Thread: finding the eqn of a straight line

A straight line crosses the x-axis at x = 3 and the y-axis at y = −3. Find the equation of the line. Please give your answer in the functional form:
y = a x + b
for some constants a and b.

I know what b is, (-3) but how do I find a? please can someone help?

Originally Posted by [nuthing]

A straight line crosses the x-axis at x = 3 and the y-axis at y = −3. Find the equation of the line. Please give your answer in the functional form: y = a x + b for some constants a and b.

If we know that neither $\displaystyle p\text{ nor }q$ is zero then the points $\displaystyle (p,0)~\&~(0,q)$ determine the line
$\displaystyle \dfrac{x}{p}+\dfrac{y}{q}=1$.

Originally Posted by [nuthing]

A straight line crosses the x-axis at x = 3 and the y-axis at y = −3. Find the equation of the line. Please give your answer in the functional form:
y = a x + b
for some constants a and b.

I know what b is, (-3) but how do I find a? please can someone help?

When x= 3 y= 0 so 0= 3a+ b. When x= 0, y= -3 so -3= 0a+ b. That second equation tells you that b= -3. Now put that into the first equation and solve for a.

the slope of the line is 1 and the y intercept is -3 so y=x-3


bjh

Originally Posted by bjhopper

the slope of the line is 1 and the y intercept is -3 so y=x-3


bjh

I would have preferred that [nuthing] be allowed to finish the problem himself.