Question

A line with slope 3 intersects a line with slope 5 at the point (10, 15). What is the distance between the x-intercepts of these two lines?​

Answer 1

The distance between the intercepts of both lines is 2 units

The equation of a line in slope intercept form is:

$y = mx + b$

Where:

$m \to$ slope

$b \to$ y intercept

For the first line, we have:

$m_1 = 3$ ---- the slope

So, the equation of the first line is:

$y = 3x + b_1$

For the second line, we have:

$m_2 = 5$ --- the slope

So, the equation of the second line is:

$y = 5x + b_2$

Both lines intersect at (10,15) means that (10,15) is a common solution to the equation of both lines

i.e.

$(x,y) = (10,15)$

Substitute these values in the first equation and solve for b

$y = 3x + b_1$

$15 = 3*10 + b_1$

$15 = 30 + b_1$

$b_1 = 15 - 30$

$b_1 = -15$

So, the equation of the first line is

$y = 3x - 15$

Repeat the same process for the second line

$y = 5x + b_2$

$15 = 5*10 + b_2$

$15 = 50 + b_2$

$b_2 = 15 - 50$

$b_2 = -35$

So, the equation of the second line is

$y =5x - 35$

The x intercept is when $y =0$

So, we substitute 0 for y and solve for x in the equations of both lines

For line 1

$y = 3x - 15$

$0 = 3x - 15$

$3x= 15$ ---- Collect like terms

$x = 5$ --- Divide both sides by 3

The x intercept of line 1 is 5

For line 2

$y =5x - 35$

$0 = 5x - 35$

$5x = 35$ --- Collect like terms

$x = 7$ -- Divide both sides by 5

The x intercept of line 2 is 7

The distance (d) between both is the difference in the calculated intercepts:

$d = 7 - 5$

$d = 2$