,y).

If we substitute the x and y co-ordinates of A and B which lie on the line, in the equation of the line, we get:

for point A: t=2*2+b ...(1)

and for point B: 2t^2+7=2*7+b ...(2)

(1) gives b=t-4

Substituting the value of b in (2) in terms of t as b=t-4 we get:

2t^2+7=14+t-4=10+t or 2t^2-t-10+7=2t^2-t-3=0

Now we have to solve the equation 2t^2-t-3=0

2t^2-t-3=2t^2-3t+2t-3=t(2t-3)+(2t-3)=(t+1)(2t-3)=0

To solve further (t+1) and (2t-3) have to be equated to 0 and the value of t derived in each case.

for t+1=0 we have t=-1

and for 2t-3=0 we have t=3/2.

Therefore the possible values of t are -1 and 3/2.

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

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

m = (yB-yA)/(xB-xA)

m = (2t^2 - t + 7)/(7-2)

m = (2t^2 - t + 7)/(5)

We know, from enunciation , that m = 2, so we'll substitute my by it's given value.

2 = (2t^2 - t + 7)/(5)

We'll cross multiply:

10 = (2t^2 - t + 7)

We'll move all terms to one side:

2t^2 - t + 7 - 10 = 0

2t^2 - t - 3 = 0

We'll apply the quadratic formula:

t1 = [1 + sqrt(1 + 24)]/4

t1 = (1+5)/4

t1 = 6/4

t1 = 3/2

t2 = (1-5)/4

t2 = -1

The values of t are: {-1 ; 3/2}

The gradient m of the line joining the points A(2,t) and B(7 , 2t^2+7) is given by:

(yB-yA)/(xB-xA) = m........(1)

Here m=2. Substitute the coordinate and gradient value and coordinates of A and B in the relation (1):

(2t^2+7- t)/(7-2) =2

2t^2-t+7 = 2*5 = 10

2t^2-t+7-10 = 0

2t^2-t-3 = 0

2t^2+2t-3t-3

(t+1)(2t-3) = 0

t+1 , 2t -3 = 0

2t-3 = 0 gives 2t=3 , t = 3/2 .

t+1 = 0 gives t = -1.

So the possible values are: t= 3/2 or t = -1.