Consider points A(1,2) and B(4,-3)

Question

A) Find an equation of the line which passes through points A and B

B) What are the x and y intercepts of the line which is perpendicular to the line from part (a) which also passes through point A? Express each as an ordered pair (x,y).

Answer 1

For this problem, the key is to use the coordinates to substitute for x and y since coordinates are both values (x, y).

 

In part A, to find the equation for the line through both points you must find what the slope is first. If you were graphing, you could draw it and then count the rise over run (the slope). Algebraically, you take the two y values and divide them by the difference between the x values. Let's say y₁ is 2, y₂ is -3, so x₁ is 1 and x₂ is 4. So the slope is: (y₂-y₁)/(x₂-x₁)= (-3-2)/(4-1)= -5/3.

 

Now that you know the slope, you need to find the y-intercept since y=mx+b and m is -5/3. Pick either coordinate to use its x and y value to solve for b. If we used (x₁, y₁) it would look like this: 2= -5/3•1 +b which is 2= (-5/3)+b. 2 as a fraction is 2/1 but we need to add both sides by 5/3 so the b is isolated thus the denominators MUST be the same. 2/1=6/3 so 6/3 +5/3= -5/3 +5/3+b which is 11/3 for b. Now the equation for part A is y= -5/3+11/3

 

For part B, you need to find the perpendicular slope to the equation you just found before finding the intercepts. The perpendicular slope to the equation is the opposite reciprocal since it must go in the opposite direction AND form a 90° angle. This means that the perpendicular slope is 3/5.

 

Again, use the coordinate values for x and y, this time you must use (x₁, y₁) to find the intercepts. Let's start with b, the y intercept. 2= 3/5•1+b so 2=3/5 +b. 2/1= 10/5 so 10/5=3/5+b thus b=7/5. For the x intercept, make y equal 0 and solve for x: 0=3/5x + 7/5 which is -7/5= 3/5x. x would be -7/3. You would write it as (-7/3, 7/5).

Answer 2

the equation through two points (x1,x2) and (y1,y2) wiith slope m=(y1-y2)/(x1-x2) is (y-y1)=m(x-x1) 

the slope of line through A,B is -5/3 and line equation is 5x+3y=11

 

the line perpendicular to a line with slope 'm' is '-1/m' therefore the perpendicular line slope is 3/5

 

the perpendicular line equation with slope 3/5 and passing through point (1,2) is 3x-5y+7=0

 

the x intercept and y intercept mentioned as points upon substitution as y=0 for x-intercept and x=0 for y-intercept are (-7/3,0) and (0, 7/5).