Answer to Question #126497 in Algebra for Deborah

Question

1. Find the equation of the straight line which has a gradient 3 and passes through (2,1).

2. Find the equation of a straight line which makes an intercept of 3 on the x-axis and 4 on the y-axis.

Answer 1

1) The general form of the equation of a line is y = mx+c.

If the line has a gradient 3 (which is m here) and passes through (2,1) which is (x,y) here, we can find c the intercept on the y-axis as

1 = 3*2+ c, or

c = 1-6 = -5.

So the equation of the line you are seeking is

y = 3x-5.

2) An intercept of 3 on the x-axis - in point (3,0)

An intercept of 4 on the y-axis - in point (0,4)

$\dfrac{x-x_a}{x_b-x_a}=\dfrac{y-y_a}{y_b-y_a} \\ \dfrac{x-3}{0-3}=\dfrac{y-0}{4-0}\\ \dfrac{x-3}{-3}=\dfrac{y}{4}\\ answer: y=-\dfrac{4}{3}x+4$