# In a linear relationship two data points are (9,3) and (33,9). If the function is y=mx+b, we have?

## (a) $m = 4$ and $b = - 33$ (b) $m = \frac{1}{4}$ and $b = \frac{3}{4}$ (c) $m = 4$ and $b = - 123$ (d) $m = \frac{1}{4}$ and $b = 24$

Jan 8, 2017

Looking at my calculations and your question's structure you have asked for the wrong thing. You need the value of $m$ and not $b$

$m = \frac{1}{4} \to \text{multiple choice a}$

#### Explanation:

Let point 1 $\to {P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(9 , 3\right)$
Let point 2 $\to {P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(33 , 9\right)$

gradient $\to m = \left(\text{change in y")/("change in x}\right)$ as you read left to right on the graph.

$\implies m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{9 - 3}{33 - 9} = \frac{6}{24} \equiv \frac{6 \div 3}{24 \div 3} = \frac{2}{8} = \frac{1}{4}$

Jan 8, 2017

#### Explanation:

As the two data points given in $\left(x , f \left(x\right)\right)$ form are $\left(9 , 3\right)$ and $\left(33 , 9\right)$

the linear relation is $\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$, where $y = f \left(x\right)$

i.e. $\frac{y - 3}{9 - 3} = \frac{x - 9}{33 - 9}$

or $\frac{y - 3}{6} = \frac{x - 9}{24}$

or $y - 3 = \frac{x - 9}{4}$

or $4 y - 12 = x - 9$

or $4 y = x - 9 + 12 = x + 3$

or $y = \frac{x}{4} + \frac{3}{4}$

i.e. $f \left(x\right) = \frac{1}{4} \times x + \frac{3}{4}$

Hence, while $m = \frac{1}{4}$, $b = \frac{3}{4}$