# How do you find t if the slope is 5/2 and the coordinates are (-1, 4) and (4, t)?

May 24, 2016

$t = \frac{41}{4} \to 10 \frac{1}{4}$

#### Explanation:

Slope (gradient) is the relationship between the change in the up/down compared to the change in along

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(- 1 , 4\right)$
Let point 2 be P_2->(x_2,y_2)(->(4,t)

The gradient is given as $\frac{5}{2}$

$\left(\text{Change in y")/("Change in x}\right) = \frac{5}{2} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

So

$\frac{5}{2} = \frac{t - 4}{4 - \left(- 1\right)} = \frac{t - 4}{5}$

$\frac{5}{2} = \frac{t - 4}{5}$

Multiply both sides by 5

$5 \times \frac{5}{2} = \left(t - 4\right) \times \frac{5}{5}$

But $\frac{5}{5} = 1$

$\frac{25}{2} = t - 4$

$t = \frac{25}{4} + 4$
$t = \frac{41}{4}$